HC Verma Solutions for Class 12 Physics Chapter 24 – Kinetic Theory of Gases

Page No 32:

Question 1:

When we place a gas cylinder on a van and the van moves, does the kinetic energy of the molecules increase? Does the temperature increase?

Answer:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas’s kinetic energy and there will be a rise in the temperature.

Question 2:

While gas from a cooking gas cylinder is used, the pressure does not fall appreciably till the last few minutes. Why?

Answer:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.

Question 3:

Do you expect the gas in a cooking gas cylinder to obey the ideal gas equation?

Answer:

No, the gas won’t obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation.

Question 4:

Can we define the temperature of (a) vacuum, (b) a single molecule?

Answer:

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities.

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules. 

Question 5:

Comment on the following statement: the temperature of all the molecules in a sample of a gas is the same.

Answer:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules.

Question 6:

Consider a gas of neutrons. Do you expect it to behave much better as an ideal gas as compared to hydrogen gas at the same pressure and temperature?

Answer:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size.

Question 7:

A gas is kept in a rigid cubical container. If a load of 10 kg is put on the top of the container, does the pressure increase?

Answer:

No, the pressure on gas won’t increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.

Question 8:

If it were possible for a gas in a container to reach the temperature 0 K, its pressure would be zero. Would the molecules not collide with the walls? Would they not transfer momentum to the walls?

Answer:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule’s collision with the walls of the container.

Question 9:

It is said that the assumptions of kinetic theory are good for gases having low densities. Suppose a container is so evacuated that only one molecule is left in it. Which of the assumptions of kinetic theory will not be valid for such a situation? Can we assign a temperature to this gas?

Answer:

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time.

Question 10:

A gas is kept in an enclosure. The pressure of the gas is reduced by pumping out some gas. Will the temperature of the gas decrease by Charles’s low?

Answer:

If the gas is ideal, there will be no temperature change. Moreover, Charles’s law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles’s law.

Question 11:

Explain why cooking is faster in a pressure cooker.

Answer:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 1000C,. Thus, cooking process gets faster.

Question 12:

If the molecules were not allowed to collide among themselves, would you expect more evaporation or less evaporation?

Answer:

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.

Question 13:

Is it possible to boil water at room temperature, say 30°C? If we touch a flask containing water boiling at this temperature, will it be hot?

Answer:

Yes, it is possible to boil water at 300C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 300C will not be hot.

Question 14:

When you come out of a river after a dip, you feel cold. Explain.

Answer:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.

Page No 33:

Question 1:

Which of the following parameters is the same for molecules of all gases at a given temperature?
(a) Mass
(b) Speed
(c) Momentum
(d) Kinetic energy.

Answer:

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

Hence, correct answer is (d).

Question 2:

A gas behaves more closely as an ideal gas at
(a) low pressure and low temperature
(b) low pressure and high temperature
(c) high pressure and low temperature
(d) high pressure and high temperature.

Answer:

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.
At high temperature, molecules move very fast. So, they tend to collide elastically and ​forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

Question 3:

The pressure of an ideal gas is written as

P=2E3V. Here E refers to
(a) translational kinetic energy
(b) rotational kinetic energy
(c) vibrational kinetic energy
(d) total kinetic energy.

Answer:

According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional.

Thus, (a) is the correct answer.

Question 4:

The energy of a given sample of an ideal gas depends only on its
(a) volume
(b) pressure
(c) density
(d) temperature.

Answer:

Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature.

Thus, (d) is the correct answer.

Question 5:

Which of the following gases has maximum rms speed at a given temperature?
(a) hydrogen
(b) nitrogen
(c) oxygen
(d) carbon dioxide.

Answer:

The rms speed of a gas is given by â€‹

3RTMo.
Since hydrogen has the lowest Mo compared to other molecules, it will have the highest rms speed.
Thus, (a) is the correct answer.

Question 6:

Figure shows graphs of pressure vs density for an ideal gas at two temperatures T1 and T2.
(a) T1 > T2
(b) T1 = T2
(c) T1 < T2
(d) Any of the three is possible.
Figure

Answer:

The straight line T1 has greater slope than T​2. This means

Pρ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by

3Pρ. This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms velocity.

So, T1 > T2.

Thus, (a) is the correct answer.

Question 7:

The mean square speed of the molecules of a gas at absolute temperature T is proportional to
(a)

1T(b)

T(c) T
(d) T2.

Answer:

Root mean squared velocity is given by

vrms=3RTM⇒vrms2=3RTM⇒vrms2 α TThus, (c) is the correct answer.

Question 8:

Suppose a container is evacuated to leave just one molecule of a gas in it. Let va and vrms represent the average speed and the rms speed of the gas.
(a) va > vrms
(b) va < vrms
(c) va = vrms
(d) vrms is undefined.

Answer:

Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

Question 9:

The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at the same temperature is about
(a) 125 m s−1
(b) 2000 m s−1
(c) 8000 m s−1
(d) 31 m s−1.

Answer:

(b) 2000 ms−1

Given,
Molecular mass of hydrogen, MH = 2
Molecular mass of oxygen, Mo = 32
RMS speed is given by,

 

vrms=3RTM ⇒3RTMO=500Now,

⇒vOrmsvHrms=3RTMO3RTMH⇒vOrmsvHrms=3RT323RT2⇒vOrmsvHrms=14⇒500vHrms=14⇒vHrms=4×500=2000 ms-1Hence, the correct answer is (b).

Question 10:

The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be
(a) 100 kPa
(b) 200 kPa
(c) 400 kPa
(d) 800 kPa.

Answer:

Let the number of moles in the gas be n.
Applying equation of state, we get

PV=nRT⇒P=nRTV⇒2×105=nRTV                      …1When half of the gas is removed, number of moles left behind =n2Let the pressure be P’.P’=n2RTVNow,P’=12×2×105=105                   From eq. 1=100 kPa

Thus, (a) is the correct answer.

Question 11:

The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become
(a) v
(b)

v2(c) 2v
(d) 4v.

Answer:

 

Given: v=3RT32Let the new rms speed be v’.Molecule dissociate, M=16v’=3R2T16=3R4T32=23RT32=2vThus, (c) is the correct answer.

Question 12:

The quantity

pVkTrepresents
(a) mass of the gas
(b) kinetic energy of the gas
(c) number of moles of the gas
(d) number of molecules in the gas.

Answer:

 

Here,PV=nRT                      …1Also,k=RN⇒R=kN                      …2Now,PV=nkNT                        From eq. 1 and eq. 2⇒nN=PVkTnN = Number of moleculesPVkT=Number of moleculesThus, (d) is the correct answer.

Question 13:

The process on an ideal gas, shown in figure, is
(a) isothermal
(b) isobaric
(c) isochoric
(d) none of these.
Figure

Answer:

According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT

⇒P=nRVTGiven: P α TThis means nRV is a constant. So, V is also a constant.Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

Question 14:

There is some liquid in a closed bottle. The amount of liquid is continuously decreasing. The vapour in the remaining part
(a) must be saturated
(b) must be unsaturated
(c) may be saturated
(d) there will be no vapour.

Answer:

As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Thus, (b) is the correct answer.

Question 15:

There is some liquid in a closed bottle. The amount of liquid remains constant as time passes. The vapour in the remaining part
(a) must be saturated
(b) must be unsaturated
(c) may be unsaturated
(d) there will be no vapour.

Answer:

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Thus, (a) is the correct answer.

Question 16:

Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel
(a) increases continuously
(b) decreases continuously
(c) first increases and then decreases
(d) first increases and then becomes constant.

Answer:

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

Question 17:

A vessel A has volume V and a vessel B has volume 2V. Both contain some water which has a constant volume. The pressure in the space above water is pa for vessel A and pb for vessel B.
(a) pa = pb
(b) pa = 2pb
(c) pb = 2pa
(d) pb = 4pa

Answer:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

Thus, (a) is the correct answer.

Question 1:

Consider a collision between an oxygen molecule and a hydrogen molecule in a mixture of oxygen and hydrogen kept at room temperature. Which of the following are possible?
(a) The kinetic energies of both the molecules increase.
(b) The kinetic energies of both the molecules decrease.
(c) kinetic energy of the oxygen molecule increases and that of the hydrogen molecule decreases.
(d) The kinetic energy of the hydrogen molecule increases and that of the oxygen molecule decreases.

Answer:

According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision.

Thus, (c) and (d) are correct answers.

Page No 34:

Question 2:

Consider a mixture of oxygen and hydrogen kept at room temperature. As compared to a hydrogen molecule an oxygen molecule hits the wall
(a) with greater average speed
(b) with smaller average speed
(c) with greater average kinetic energy
(d) with smaller average kinetic energy.

Answer:

The average speed of molecules is given by

8kTπm. We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

Question 3:

Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
(a) Kinetic energy
(b) Momentum
(c) Density
(d) Speed.

Answer:

The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

Question 4:

Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?
(a) Rms speed of a molecule
(b) Density
(c) Pressure
(d) Average magnitude of momentum.

Answer:

Pressure of an ideal gas is given by PV =

13mnu2. We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

Thus, (c) is the correct answer.

Question 5:

The average momentum of a molecule in a sample of an ideal gas depends on
(a) temperature
(b) number of moles
(c) volume
(d) none of these.

Answer:

Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

Question 6:

Which of the following quantities is the same for all ideal gases at the same temperature?
(a) The kinetic energy of 1 mole
(b) The kinetic energy of 1 g
(c) The number of molecules in 1 mole
(d) The number of molecules in 1 g

Answer:

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases.
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro’s law.
Thus, (a) and (c) are correct answers.

Question 7:

Consider the quantity

MkTpVof an ideal gas where M is the mass of the gas. It depends on the
(a) temperature of the gas
(b) volume of the gas
(c) pressure of the gas
(d) nature of the gas.

Answer:

 

In an ideal gas, the equation of state is given byPV=nRT⇒PV=nNARNAT⇒PV=nNAkT⇒1nNA=kTPVMultiplying both sides by mass of the gas M, we getMnNA=MkTPVNow, nNA gives the total number of molecules of the gas.Also, MnNA gives the mass of a single molecule.Hence,MkTPV is the mass of a single molecule of the gas, Molecular mass is a property of the gas.Thus, (d) is the correct answer.

Question 1:

Calculate the volume of 1 mole of an ideal gas at STP.

Answer:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325

×105 Pa
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

V =

RTP

V=

8.314×2731.01325×105=0.0224 m3

Question 2:

Find the number of molecules of an ideal gas in a volume of 1.000 cm3 at STP.

Answer:

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022

×1023
Number of molecules in 22.4​

×103 cmof ideal gas at STP = ​6.022

×1023

Now,
Number of molecules in 1 cm3 of ideal gas at STP =

6.022×102322.4×103=2.688×1019

Question 3:

Find the number of molecules in 1 cm3 of an ideal gas at 0°C and at a pressure of 10−5 mm of mercury.

Answer:

Given:
Volume of ideal gas, V = 1 cm3 = 10-6 m​3
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10−8 m of Hg
Density of ideal gas, ρ = 13600 kgm-3
Pressure

Pis given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get
n=PVRT⇒n=ρgh×VRT⇒n=13600×9.8×10-8×10-68.31×273⇒n=5.87×10-13Number of molecules = N × n
= 6.023 × 1023 × 5.874 × 10−13
= 35.384 × 1010
= 3.538 × 1011

Question 4:

Calculate the mass of 1 cm3 of oxygen kept at STP.

Answer:

We know that 22.4 L of O2 contains 1 mol O2 at STP. Thus,

22.4×103 cm3 of O2=1 mol O21 cm3 of O2 = 122.4×103  mol O21 mol of O2 = 32 g122.4×103 mol of O2 =3222.4×103 =1.43×10-3 g                                                                 = 1.43 mg

Question 5:

Equal masses of air are sealed in two vessels, one of volume V0 and the other of volume 2V0. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

Answer:

Let the pressure and temperature for the vessels of volume V0 and 2V0 be P​1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gas, n1 = n2 = n.

T1=300 KT2 = 600 KUsing the equation of state for perfect gas, we getPV=nRTFor the vessel of volume Vo:P1Vo=nRT1                                   …1For the vessel of volume 2Vo:P22Vo=nRT2                               …2Dividing eq. 2 by eq. 1, we get2P2P1=T2T1=600300=2⇒P2P1=1⇒P2:P1=1:1

Question 6:

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10−3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 1023 mol−1, density of mercury = 13600 kg m−3 and g = 10 m s−2.

Answer:

Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27  + 273  = 300 K
Height of mercury, h = 10−3 mm
Density of mercury,

ρ=  13600 kgm−3
Avogadro constant, N = 6 × 1023 mol−1
Pressure

Pis given by
P =

ρghUsing the ideal gas equation, we get

PV=nRTPV = nRT⇒n =PVRT⇒n=ρghVRT⇒n =10-6×13600×10×250×10-6 8.314×300Now, number of molecules =nN=10-6×13600×10×250×10-6 8.314×300×6×1023=8×1015

Question 7:

A gas cylinder has walls that can bear a maximum pressure of 1.0 × 106 Pa. It contains a gas at 8.0 × 105 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

Answer:

Given:
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K

Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus,                  (Given)
V1= V2 = V

Applying the five variable gas equation, we get

P1VT1=P2VT2     (∵V1=V2)⇒P1T1=P2T2⇒ T2=P2×T1P1⇒T2=1.0×106×3008.0×105=375 K

Question 8:

2 g of hydrogen is sealed in a vessel of volume 0.02 m3 and is maintained at 300 K. Calculate the pressure in the vessel.

Answer:

Given:
Mass of hydrogen, m = 2 g
Volume of the vessel, V = 0.02 m3
Temperature in the vessel, T = 300 K
Molecular mass of the hydrogen, M = 2 u

No of moles, n =

mM=22= 1 mole

Rydberg’s constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT

⇒P=nRTV⇒P=1×8.3×3000.02

P = 1.24 × 105 Pa

Question 9:

The density of an ideal gas is 1.25 × 10−3 g cm−3 at STP. Calculate the molecular weight of
the gas.

Answer:

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas,

ρ= 1.25 × 10−3 gcm−3 =1.25 kgm−3
Pressure, P = 1.01325

×105 Pa   (At STP)
Temperature, T = 273 K    (At STP)

Using the ideal gas equation, we get

PV = nRT                    …(1)n = mM                        …(2)∴ PV = mMRT⇒M=mVRTP⇒M=ρRTP⇒M=1.25×8.31×273105⇒M=2.83×10-2          =28.3 g-mol-1

Question 10:

The temperature and pressure at Simla are 15.0°C and 72.0 cm of mercury and at Kalka these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla.

Answer:

Here,
Temperature in Simla, T1= 15 + 273 = 288 K
Pressure in Simla, P​1 = 0.72  m of Hg 
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, P​2 = 0.76  m of Hg
Let density of air
at Simla and Kalka be

ρ1 and

ρ2, respectively. Then,

PV=mMRT⇒mV=PMRT⇒ρ=PMRTThus,

ρ1=P1MRT1

ρ2=P2MRT2Taking ratios, we get

ρ1ρ2=P1T1×T2P2⇒ρ1ρ2=0.72288×3080.76⇒ρ2ρ1=0.987

Question 11:

Figure shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts of the vessel.
Figure

Answer:

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
n= n2= n

​Volume of the first part = V
Volume of the second part =​3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
T1 = T2 = T
Let pressure of first and second parts be P1 and P2, respectively.

For first part:Applying equation of state, we getP1V=nRT                                        …1For second part:Applying equation of state, we getP23V=nRT                                   …2    Dividing eq. 1 by eq. 2, we getP1VP23V=1⇒P1P2=31⇒P1:P2=3:1

Question 12:

Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which the rms speed is double the speed calculated in the previous part.

Answer:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M​0 = 2 g/mol=0.002 kg /mol
We know,

C=3RTM0⇒C=3×8.3×3000.002⇒C=1932.6 ms-1In the second case, let the required temperature be T.
​
Applying the same formula, we get

3×8.3T0.002=2×1932.6⇒T=1200 K

Question 13:

A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.

Answer:

Here,
V = 10-3 m3
Density = 0.177 kgm-3
P = 105pa

C=3Pρ=3×1050.177                    =1301.9 ms-1

Question 14:

The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10−19 J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10−23 J K−1.

Answer:

We know from kinetic theory of gases that the average translational energy per molecule is

32kT.
Now,
Eavg= 0.040 eV =

0.040×1.6×10-19=6.4×10-21J

6.40×10-21=32×1.38×10-23×T⇒T=23×6.40×10-211.38×10-23=309.2 K

Question 15:

Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.

Answer:

Here,Vavg=8RTπM=8×8.83×3003.14×0.032=445.25 m/sWe know,T=DistanceSpeed=6400000×2445.25=28747.83 h3600=7.985 h=8 h

Question 16:

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64 × 10−27 kg and Boltzmann constant = 1.38 × 10−23 J K−1.

Answer:

Here,
​m = 6.64 × 10−27 kg
T = 273 K

Average speed of the He atom is given byVavg=8kTπm=8×1.38×10-23×2733.14×6.64×10-27=1202.31We know,
Momentum = m × Vavg
= 6.64 × 10−27 × 1201.35
= 7.97 × 10−24
= 8 × 10−24 kg-m/s

Question 17:

The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Answer:

Mean velocity is given by

Vavg=8RTπMLet temperature for H and He respectively be  T​1 and T2, respectively.
For hydrogen:
MH = 2g = 2

×10-3 kg
For helium:
MHe= 4 g = 4

×10-3 kg

Now,
A/q

8RT1πMH=8RT2πMHe⇒8RT12×10-3π=8RT2π×4×10-3⇒T12=T24⇒T1T2=12⇒T1:T2=1:2

Question 18:

At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth?

Answer:

Mean speed of the molecule is given by

8RTπMFor H molecule, M =2×10-3kg =4RT×103πFor escape velocity of Earth:

 

Let r be the radius of Earth.v=2GMrMultiplying neumerator and denominator by R, we getvc=GMr22rg=GMr2vc=2gr4RT×103π=2gr⇒2×8.314×T×1033.142=9.8×6.37×106⇒T≈11800 K

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Question 19:

Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

Answer:

We know,

Vavg=8RTπMMolar mass of H2 = MH = 2

×10-3 kg

Molar mass of N2 = MN = 28

×10-3 kg
Now,

<V>H=8RTπMH<V>N=8RTπMN<V>H<V>N=MNMH=282=14=3.74

Question 20:

Figure shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.
Figure

Answer:

Let the temperature of gas in both the chambers be T.
Let the molar mass of gas in the left chamber and right chamber be M​1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

A/Qvrms=3kTm1=8kTπm2        ⇒3m1 = 8m23.14        ⇒m2m1=1.18

Question 21:

Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10−5 cm.

Answer:

Here,

λ=1.38×10-8 mT = 273 K
M =

2×10-3 kgAverage speed of the H molecules is given  by

vavg=8RTπM=8×8.31×2733.14×2×10-3=1700 ms-1The time between two collisions is given by

t=λvavg⇒t=1.38×10-81700⇒t=8×10-12 sNumber of collisions in 1 s = 18.11×10-12=1.23×1011

Question 22:

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Answer:

Here,
P = 105 Pa
T = 300 K
For H2
M = 2×10-3 kg

a Mean speed is given by<v>=8RTπM=8×8.3×300×72×10−3×22=1780 ms−1Let us consider a cubic volume of 1 m3.V=1 m3Momentum of 1 molecule normal to the striking surface before collision = musin450Momentum of 1 molecule normal to the striking surface after collision = −musin450Change in momentum of the molecule = 2musin450=2muChange in momentum of n molecules = 2mnusin450=2mnuLet Δt be the time taken in changing the momentum.Force per unit area due to one molecule = 2muΔt=2muΔtObserved pressure due to collision by n molecules =2mnuΔt=105n=2mnuΔt2muΔt=1052mu6.0×1023 molecules = 2×10−3kg1 molecule = 2×10−36×1023=3.3×10−27kg⇒n=1052×3.3×10−27×1780=1.2×1028

Question 23:

Air is pumped into an automobile tyre’s tube up to a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day the temperature rises to 40°C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.

Answer:

Here,P1=2×105 PaP2=?T1=293 KT2=313 KV2=V1+0.02V1=V1(1.02)Now,P1V1T1=P2V2T2⇒2×105V1293=P2V1(1.02)313⇒P2=2×105×313293×1.02=209 kPa

Question 24:

Oxygen is filled in a closed metal jar of volume 1.0 × 10−3 m3 at a pressure of 1.5 × 105 Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 105 Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

Answer:

Here,V1=1.0×10−3m3T1=400KP1=1.5×105 PaP2=1.0×105 PaT2=300M=32 gNumber of moles in the jar before n1=P1V1RT1Volume of the gas when pressure becomes equal to external pressure is given byP1V1T1=P2V2T2⇒V2=P1V1T2P2T1⇒V2=1.5×105×1.0×10−3×3001.0×105×400=1.125×10−3Net volume of leaked gas = V2−V1=1.125×10−3−1.0×10−3=1.25×10−4m3Let n2 be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get n2=P2V2RT2=1.0×105×1.25×10−48.3×300=0.005Mass of leaked gas= 32×0.005=0.16 g

Question 25:

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 × 105 Pa and density of water = 1000 kg m−3.

Answer:

Here,V1=43π(2.0×10−3)3h=3.3 mP1=Po+ρgh⇒P1=1.0×105+1000×9.8×3.3⇒P1=1.32×105 PaP2=1.0×105 PaSince temperature remains the same, applying Boyle’s law we getP1V1=P2V2⇒V2=P1V1P2⇒V2=1.32×105×43π(2.0×10−3)31.0×105Let R2 be the new radius. Then,43πR23=1.32×105×43π(2.0×10−3)31.0×105⇒R23=1.32×105×(2.0×10−3)31.0×105⇒R3=1.32×105×(2.0×10−3)31.0×1053⇒R3=2.2×10−3m

Question 26:

Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m3. One of the tubes gets punctured and the volume of the tube reduces to 0.0005 m3. How many moles of air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.

Answer:

Here,P1=2×105paV1=0.002m3V2=0.0005m3T1=T2=300KNumber of moles initially, n1=P1V1RT1⇒n1=2×105×0.0028.3×300⇒n1=0.16Applying equation of state, we get P2V2=n2RTAssuming the final pressure becomes equal to the atmospheric pressure, we getP2=1.0×105pa⇒n2=P2V2RT⇒n2=1.0×105×0.00058.3×300⇒n2=0.02Number of leaked moles = n2−n1                                      =0.16−0.02                                      =0.14

Question 27:

0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.

Answer:

Here,m=0.040gM=4gn=0.0404=0.01T1=100+273 K=373 KHe is a monoatomic gas. Thus,Cv=3×(12R)⇒Cv=1.5×8.3=12.45Let the initial internal energy be U1.Let the final internal energy be U2.U2−U1=nCv(T2−T1)⇒0.01×12.45(T2−373)=12⇒T2=469 KThe temperature in 0C  can be obtained as follows: 469 – 273 = 1960 C

Question 28:

During an experiment, an ideal gas is found to obey an additional law pV2 = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.

Answer:

 

Applying equation of state of an ideal gas, we getPV=nRT⇒P=nRTV                               …1Taking differentials, we get⇒PdV+VdP=nRdT               …2Applying the additional law, we getPV2=cV2dP+2VPdV=0⇒VdP+2PdV=0                    …3Subtracting eq. 3 from eq. 2, we getPdV=−nRdT⇒dV=−nRPdTNow,⇒dV=−VTdT                From eq. 1⇒dVV=−dTTIntegrating between T2 and T1, we get⇒∫V12V=−∫T1T2⇒ln(2V)−ln(V)=ln(T1)−ln(T2)⇒ln(2VV)=ln(T1T2)⇒T2=T12

Question 29:

A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture.

Answer:

Here, V = 0.166 m3T=300 KMass of O2=1.60 gMO=32 gnO=1.6032=0.05Mass of N2=2.80 gMN=28 gnN=2.8028=0.1Partial pressure of O2 is given byPO=nORTV=0.05×8.3×3000.166=750Partial pressure of N2 is given byPN=nNRTV=0.1×8.3×3000.166=1500Total pressure is sum of the partial pressures.⇒P=PN+PO=750+1500=2250 Pa

Question 30:

A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.

Answer:

Here,h=1 mP1=0.75mHg = 0.75ρg Paρ=13500 kg/m3Let h be the height of the mercury above the piston.P2=P1+hρgLet the CSA be A.V1=Ah=AV2=(1−h)AApplying Boyle’s law, we getP1V1=P2V2⇒0.75ρgA=P2(1−h)A⇒0.75ρg=(0.75ρg+hρg)(1−h)⇒0.75=(0.75+h)(1−h)⇒h=0.25 mh = 25 cm

Question 31:

Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy

pT=12pATA+pBTBwhen equilibrium is achieved.
Figure

Answer:

Let the partial pressure of the gas in chamber A and B be PA’ and PB’, respectively.Applying equation of state for gas A, we getPAVTA=PA’2VT⇒PA’=PAT2TASimilarly, for gas B:PB’=PBT2TBTotal pressure is the sum of the partial pressures. It is given byP=PA’+PB’=PAT2TA+PBT2TB⇒P=T2(PATA+PBTB)⇒PT=12(PATA+PBTB)

Question 32:

A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.

Answer:

a Here,V1=5×10−5m3P1=105 PaT1=273KM = 28.8 gP1V1=nRT1⇒n=P1V1RT1⇒mM=105×5×10−58.3×273⇒m=105×5×10−5×28.88.3×273⇒m=0.0635 gb Here,V1=5×10−5m3P1=105 PaP2=105 PaT1=273KT2=373KM = 28.8 gP1V1T1=P2V2T2⇒ 5×10−5273=V2373⇒V2= 5×10−5×373273⇒V2=6.831×10−5Volume of expelled air = 6.831×10−5−5×10−5=1.831×10−5Applying equation of state, we getPV=nRT⇒mM=PVRT=105×1.831×10−58.3×373⇒m=28.8×105×1.831×10−58.3×373=0.017Thus, mass of expelled air = 0.017 gAmount of air in the container = 0.0635−0.017                                                    =0.0465g c Here,T=273KP=105PaV=5×10−5 m3Applying equation of state, we getPV=nRT⇒P=nRTV=0.0465×8.3×27328.8×5×10−5⇒P=0.731×105≈73kPa

Question 33:

A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.

Answer:

Let the CSA of the tube be A.Initial volume of air, V1 = 20A cm = 0.2A Length of mercury, h = 0.1 m Let the pressure of the trapped air when the tube is inverted and vertical be P1.Now, pressure of the mercury and trapped air balances the atmospheric pressure. Thus,  P1+0.1ρg=0.75ρg⇒P1=0.65ρgWhen the tube is inverted with the closed end down, the pressure acting upon the trapped air is given byAtmospheric pressure + Mercury column pressureNow,Pressure of trapped air = Atmospheric pressure + Mercury column pressure             In equilibriumP2=0.75ρg+0.1ρg=0.85ρgApplying the Boyle’s law when the temperature remains constant, we getP1V1=P2V2Let the new height of the trapped air be x.⇒0.65ρg0.2A = 0.85ρgxA⇒x=0.15 m = 15 cm

Question 34:

A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer:

Let CSA of the tube be A.On the colder side: P1=0.76 m HgT1=300KV1=V T2=273KV2=AxP1VT1=P2AxT2⇒P2=P1VT2T1AxOn the hotter side:P1=0.76 m HgT1=300KV1’=V T2’=400KV2’=AyP1’VT1=P2’AyT2’⇒P2’=P1VT2’T1AyIn equilibrium, the pressures on both side will balance each other.⇒P2’=P2⇒P1VT2’T1Ay=P1VT2T1Ax⇒T2’y=T2xFrom the length of the tube, we getx+y+0.1=1⇒y=0.9−x400(0.9−x)=273x⇒x=0.365 m⇒ x=36.5 cm

Question 35:

An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.

Answer:

Here,Initial pressure = Atmospheric pressure + Pressure due to mercury ⇒P1=Po+PHgLet the CSA of the tube be A.P1=0.76+0.2 =0.96 m HgT1=T2=TV1=0.43 A​If the tube is slanted, then the atmospheric pressure Po​remains the same. Only the PHg changes. ​P2=Po+PHgcos600=0.76+0.2×0.5=0.86 P1V1=P2V2⇒V2=P1V1P2=0.96×0.43A0.86 Let the length of the air column be l.⇒Al=P1V1P2=0.96×0.43A0.86 ⇒l=0.48 m⇒l=48 cm

Page No 36:

Question 36:

Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.
Figure

Answer:

Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.

​Let the CSA be A. VA1=0.20 ATA1=400 KVB1=0.1 ATB1=100 KAt first equilibrium, both side pressures will be the same. ⇒PA1=PB1Let the final temperature at equilibrium be T. Then,PA1VA1TA1=PA2VA2T⇒PA10.2A400=PA2VA2T⇒PA2=PA10.2AT400VA2                         …1For second chamber:PB1VB1TB1=PB2VB2T⇒PB10.1A100=PB2VB2T⇒PB2=PB10.1AT100VB2                      …2At second equilibrium, pressures on both sides will be the same.⇒PA2=PB2⇒PA10.2AT400VA2=PB10.1AT100VB2⇒PA12VA2=PB1VB2⇒VB2=2VA2                                  …3Now,VB2+VA2=0.3A⇒3VA2=0.3A⇒VA2=0.1ALet VA2 be lA.⇒l=0.1 m⇒ l=10 cm

Question 37:

A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out.

Answer:

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t.  Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP.Applying equation of state to the gas inside the vessel, we get(P−dP)Vo=(n−dn)RT⇒PVo−dPVo=nRT−dnRTBut PVo=nRT⇒VodP=dnRT                              …1The pressure of the gas taken out is equal to the inner pressure.Applying equation of state, we get (P−dP)dV =dnRT⇒PdV=dnRT                                 …2From eq. 1 and eq. 2, we getVodP=PdV⇒dPP=dVVodVdt=r⇒dV=rdt⇒dV=−rdt                                      …3                 Since pressures decreases, rate is negativeNow,dPP=−rdtVo                          From eq. 3   aIntegrating the equation P = P0 to P = P and time t = 0 to t = t, we get∫PoP=∫0t⇒lnP−lnPo=−rtVo⇒ln(PPo)=−rtVo⇒P=Poe−rtVobP = Po2Po2=Poe−rtVo⇒ertVo=2⇒rtVo=ln2⇒t=Voln2r

Question 38:

One mole of an ideal gas undergoes a process

p=p01+(V/V0)2where p0 and V0 are constants. Find the temperature of the gas when V = V0.

Answer:

Given: p=po1+(VVo)2Multiplying both sides by V, we getpV=poV1+(VVo)2pV=RT                        From eq. 1Now,RT=poV1+(VVo)2T=1R(poVo1+(VoVo)2)           V=Vo⇒T=poVo2R

Question 39:

Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows, doors, etc.

Answer:

We know internal energy at a particular temperatureU=nCvTAir in home is are chiefly diatomic molecules, so Cv=52R ∴U=5n2RTNow by eqn. of state nRT=PVU=52(nRT)=>U=52PVNow pressure P is constant also V of the room = constantThus,U=constant

Question 40:

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate

dNdl.
Figure

Answer:

Here,P1=105 PaA=π(0.05)2L=0.2 mV=AL=0.0016 m3T1=300 KT2=600 Kμ=0.20Applying 5 variable equation of state, we getP1VT1=P2VT2⇒P1T1=P2T2⇒P2=T2T1×P1=600300×105⇒P2=2×105Net pressure, P = P2−P1 = 2×105−105=105Total force acting on the stopper = PA =105×π×(0.05)2Applying law of friction, we getF=μN=0.2N⇒N=Fμ=105×π×(0.05)20.2dNdl=N2πr=105×π×(0.05)20.2×2π×(0.05)=0.125×105⇒dNdl=1.25×104​ N/m

Question 41:

Figure shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T0 and its pressure is p0 which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T0?
Figure

Answer:

aSince pressure from outside and inside the cylinder is the same, there is no net pressureacting on the pistons. So, tension will be zero. bT1=ToT2=2ToP1=po=105 PaCSA=ALet the pistons be L distance apart.V=ALApplying five variable gas equation, we getP1VT1=P2VT2⇒105T0=P22To⇒P2=2×105=2PoNet force acting outside = 2P0−P0=P0Force acting on a piston F= PoABy the free body diagram, we get F-T=0T = PoA

Question 42:

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h2 of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.
Figure

Answer:

a Pressure of water above the water level of the bigger tank is given by P =(h2+ho)ρgLet the atmospheric pressure above the tube be Po.Total pressure above the tube = P0+P=(h2+ho)ρg+PoThis pressure initially is balanced by pressure above the tank 2Po.⇒2Po=(h2+ho)ρg+Po⇒h2=Poρg−hob Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.Total pressure above the outlet = 2Po+(h1−ho)ρgApplying Bernouli’s law, we getLet the velocity of efflux be v1 and the velocity with which the level of the tank falls be v2.​ Pressure above the outlet is Po. Then, 2Po+(h1−ho)ρgρ+gz+v222=Poρ+gz+v222Now, let the reference point of the liquid be the level of the outlet. Thus,z =0⇒Po+(h1−ho)ρgρ+v222=v122Again, the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux. Thus,v2=0⇒Po+(h1−ho)ρgρ=v122⇒v1=[2ρ(Po+(h1−ho)ρg)]12

c)Water maintains its own level, so height of the water of the tankwill be h1​when water will stop flowingThus height of water in the tube below the tank height will be = h1Hence height of the water above the tank height will be = −h1

Question 43:

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg (figure). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.
Figure

Answer:

Atmospheric pressure inside the cylinderical vessel, P0=105 PaA=10cm2=10×10−4 m2Pressure due to the weight of the piston = mgA=1×9.810×10−4P1=105+9.8×103V1=0.2×10×10−4=2×10−4After evacution, external pressure above the piston =0P2=0+9.8×103Now,P1V1=P2V2Let L be the final length of the gas column. Then,V2=10×10−4L ⇒(105+9.8×103)×0.2×10×10−4=9.8×103×10×10−4L ⇒L=2.2 m

Question 44:

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional
area 10 cm2 and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder.

Answer:

Here,V=50m3T=273+15=288KRH=40%a)Let x be the amount of water that evaporates.P=RH×SVP⇒P=0.4×1600=640PV=mMRT⇒m=PVMRT=640×50×188.3×288=241gWater will evaporate until VP becomes equal to SVP.⇒1600×50=m+xMRT⇒1600×50=241+x18×8.3×288⇒x≈361 gb)T=288+5=293KSVP=2400PaDifference in pressure = 2400−1600=800 PaLet x be the amount of water that evaporates.PV=xMRT⇒800×50=x18×8.3×293⇒x≈296 g

Question 45:

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

Answer:

 

Here,P1=0.76 m HgP2=PT1=273 KT2=335KLet each of the bulbs have n1 moles initially.Let the number of moles left in second bulb after its pressure reached P be n2.Applying equation of state, we getP1Vn1T1=PVn2T2⇒0.76273n1=P335n2⇒n2=273P335×0.76n1Number of moles left in the second bulb after the temperature rose = n1−n2=n1−273P335×0.76n1Let n3 moles be left when pressure reached P. Applying equation of state in the first bulb, we getP1Vn1T1=PVn3T1⇒0.76n1=Pn3⇒n3=Pn10.76n3=​its own n1 moles + the it recieved from the firstn3=n1+(n1−n2)⇒Pn10.76=n1+n1−273P335×0.76n1⇒P0.76=2−273P335×0.76⇒P=0.8375⇒P=84 cm of Hg

Page No 37:

Question 46:

The weather report reads, “Temperature 20°C : Relative humidity 100%”. What is the dew point?

Answer:

Here,
Relative humidity = 100%

RH=Vapour pressure of airSVP at the same temperature=1⇒Vapour pressure of air=SVP at the same temperatureSo, the air is saturated at 200C​. So, dew point is 200C.

Question 47:

The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.

Answer:

Here,T = 298KRH = 60%P = 1.04×105 paRH=Vapour pressure of water vapourSaturated vapour pressure     =0.6Saturated vapour pressure = 3.2×103 Pa⇒Vapour pressure of water vapour (VP)= 0.6×3.2×103 =1.92×103 PaIf the water vapour is completely removed from the air, then net pressure =1.04×105−1.92×103                                                                                                                              =1.02×105 Pa                                                                                                                                =102 kPa

Question 48:

The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point?

Answer:

Here,Temperature = 20°CDew point = 10°CAir becomes saturated at 10°C. But if the room temperature is lowered to 15°C, the dew point will still be at 10°C.

Question 49:

Pure water vapour is trapped in a vessel of volume 10 cm3. The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

Answer:

Here,RH=40%V1=10×10−6m3RH=VPSVP=0.4Let SVP = PoCondensation occurs when VP = Po.⇒P1=0.4Po⇒P2=PoSince the process is isothermal, applying Boyle’s law we getP1V1=P2V2⇒V2=P1V1P2⇒V2=0.4Po×10×10−6Po⇒V2=4.0×10−6⇒V2=4.0cm3Thus water vapour condenses at volume 4.0 cm3.

Question 50:

A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapour pressure at the room temperature is 1.0 cm.

Answer:

Here,Atmospheric pressure, P=0.76 m HgPressure due to water vapour inside, P’ =0.754 mHgVapour pressure = P−P’=0.76−0.754=0.006 mHgSVH = 0.01 mHgRH=Vapour pressureSVH ×100%=0.0060.01 ×100%=60%

Question 51:

Using figure of the text, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm.

Answer:


We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 650 of methyl alcohol.

For 0.5 atm pressure, corresponding pressure in mm Hg will be 375 mm. We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 375 mm. This gives the boiling point 480 of methyl alcohol.

Question 52:

The human body has an average temperature of 98°F. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure of the text for the vapour pressures.

Answer:


Here,
T = 980F

When we convert the temperature to 0C, we get59(F−32)=59(98−32)=36.70CWe drop perpendicular corresponding to a temperature of 36.70C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg.

Question 53:

A glass contains some water at room temperature 20°C. Refrigerated water is added to it slowly. when the temperature of the glass reaches 10°C, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20°C and at 8.9 mm of mercury it is 10°C.

Answer:

Here,Dew point = 100C          ∵ Dew appears at 100CAt boiling point, SVP equals atmospheric pressure.At 200C, SVP = 17.5 mmHgAt dew point, SVP= 8.9 mmHgRH=SVP at dew pointSVP at air temperature×100%=8.917.5×100%=51%

Question 54:

50 m3 of saturated vapour is cooled down from 30°C to 20°C. Find the mass of the water condensed. The absolute humidity of saturated water vapour is 30 g m−3 at 30°C and 16 g m−3 at 20°C.

Answer:

We know that 1 m3 of air contains 30 g of water vapour at 300C. So, amount of water vapour in 50 m3 of air at 300C = 30×50 g=1500 gAlso, 1 m3 of air contains 16 g of water vapour at 200C. Amount of water vapour in 50 m3 of air at 200C = 16×50 g=800 gAmount of water vapour condensed= 1500 − 800 g=700 g

Question 55:

A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapour pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value.

Answer:

Atmospheric pressure = 76 cm HgSVP = 0.80 cm HgWhen water is introduced into the barometer, water evaporates. Thus, it exerts its vapour pressure over the mercury meniscus.As more and more water evaporates, the vapour pressure increases thatforces down the mercury level further.Finally, when the volume is saturated with the vapour at the atmospheric temperature, the highest vapour pressure, i.e. SVP is observed and the fall of mercury levelreaches its minimum. Thus,Net pressure acting on the column = 76 −0.80 cmHgNet length of Hg column at SVG = 75.2 cm

Question 56:

50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27°C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

Answer:

Here,Atmospheric pressure, Po=99.4×103 PaSVP at 270C Pw = 3.4×103 PaT=300KV=50×10−6 m3Now,Pressure inside the jar = Pressure outside the jar           ∵ Level of water is same inside and outside of the jarPressure outside the jar = Atmospheric pressurePressure inside the jar = VP of oxygen + SVP of water at 270C⇒P0=P+Pw⇒P=Po−Pw=99.4×103−3.4×103=96×103Applying equation of state, we getPV=nRT⇒96×103×50×10−6=n×8.3×300⇒n=1.9277×10−3 mol ≈1.93×10−3 mol

Question 57:

A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir.

Answer:

Given:Let the CSA be A.Case 1:V1=(x−74)ASVP = 1 cm HgAtmospheric pressure, Po=76 cm HgMercury column height = 74.0 cmLet P be the air pressure above the barometer. Then,Atmospheric pressure =SVP + Air pressure above the barometer mercury level + Mercury column height⇒1+P +74 =76 ⇒P = 1 cmCase 2:Atmospheric pressure, P0’=74.0 cm HgLet P’​ be the air pressure. Then,P’+72.10+1=76⇒P’=0.9V2=(x−72.1)AApplying Boyle’s law, we get PV1=P’V2⇒1×(x−74)A=0.9×(x−72.1)A⇒x=91.1 cmLength of the tube = 91.1 cm

Question 58:

On a winter day, the outside temperature is 0°C and relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapour pressure at 0°C is 4.6 mm of mercury and at 20°C it is 18 mm of mercury.

Answer:

Given: SVP at 0oC = 4.6 mm HgRH =40%⇒VP of air (P1)SVP at the same temperature=0.4⇒P14.6=0.4⇒P1=1.84 mmHgHere, V is constant.T1=273KT2=293KApplying equationof state, we getP1VT1=P2VT2⇒P2=P1T2T1⇒P2=1.84×293273⇒P2=1.97RH inside home=VP of air (P2)SVP at the same temperature×100%SVP at the same temperature=18mmHg⇒RH=1.9718×100=10.9%

Question 59:

The temperature and humidity of air are 27°C and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at 27°C = 3600 Pa.

Answer:

Here,SVP=3600 PaT = 273 + 27 = 300KV = 1m3M = 18 g for waterRH=50%⇒VPSVP=0.5⇒VP=0.5×3600=1800Let m1 be the mass of water present in the 50% humid air.PV = nRT⇒PV=m1MRT⇒1800=m118×8.3×300⇒m1=13gRequired pressure for saturation=3600 PaLet m2 be the amount of water required for saturation.⇒3600=m2MRT⇒m2=3600×188.3×300=26gTotal excess water vapour that has to be added = m2−m1=36−13=13g

Question 60:

The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m3. The saturation vapour pressure at 300 K 3.3 kPa. Calculate the mass of the water vapour present in the room.

Answer:

Here,T=300KSVP=3300 Pa at 300 KRH=20%⇒PSVP=0.2⇒P=0.2×SVP=0.2×3300=660V=50 m3M=18 g Now,PV = nRT⇒ PV = mMRT⇒660×50=m18×8.3×300⇒m=238.55 g≈238 g

Question 61:

The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m3. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa.

Answer:

Here,M=18g for waterm=500gV=50 m3T=300KSVP=3300PaRH=20%VPSVP=0.2⇒VP=P1=0.2×3300=660PaPartial pressure P2 for evaporated water is given byP2V=mMRT⇒P2=50018×50×8.31×300⇒P2=1385PaTotal pressure, P=P1+P2=1385+660=2045PaRH=PSVP×100=20453300×100%=61.9≈62%

Question 62:

A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5°C, how much more water will evaporate? The saturation vapour pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.

Answer:

(a) Relative humidity is given by

VPSVP at 15°C⇒0.4=VP1.6×103⇒ VP = 0.4 × 1.6 × 103

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change = 1.6 × 103 − 0.4 × 1.6 × 103
= (1.6 − 0.4 × 1.6) 103
= 0.96 × 103

Let the mass of water evaporated be m. Then,

⇒0.96×103×50=m×8.3×28818⇒m=0.96×50×18×1038.3×288=361.45≈361 g(b) At 20°C, SVP = 2.4 KPa
At 15°C, SVP = 1.6 KPa

Net pressure change = (2.4 − 1.6) × 103 Pa
= 0.8 × 103 Pa

Mass of water evaporated is given by

m=m’×8.3×29318⇒m’=0.8×50×18×1038.3×293= 296.06 ≈ 296 g

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2