# H C VERMA Solutions for Class 12-science Physics Chapter 2 - Kinetic Theory of Gases

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## Chapter 2 - Kinetic Theory of Gases Exercise 34

Question 1

Use R=8.3 J/mol-K wherever required.

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

Solution 1

Pa

n=1 mol

T=273 K

R=8.3 J/mol-K

PV=nRT

Question 2

Use R=8.3 J/mol-K wherever required.

Find the number of molecules of an ideal gas in a volume of 1.00  at STP.

Solution 2

Pa; ; T=273 K; R=8.3 J/mol-K

PV=nRT

PV=

N=

=

Question 3

Use R=8.3 J/mol-K wherever required.

Find the number of molecules in 1 of an ideal gas at 0 and at a pressure of  mm of mercury.

Solution 3

Pa

T=273 K

Question 4

Use R=8.3 J/mol-K wherever required.

Calculate the mass of 1 of oxygen kept at STP.

Solution 4

At STP, 22.4 ltr. of oxygen contains 32gm of oxygen.

gm

mg

Question 5

Use R=8.3 J/mol-K wherever required.

Equal of masses of air are sealed in two vessels, one of volume  and the other of volume 2. 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.

Solution 5

Since mass is same,

Ideal gas equation

PV=nRT

Question 6

Use R=8.3 J/mol-K wherever required.

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of mm of mercury at 27. Compute the number of air molecules contained in the bulb. Avogadro constant= 6 per mol, density of mercury= 13600 kg/ and g=10 m/.

Solution 6

PV=nRT

Question 7

Use R=8.3 J/mol-K wherever required.

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

Solution 7

Ideal gas equation:

PV=nRT

PT

375 K

Question 8

Use R=8.3 J/mol-K wherever required.

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

Solution 8

mol

PV=nRT

Pa

Question 9

Use R=8.3 J/mol-K wherever required.

The density of an ideal gas is  at STP. Calculate the molecular weight of the gas.

Solution 9

Question 10

Use R=8.3 J/mol-K wherever required.

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

Solution 10

0.987

Question 11

Use R=8.3 J/mol-K wherever required.

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 in the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now solid 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.

Solution 11

Initially,

So,

Now,

(diathermic separator)

Since,

Question 12

Use R=8.3 J/mol-K wherever required.

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.

Solution 12

m/s

m/s

Now,

1200 K

Question 13

Use R=8.3 J/mol-K wherever required.

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

Solution 13

=1300 m/s

Question 14

Use R=8.3 J/mol-K wherever required.

The average translational kinetic energy of air molecules is 0.040eV (1eV=J). Calculate the temperature of the air. Boltzmann constant k= J/K.

Solution 14

310K

Question 15

Use R=8.3 J/mol-K wherever required.

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.

Solution 15

445.25 m/s

We know,

[R=6400 km]

sec

Question 16

Use R=8.3 J/mol-K wherever required.

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= kg and Boltzmann constant= J/K.

Solution 16

1201.35 m/s

Momentum=m

=

=

Kg-m/s

Question 17

Use R=8.3 J/mol-K wherever required.

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

Solution 17

Question 18

Use R=8.3 J/mol-K wherever required.

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

Solution 18

11800 K

## Chapter 2 - Kinetic Theory of Gases Exercise 35

Question 19

Use R=8.3 J/mol-K wherever required.

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

Solution 19

Question 20

Use R=8.3 J/mol-K wherever required.

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. Calculate the ratio of the mass of a molecule molecule in the left part of the mass of a molecule in the right part.

Solution 20

1.18

Question 21

Use R=8.3 J/mol-K wherever required.

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) = cm.

Solution 21

=1687 m/s

Time between two collisions=sec

Number of collision in 1 sec=

Question 22

Use R=8.3 J/mol-K wherever required.

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  with it. How many molecules strike each square meter of the wall per second?

Solution 22

(a)

=1781 m/s

1780 m/s

(b)

Change in momentum of molecule=mv (mv)

= mv

Total DP for N molecules=

Mass of a molecule      Kg So,
Question 23

Use R=8.3 J/mol-K wherever required.

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

Solution 23

PV=nRT

Question 24

Use R=8.3 J/mol-K wherever required.

Oxygen is filled in a closed jar of volume  at a pressure of  and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is  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 equalize with the surrounding.

Solution 24

Initially,

Now

Mass of gas leaked=

= 1.446-1.285

= 0.16 gm

Question 25

Use R=8.3 J/mol-K wherever required.

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= and density of water=1000 kg/

Solution 25

Since temperature and number of moles are constant.

Question 26

Use R=8.3 J/mol-K wherever required.

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. One of the tubes gets punctured and the volume of the tube reduces to 0.0005. 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.

Solution 26

Initially,

atm;

Now,

No. of moles leaked=

=0.16-0.02

=0.14

Question 27

Use R=8.3 J/mol-K wherever required.

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

Solution 27

T-100=96.38

T=196.38

T196

Question 28

Use R=8.3 J/mol-K wherever required.

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

Solution 28

=constant

=constant

TV=constant

Question 29

Use R=8.3 J/mol-K wherever required.

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

Solution 29

2250 Pa

Question 30

Use R=8.3 J/mol-K wherever required.

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.

Solution 30

No of moles initially=No of moles finally [ Container is closed]

h=0.25 m

h=25 cm

Question 31

Use R=8.3 J/mol-K wherever required.

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

When equilibrium is achieved.

Solution 31

No. of moles initially= No. of moles finally

Question 32

Use R=8.3 J/mol-K wherever required.

A container of volume 50 contains air (mean molecular weight=28.8 g) and is open to atmosphere when the pressure is 100 kPa. The container is kept in a bath containing melting ice (0). (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). 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.

Solution 32

(a)

(b)

(c)

P73.16 kPa

P73 kPa

Question 33

Use R=8.3 J/mol-K wherever required.

A uniform tube closed at one end, contains a pallet 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.

Solution 33

No. of moles initially=No. of moles finally

h=15.3 cm

h15 cm

Question 34

Use R=8.3 J/mol-K wherever required.

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 and at a pressure 76 cm of mercury. The air column on one side is maintained at 0 and the other side is maintained at 127. Calculated the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Solution 34

No of moles in left side=No of moles in Right side

[Initially P,V,R,Tis same on both sides.]

Now,

400l=273(90-l)

673l=273×90

l=36.5 cm

Question 35

Use R=8.3 J/mol-K wherever required.

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 titled slowly in a vertical plane through an angle of 60? Assume the temperature to remain constant.

Solution 35

cm of Hg

cm of Hg

cm of Hg

cm of Hg

No. of moles initially=No. of moles finally

l=48 cm

## Chapter 2 - Kinetic Theory of Gases Exercise 36

Question 36

Use R=8.3 J/mol-K wherever required.

Figure shows a cylindrical tube of length 30cm 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.

Solution 36

Initially,

By ideal gas equation

divide

- (i)

Now,

By ideal gas equation,

divide

(ii)

From (i) and (ii)

from left end.

Question 37

Use R=8.3 J/mol-K wherever required.

A vessel of volume  contains an ideal gas at pressure  and temperature T. Gas is continuously pumped out of this vessel at a constant-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 gas, (b) the time taken before half the original gas is pumped out.

Solution 37

Let at any time t, pressure and no. of moles be P and n respectively.

So,

In dt time, dn moles are taken out and pressure decreases by dP.

-(i)

The pressure of gas taken out is equal to the inner pressure.

-(ii) [Here dP.dV0]

From (i) and (ii)

(b) When half gas is leaked out, pressure will become half of initial.

So,

Taking in on both the sides.

Question 38

Use R=8.3 J/mol-K wherever required.

One mole of an ideal gas undergoes a process

Where and  are constants. Find the temperature of gas when.

Solution 38

[Using PV=nRT]

Now,

[Here n=1]

Question 39

Use R=8.3 J/mol-K wherever required.

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 etc.

Solution 39

Internal energy=

Since pressure P is constant and V=Volume of room is also constant.

So, U=Constant.

Question 40

Use R=8.3 J/mol-K wherever required.

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction co-efficient 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.

Solution 40

No. of moles at 300 K=No. of moles at 600 K(just before cork comes out)

Pascal

Net force on cork by gas and atmosphere at 600 K=

Newton

Just before cork comes out

Now,

Question 41

Use R=8.3 J/mol-K wherever required.

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  and its pressure is  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 2?

Solution 41

(a) The pressure outside and inside is same,

So, net pressure is zero on the pistons.

So, Tension=0

(b)

Net force on piston is zero

Question 42

Use R=8.3 J/mol-K wherever required.

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height  and pressure 2 where  is the atmospheric pressure. There is a hole in the wall of the tank at a depth  below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height  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.

Solution 42

(a)

(b) Applying Bernoulli's at A and C

(c) If water stops coming out then pressure on horizontal line passing through hole will be . So, in long vertical no water must be present above this line as it is open to atmosphere. Height of water in long tube above the top is

Question 43

Use R=8.3 J/mol-K wherever required.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 and weight 1 kg. 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.

Solution 43

No. of moles initially=No. of moles finally

l=2.2m

Question 44

Use R=8.3 J/mol-K wherever required.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10  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.

Solution 44

No. of moles initially=No. of moles finally

cm

Question 45

Use R=8.3 J/mol-K wherever required.

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

Solution 45

Here ,

P=84 cm of Hg

## Chapter 2 - Kinetic Theory of Gases Exercise 37

Question 46

Use R=8.3 J/mol-K wherever required.

The weather report reads, "Temperature 20 : Relative humidity 100%". What is the dew point?

Solution 46

Temperature is 20 and relative humidity is 100%, So air is in saturated condition.

Dew point is the temperature at which SVP is equal to present vapour pressure.

So, 20 is the dew point.

Question 47

Use R=8.3 J/mol-K wherever required.

The condition of air in a closed room is described as follows. Temperature=25, 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=3.2 kPa.

Solution 47

Pa

When vapours are removed VP reduces to zero.

Net pressure inside the room now=

= KPa

102 KPa

Question 48

Use R=8.3 J/mol-K wherever required.

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

Solution 48

Air becomes saturated at 10.

If room temperature falls to 15 then also dew point=10.

Question 49

Use R=8.3 J/mol-K wherever required.

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

Solution 49

The point where the vapor starts, condensing VP=SVP.

Since, the process is isothermal.

Question 50

Use R=8.3 J/mol-K wherever required.

A barometer tube is 80 cm long (above 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 vapor pressure at the room temperature is 1.0 cm.

Solution 50

76 cm of Hg

Now, when water is introduced the water vapor exerts some pressure against atmospheric pressure.

Pressure of vapor=76-75.4

=0.6 cm of Hg

Question 51

Use R=8.3 J/mol-K wherever required.

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

Solution 51

At pressure of 760 mm we drop perpendicular on temperature axis, So, T=65

Similarly, at 0.5 atm, T=48

Question 52

Use R=8.3 J/mol-K wherever required.

The human body has an average temperature of 98. Assume that 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.

Solution 52

Temperature of body=

=

=36.7

From graph, pressure corresponding to temperature 36.7 is 50 mm of Hg.

Question 53

Use R=8.3 J/mol-K wherever required.

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

Solution 53

Dew point=10

At 20, SVP=17.5 mm of Hg

At 10 (dew point), SVP=8.9 mm of Hg

=51%

Question 54

Use R=8.3 J/mol-K wherever required.

of saturated vapour is cooled down from 30 to 20. Find the mass of the water condensed. The absolute humidity of saturated water vapour is  at 30 and  at 20

Solution 54

At 30,  of air contains 30 gm of water vapour

In  air, water vapour=

=1500 gm

At 20,  of air contains 16 gm of water vapour

In  air, water vapour=

=800 gm

Amount of water vapour condensed=1500-800=700 gm

Question 55

Use R=8.3 J/mol-K wherever required.

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 atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value.

Solution 55

76 cm of Hg

SVP=0.8 cm of Hg

When water is introduced into the barometer, water evaporates and exerts pressure on mercury meniscas.

Pressure is minimum when the vapour reaches at saturation.

Net length of Hg column at SVG=76-0.8

=75.2 cm

Question 56

Use R=8.3 J/mol-K wherever required.

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. The water level in the jar is same as the level outside. The saturation vapour pressure at 27 is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

Solution 56

Pressure exerted by  vapour=V.P

=99.4KPa-3.4KPa

=96kPa

Using, PV=nRT

96×

Question 57

Use R=8.3 J/mol-K wherever required.

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

Solution 57

Let length of barometer tube be x

Initially,

1 cm of Hg

Later

0.9 cm of Hg

Now,

X=91.1 cm

Question 58

Use R=8.3 J/mol-K wherever required.

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

Solution 58

At 0,

1.84 mm of Hg

Using PV=nRT;

1.97

RH=10.9%

Question 59

Use R=8.3 J/mol-K wherever required.

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

Solution 59

VP=1800 Pa

Extra pressure required for saturation=3600-1800=1800 Pa

Let mass of vapour required be m for extra pressure

m=13 gm

Question 60

Use R=8.3 J/mol-K wherever required.

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

Solution 60

VP=660 Pa

Now,

PV=nRT

m=238.5 gm

m238 gm

Question 61

Use R=8.3 J/mol-K wherever required.

The temperature and relative humidity are 300 K and 20% in a room of volume . 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.

Solution 61

VP=660 Pa

Pressure for evaporated water is given by

1385 Pa

Net pressure=VP+

=660+1385

=2045 Pa

= 61.9%

R.H62%

Question 62

Use R=8.3 J/mol-K wherever required.

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

Solution 62

(a)

VP=0.64 kPa

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change= (1.6-0.64) kPa

= 0.96 kPa

Let mass of water evaporated be m. Then,

PV=nRT

m=361.45 gm

m361 gm

(b) At 20, SVP=2.4 KPa

At 15, SVP=1.6 KPa

Net pressure change=2.4-1.6=0.8 KPa

Mass of water evaporated

PV=

m=296 gm

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