H C VERMA Solutions for Class 12-science Physics Chapter 5 - Specific Heat Capacities of Gases

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Chapter 5 - Specific Heat Capacities of Gases Exercise 77

Question 1

A vessel containing one mole of a monatomic ideal gas (molecular weight =20g/mol) is moving on a floor at a speed of 50 m/s. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Solution 1

Mechanical energy is converted into internal energy. Thus,

  

  

  

Question 2

5g of a gas is contained in a rigid container and is heated from 15°C to 25°C. Specific heat capacity of the gas at constant volume is 0.172cal/g-°c and the mechanical equivalent of heat is 4.2 J/cal. Calculate the change in the internal energy of the gas.

Solution 2

We know,

dQ=dU+dW

dQ=dU (As dW=0)

  

  

Q=8.6cal

Or Q=36.12J

Question 3

Figure. Shows a cylindrical container containing oxygen (𝛾=1.4) and closed by aa 50kg frictionless piston. The area of cross section is 100cm3, atmospheric pressure is 100kPa and g is 10m/s2. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20cm.

  

Solution 3

We know,

dW=PdV

  

  

dW=300J

Also, dW=nRdt

  

And   

  

  

dQ=1050J

Question 4

The specific heat capacities of hydrogen at constant volume and at constant pressure are 2.4 cal/g-°c and 3.4 cal/g-°c respectively, The molecular weight of hydrogen is 2g/mol and the gas constant R =8.3×107 erg/mol-°c. Calculate the value of J.

Solution 4

We know,

  

So

  

  

Also,

  

  

  

Question 5

The ratio of the molar heat capacities of an ideal gas is Cp/Cv=7/6. Calculate the change in internal energy of 1.0 mol of the gas when its temperature is raised by 50K (a) keeping the pressure constant, (b) keeping the volume constant and (c) adiabatically.

Solution 5

a) We know,

dQ=dU+dW

dU=dQ-dU

  

  

  

=6RdT

  

dU=2490J

b) dU=dQ

  

  

  

dU=2490J

c) Adiabatically   

  

  

  

  

dU=2490J

Chapter 5 - Specific Heat Capacities of Gases Exercise 78

Question 6

A sample of air weighing 1.18g occupies 1.0×103cm 3 when kept at 300 k and 1.0×105 Pa. When 2.0cal of heat is added to it at constant volume, its temperature increase the temperature of air by 1°c t constant pressure , if the temperature of air by 1°c at constant pressure, if the mechanical equivalent of heat is 4.2×107 erg/cal. Assume that air behave as an ideal gas.

Solution 6

We know,

PV = nRT

  

  

  

Also,

  

  

  

Also,

  

  

=51.18

  

  

=2.08cal

Question 7

An ideal gas expands from 100cm3 to 200cm3 at a constant pressure of 2.0×105 Pa when 50J of heat is supplied to it. Calculate (a) the change in internal energy of the gas, (b) the number of moles in the gas if the initial temperature is 300K, (c) the molar heat capacity Cp at constant pressure and (d) the molar heat capacity Cv at constant volume.

Solution 7

a) We know,

dQ=dU+dW

dU=dQ-dW

=dQ-PdV

  

dU=30J

b) For monoatomic gas

  

  

n=0.008

c) We know,

  

  

  

And   

=12.5+8.3=20.3

d) From equation 1

  

Question 8

An amount Q of heat is added to a monatomic ideal gas in a process in which the gas performs a work Q/2 on its surrounding. Find the molar heat capacity for the process.

Solution 8

We know,

  

  

  

  

Also,

  

  

Q=3nRT

Or

  

  

Question 9

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p=kV. Show that the molar heat capacity of the gas for the process is given by C=Cv+ .

Solution 9

As given:

P=kV

  

  

RdT=2kVdV

  

We know,

dQ=dU+dW

  

  

  

  

Question 10

An ideal gas (c9/cv=𝛾) is taken through a process in which the pressure and the volume vary as p=aVb. Find the value of b for which the specific heat capacity in the process is zero.

Solution 10

As given

  

  

  

  

  

  

Question 11

Two ideal gases have the same value of (c9/cv=𝛾). What will be the value of this ratio for a mixture of the two gases in the ratio 1:2?

Solution 11

For   ideal gas:

  

For   ideal gas:

  

As given   

i.e.   

  

When we mix the gas

  

  

  

  

  

Also, 

  

Taking ratio of equation 1 and 2

  

Question 12

A mixture contains 1 mole of helium (Cp=2.5R, Cv=1.5R) and 1 mole of hydrogen (Cp=3.5R, Cv=2.5R). Calculate the values of Cp,Cv and 𝛾 for the mixture.

Solution 12

We know,

  

  

  

  

Also,

  

  

Ratio   

  

Question 13

Half mole of an ideal gas (𝛾 = 5/3) is taken through the cycle abcda as shown in fig. Take R=  (a) Find the temperature of the gas in the states a,b,c and d. (b) Find the amount of heat supplied in the processes ab and bc (c) Find the amount of heat liberated in the processes cd and da.

 

  

Solution 13

a) We know,

  

  

  

  

b) We know,

  

  

  

=1250J

Also, dQ=dU+dW [[For bc process, dW=0]

  

  

  

  

c) Heat liberated,   

  

  

=750J

d) Also, Heat libeated,  (for cd process)

  

  

=2500J

Question 14

An ideal gas (𝛾=1.67) is taken through the process abc shown in fig. The temperature at the point a is 300K. Calculate (a) the temperature at b and c, (b) the work done in the process, (c) the amount of heat supplied in the path ab and in the path bc and (d) the change in the internal energy of the gas in the process.

 

  

Solution 14

a) In case of ab ,   is constant

Thus,   

  

  

In case of bc, Pis constant

Thus   

  

  

 

b) Work done=Area under graph

  

c) We know,

  

  

  

Q=14.9J

And also,

  

  

  

Q=24.9J

d) We know,

  

  

=(24.9+14.9)-1 [from1 and 2]

  

Question 15

In Joly's differential steam calorimeter, 3g of an ideal gas is contained in a rigid closed sphere at 20°c. The sphere is heated by steam at 100°c and it is found that an extra 0.095g steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J/g-K. The latent heat of vaporization of water=540 cal/g.

Solution 15

We know

  

  

  

Question 16

The volume of an ideal gas (𝛾 = 1.5) is changed adiabatically from 4.00 liters to 3.00 liters. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.

Solution 16

a) We know

  

  

  

b) We know,

  

  

  

  

Question 17

An ideal gas at pressure 2.5×105 Pa and temperature 300K occupies 100cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure, (b) the final temperature and (c) the work done by the gas in the process. Take 𝛾 = 1.5.

Solution 17

a)   

  

  

  

b)

  

  

  

  

c) Workdone is given as:

  

  

W=21J

Question 18

Air (𝛾 = 1.4) is pumped at 2atm pressure in a motor tyre at 20°c. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmospheric air.

Solution 18

A diabetic process

Thus,

  

  

  

  

  

Question 19

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100kPa, 400cm3 and 300K respectively. The ratio of the specific heat capacities of the gas is . Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to 100cm3.

Solution 19

a) We know

  

  

  

  

Also,

  

  

  

b) When the container is slowly compressed then also heat transferred is 0 because walls are adiabatic and thus

Pressure p=800kPa

And temperature T=600K

Question 20

The initial pressure and volume of a given mass of a gas (Cp/Cv=𝛾) are p0 and V0. The gas can exchange heat with the surrounding. (a) It is slowly compressed to V0/4. Find the final pressure. (b) If the gas is suddenly compressed from the volume V0 to V0/2 and then slowly compressed to V0/4, what will be the final pressure?

Solution 20

a) First slowly compressed Isothermal compression.

So,

  

  

Now, suddenly compressed Adiabatic compression.

  

  

  

  

b) First gas is compressed suddenly Adiabatic compressin

  

  

  

  

Now, gas is compressed slowly Isothermal compression.

  

  

  

Question 21

Consider a given sample of an ideal gas (Cp/Cv=𝛾) having initial pressure p0 and volume V0. (a) The gas is isothermally taken to a pressure p0/2 and from there adiabatically to a pressure p0/4. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure p0/2 and from there isothermally to a pressure p0/4. Find the final volume.

Solution 21

a) Isothermal condition

  

  

  

Now, adiabatic condition. Thus,

  

  

  

b) Adiabatic condition:

  

  

Now, isothermal condition,

  

  

Question 22

A sample of an ideal gas (𝛾=1.5) is compressed adiabatically from a volume of 150 cm3 to 50 cm3. The initial pressure and the initial temperature are 150kPa and 300K. Find (a) the number of moles of the gas in the sample, (b) the molar heat capacity at constant volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas.

Solution 22

a) We know,

PV=nRT

  

n=0.009moles

b) We know,

  

  

  

c) Adiabatic process  

  

  

  

Or

  

d) We know,

  

  

  

W=-33J

e) We know, internal energy

  

  

W=33J

Chapter 5 - Specific Heat Capacities of Gases Exercise 79

Question 23

Three samples A, B and C of the same gas (𝛾=1.5) have equal volume and temperatures. The volume of each sample is doubled, the process being isothermal for A, adiabatic for B and isobaric for C. If the final pressure are equal for the three samples, find the ratio of the initial pressures.

Solution 23

A is isothermal, Thus.

  

  

  

B is adiabatic. Thus,

  

  

  

C is isobaric process. Thus,

  

  

And   

As given

  

  

Ratio is given as:

  

  

Question 24

Two samples A and B of the gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that   satisfies the equation  ln2.

Solution 24

Work done is given as:

  

And  

As work done is equal for both cases, Thus,

  

  

Also we know,

  

  

Substituting above equation in 3 we get

  

  

Question 25

1 liter of an ideal gas (𝛾=1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 liter. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

Solution 25

a) Adiabatic process, Thus,

  

  

  

b) We know,

  

  

  

And also

  

  

  

Now

  

  

  

c) We know,

dQ=dU+dW

dU=-dW [As dQ=0]

dU=+82J

d) from equation 1

  

  

e) Isobaric process as pressure is constant. Thus,

  

  

f) Here,   

  

  

For isothermal condition,

  

  

  

g) Net work done,   

=-82-41.4+103

=-20.4J

Question 26

Fig. shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (𝛾=1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1:3. Find the ratio of the temperatures in the two parts of the vessel.

 

  

Solution 26

In this, adiabatic process is going on. So

  

 

  

 

So,   

  

Question 27

Figure shows two rigid vessels A and B each of volume 200 cm3 containing an ideal gas (Cv=12.5 J/mol-K). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in the vessel. (b) 5.0 J of heat is supplied to the gas in the vessel A and 10J to the gas in the vessel B. Assuming no appreciable transfer of heat from A to B calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant R=8.3 J/mol-K.

 

  

Solution 27

a) We know,

PV=nRT

  

n=0.008

b) We know,

dQ=dU+dW

dQ=dU (dW=0,as no change in volume)

  

  

For A,

  

And   

  

  

For B,

  

And   

  

  

Distance moved by mercury =  

=25-12.5

  

Question 28

Figure shows two vessels with adiabatic walls, one containing 0.1 g of helium (𝛾=1.67) M=4g/mol) and the other containing some amount of hydrogen (𝛾=1.4, M=2g/mol). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to the two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.

  

Solution 28

Adiabatic process, Thus.

For  ,   

  

For  ,   

  

Where m is mass of H2

Equating 1 and 2 we get,

  

m=0.029

m=0.03g

Question 29

Two vessels A and B of equal V0 are connected by a narrow tube which can be closed by a valve. The vessels are fitted with pistons which can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cp/Cv=𝛾) at atmospheric pressure p0 and atmospheric temperature T0. The walls of the vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperature and pressure in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and the pressure.

Solution 29

a) Temperature remains constant as A is diathermic

Thus,   

  

  

And   

Now as B is adiabatic,

  

  

Solving we get

  

Also,

  

  

  

  

b) Here, temperature remains constant because value is open.

  

And

  

A is diathermic thus, T and V are constant and so pressure also remains constant.

  

B is adiabatic. Thus,

  

  

Or   

Question 30

Fig shows an adiabatic cylindrical tube of volume V0 divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part

(Cp/Cv=𝛾) is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts, (b) the heat given to the gas in the left part and (c) the final common pressure of the gases.

 

  

Solution 30

a) Adiabatic process,

  

And

  

 

  

 

  

When equilibrium is reached then

 

  

  

Solving we get

  

And

  

b) Process is adiabatic so no heat is transferred to left part.

Q=0

c) From equation 1

  

And from equation 3

  

Thus, equation 5 becomes:

  

  

Question 31

An adiabatic cylindrical tube of cross-sectional area 1cm3 is closed at one end and fitted with a piston at the other end. The tube contains 0.03g of an ideal gas. At 1 atm pressure and at the temperature of the surrounding, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure of the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.

Solution 31

Process is adiabatic. Thus,

  

  

  

  

Also,

  

V=447m/s

Chapter 5 - Specific Heat Capacities of Gases Exercise 80

Question 32

The speed of sound in hydrogen at 0°C is 1280 m/s. The density of hydrogen at STP is 0.089 kg/m3. Calculate the molar heat capacities Cp and Cv of hydrogen.

Solution 32

We know,

Spedd of sound,   

  

  

Also,

  

=18.0J/mol-x

Also,

  

  

=26.3J/mol-K

Question 33

4.0g of helium occupies 22400cm3 at STP. The specific heat capacity of helium at constant pressure is 5.0 cal/mo;-K. Calculate the speed of sound in helium at STP.

Solution 33

We know,

  

  

  

Also, speed od sound V=  

  

V=960m/s

Question 34

An ideal gas having density 1.7×10-3g/cm3 at a pressure 1.5×105 Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0cm. Calculate the molar heat capacities Cp and Cv of the gas.

Solution 34

In kundt's tube, node separation is given as:

  

  

Speed of sound   

  

=360m/s

Also,

  

Thus,

  

  

Now, we know,

  

  

Also,

  

  

  

Question 35

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K, The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities Cp and Cv of the gas.

Solution 35

Speed of sound   

  

=330m/s

Also,

  

{PV=nRT

 }

  

So,   

  

  

We know,

  

  

And