Please copy and paste this embed script to where you want to embed

Chapter # 27

Specific Heat Capacities of Gases

SOLVED EXAMPLES

EXAMPLE 27.1 0.32 g of oxygen is kept a rigid container and is heated. Find the amount of heat needed to raise the temperature from 250 C to 350 C. The molar heat capacity of oxygen at constant volume is 20 J/mol–K. Sol. The moleculer weight of oxygen = 32 g/mol.

0.32 g n = 32 g/mol = 0.01 mol . The amount of heat needed is Q = n Cv T = ( 0.01 mol ) ( 20 J/mol–K ) (10 K) = 2.0 J. EXAMPLE 27.2 A tank of volume 0.2 m 3 contains helium gas at a temperature of 300 K and pressure 1.0 × 10 5 N/m 2 . Find the amount of heat required to raise the temperature to 400 K. The molar heat capacity of helium at constant volume is 3.0 cal/mol–K. Neglect any expansion in the volume of the tank. Sol. The amount of the gas in moles is n=

=

pV RT (1.0 10 5 N/m 2 ) (0.2 m 3 ) = 8.0 mol. (8.31 J / mol K )(300 K)

The amount of heat required is Q = n C v T = ( 8.0 mol ) ( 3.0 cal/mol–K ) (100 K ) = 2400 cal. EXAMPLE 27.3 The molar heat capacity of a gas at constant volume is found to be 5 cal/mol–K. Find the ratio = CP – CV for the gas. The gas constant R = 2 cal/mol–K. Sol. We have Cv = 5 cal/mol–K. Thus, CP = Cv + R = 5 cal/mol – K = 7 cal/mol– K

or,

Cp Cv

=

7 = 1.4 . 5

EXAMPLE 27.4 Dry air at 150 C and 10 atm is suddenly released at atmosphere pressure. Find the final temperature of the air [ CP / Cv = 1.4 ]. Sol. As the air is suddenly released, it does not get time to exchange heat with the surrounding. Thus the process is adiabatic. Assuming the process to be reversible,

T

= constant

p 1

or ,

T1

1

p1

or,

or,

=

T2 p2

1

T2 p = 2 T1 p1

1

T 2 = T 1 p2 p 1

1 1

Talking p1 = 10 atm, p2 = 1 atm, = 1.41 and T 1 = ( 273 + 15 ) K = 288 K, the final temperature is T 2 = 148 K. manishkumarphysics.in

Page # 1

Chapter # 27 Specific Heat Capacities of Gases EXAMPLE 27.5 Calculate the internal energy of 1 g of oxygen at STP. Sol. Oxygen is a diatomic gas. The average energy per molecules is, therefore, mole is

5 kT and the average energy per 2

5 RT. As the molecular weight of oxygen is 32 g/mol, 1 g of oxygen has 2

1g 1 n = 32 g/mol = mol. 32 The temperature of oxygen is 273 K. Thus, the internal energy is 5 U = n RT 2 1 5 mol ( 8.31 J/mol–K) ( 273 K ) = 32 2 = 177 J.

QUESTIONS

FOR

SHORT

ANSWER

1.

Does a gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?

2.

Can we define specific heat capacity at constant temperature?

3.

Can we define specific heat capacity for an adiabatic process?

4.

Does a solid also have two kinds of molar heat capacities Cp and Cv ? If yes, do we have Cp > Cv ? Cp – Cv = R ?

5.

In a real gas the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Looking into the derivation of Cp – Cv = R, find whether Cp – Cv will be more than R, less than R or equal to R for a real gas.

6.

Can a process on an ideal gas be both adiabatic and isothermal?

7.

Show that the slope of p–V diagram is greater for an adiabatic process as compared to an isothermal process.

8.

Is a slow process always isothermal? Is a quick process always adiabatic?

9.

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?

10.

The ratio Cp/Cv for a gas is 1.29. What is the degree of freedom of the molecules of this gas?

Objective - I 1.

Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes. fdlh izØe A esa ,d vkn'kZ xSl }kjk fd;k x;k dk;Z fdlh nwljs izØe B esa fd;s x;s dk;Z dk nqxquk gSA nksuksa izØeksa esa rki leku ek=kk esa c Cv

(D*) C = 0

8.

In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about fdlh vkn'kZ xSl a ij ,d lerkih izØe esa nkc 0.5% ls c pB (B) pA = pB (C*) pA < pB (D) The relation between pA and pB cannot be deduced. (D) pA o pB ds chp lEcU/k fuxZfer ugha fd;k tk ldrk gSA 11.

Let Ta and Tb be the final temperature of the samples A and B respectively in the previous question then : ekuk Åij fn;s x;s iz'u esa nksuksa uequksa A o B ds vfUre rki Øe'k% Ta o Tb gSA (A) Ta < Tb (B) Ta = Tb (C*) Ta > Tb (D) The relation between Ta and Tb cannot be deduced. (D) Ta o Tb ds chp lEcU/k fuxZfer ugha fd;k tk ldrk gSA

12.

Let W a and W b be the work done by the systems A and B respectively in the previous question then : ekuk Åij fn;s x;s iz'u esa fudk;ksa A o B }kjk fd;k x;k dk;Z Øe'k% W a o W b gSA (A) W a > W b (B) W a = W b (C*) W a < W b (D) The relation between W a and W b cannot be deduced. (D) W a o W b ds chp lEcU/k izkIr ugha fd;k tk ldrk gSA

13.

The molar heat capacity of oxygen gas at STP is nearly 2.5 R. As the temperature is increased, it gradually increases and approaches 3.5 R. The most appropriate reason for this behaviour is that at high temperatures (A) oxygen does not behave as an ideal gas (B) oxygen molecules dissociate in atoms (C) the molecules collides more frequently (D*) molecular vibration gradually become effective ekud rki ,oa nkc (STP) ij vkWDlhtu xSla dh eksyj Å"ek /kkfjrk yxHkx 2.5 R. gSA tc rki c

View more...
Specific Heat Capacities of Gases

SOLVED EXAMPLES

EXAMPLE 27.1 0.32 g of oxygen is kept a rigid container and is heated. Find the amount of heat needed to raise the temperature from 250 C to 350 C. The molar heat capacity of oxygen at constant volume is 20 J/mol–K. Sol. The moleculer weight of oxygen = 32 g/mol.

0.32 g n = 32 g/mol = 0.01 mol . The amount of heat needed is Q = n Cv T = ( 0.01 mol ) ( 20 J/mol–K ) (10 K) = 2.0 J. EXAMPLE 27.2 A tank of volume 0.2 m 3 contains helium gas at a temperature of 300 K and pressure 1.0 × 10 5 N/m 2 . Find the amount of heat required to raise the temperature to 400 K. The molar heat capacity of helium at constant volume is 3.0 cal/mol–K. Neglect any expansion in the volume of the tank. Sol. The amount of the gas in moles is n=

=

pV RT (1.0 10 5 N/m 2 ) (0.2 m 3 ) = 8.0 mol. (8.31 J / mol K )(300 K)

The amount of heat required is Q = n C v T = ( 8.0 mol ) ( 3.0 cal/mol–K ) (100 K ) = 2400 cal. EXAMPLE 27.3 The molar heat capacity of a gas at constant volume is found to be 5 cal/mol–K. Find the ratio = CP – CV for the gas. The gas constant R = 2 cal/mol–K. Sol. We have Cv = 5 cal/mol–K. Thus, CP = Cv + R = 5 cal/mol – K = 7 cal/mol– K

or,

Cp Cv

=

7 = 1.4 . 5

EXAMPLE 27.4 Dry air at 150 C and 10 atm is suddenly released at atmosphere pressure. Find the final temperature of the air [ CP / Cv = 1.4 ]. Sol. As the air is suddenly released, it does not get time to exchange heat with the surrounding. Thus the process is adiabatic. Assuming the process to be reversible,

T

= constant

p 1

or ,

T1

1

p1

or,

or,

=

T2 p2

1

T2 p = 2 T1 p1

1

T 2 = T 1 p2 p 1

1 1

Talking p1 = 10 atm, p2 = 1 atm, = 1.41 and T 1 = ( 273 + 15 ) K = 288 K, the final temperature is T 2 = 148 K. manishkumarphysics.in

Page # 1

Chapter # 27 Specific Heat Capacities of Gases EXAMPLE 27.5 Calculate the internal energy of 1 g of oxygen at STP. Sol. Oxygen is a diatomic gas. The average energy per molecules is, therefore, mole is

5 kT and the average energy per 2

5 RT. As the molecular weight of oxygen is 32 g/mol, 1 g of oxygen has 2

1g 1 n = 32 g/mol = mol. 32 The temperature of oxygen is 273 K. Thus, the internal energy is 5 U = n RT 2 1 5 mol ( 8.31 J/mol–K) ( 273 K ) = 32 2 = 177 J.

QUESTIONS

FOR

SHORT

ANSWER

1.

Does a gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?

2.

Can we define specific heat capacity at constant temperature?

3.

Can we define specific heat capacity for an adiabatic process?

4.

Does a solid also have two kinds of molar heat capacities Cp and Cv ? If yes, do we have Cp > Cv ? Cp – Cv = R ?

5.

In a real gas the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Looking into the derivation of Cp – Cv = R, find whether Cp – Cv will be more than R, less than R or equal to R for a real gas.

6.

Can a process on an ideal gas be both adiabatic and isothermal?

7.

Show that the slope of p–V diagram is greater for an adiabatic process as compared to an isothermal process.

8.

Is a slow process always isothermal? Is a quick process always adiabatic?

9.

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?

10.

The ratio Cp/Cv for a gas is 1.29. What is the degree of freedom of the molecules of this gas?

Objective - I 1.

Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes. fdlh izØe A esa ,d vkn'kZ xSl }kjk fd;k x;k dk;Z fdlh nwljs izØe B esa fd;s x;s dk;Z dk nqxquk gSA nksuksa izØeksa esa rki leku ek=kk esa c Cv

(D*) C = 0

8.

In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about fdlh vkn'kZ xSl a ij ,d lerkih izØe esa nkc 0.5% ls c pB (B) pA = pB (C*) pA < pB (D) The relation between pA and pB cannot be deduced. (D) pA o pB ds chp lEcU/k fuxZfer ugha fd;k tk ldrk gSA 11.

Let Ta and Tb be the final temperature of the samples A and B respectively in the previous question then : ekuk Åij fn;s x;s iz'u esa nksuksa uequksa A o B ds vfUre rki Øe'k% Ta o Tb gSA (A) Ta < Tb (B) Ta = Tb (C*) Ta > Tb (D) The relation between Ta and Tb cannot be deduced. (D) Ta o Tb ds chp lEcU/k fuxZfer ugha fd;k tk ldrk gSA

12.

Let W a and W b be the work done by the systems A and B respectively in the previous question then : ekuk Åij fn;s x;s iz'u esa fudk;ksa A o B }kjk fd;k x;k dk;Z Øe'k% W a o W b gSA (A) W a > W b (B) W a = W b (C*) W a < W b (D) The relation between W a and W b cannot be deduced. (D) W a o W b ds chp lEcU/k izkIr ugha fd;k tk ldrk gSA

13.

The molar heat capacity of oxygen gas at STP is nearly 2.5 R. As the temperature is increased, it gradually increases and approaches 3.5 R. The most appropriate reason for this behaviour is that at high temperatures (A) oxygen does not behave as an ideal gas (B) oxygen molecules dissociate in atoms (C) the molecules collides more frequently (D*) molecular vibration gradually become effective ekud rki ,oa nkc (STP) ij vkWDlhtu xSla dh eksyj Å"ek /kkfjrk yxHkx 2.5 R. gSA tc rki c

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.