Physics for Scientists and Engineers, Volume 1
Physics for Scientists and Engineers, Volume 1
9th Edition
ISBN: 9781133954156
Author: Raymond A. Serway
Publisher: CENGAGE L
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Chapter 21, Problem 21.51AP

A certain ideal gas has a molar specific heat of Cv = 7 2 R. A 2.00-mol sample of the gas always starts at pressure 1.00 × 105 Pa and temperature 300 K. For each of the following processes, determine (a) the final pressure, (b) the final volume, (c) the final temperature, (d) the change in internal energy of the gas, (e) the energy added to the gas by heat, and (f) the work done on the gas. (i) The gas is heated at constant pressure to 400 K. (ii) The gas is heated at constant volume to 400 K. (iii) The gas is compressed at constant temperature to 1.20 × 105 Pa. (iv) The gas is compressed adiabatically to 1.20 × 105 Pa.

(a)

Expert Solution
Check Mark
To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 The final pressure is 100kPa , final volume is 66.5L , final temperature is 400K , change in the internal energy is 5.82kJ , the energy added to the gas by heat is 7.48kJ and the work done on the gas is 1.66kJ .

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is 72R , the amount of the gas is 2.00mol , the initial pressure of the gas is 1.00×105Pa and the initial temperature of the gas is 300K . The final temperature of the gas is 400K .

The value of universal gas constant is 8.314J/molK   .

The gas is heated at constant pressure.

Write the expression for molar specific heat of the gas at constant pressure.

Cp=R+Cv

Here,

R is the universal gas constant .

Cv is the molar specific heat of the gas at constant volume.

Substitute 72R for Cv in the above expression.

Cp=R+72R=92R

The gas is heated at constant pressure, therefore the final pressure of the gas is equal to the initial pressure of the gas.

The expression for the final pressure of the gas is,

P2=P1

Here,

P1 is the initial pressure of the gas.

P2 is the final pressure of the gas.

Substitute 1.00×105Pa for P1 in the above expression.

P2=1.00×105Pa=100kPa

The final pressure of the gas is 100kPa .

The final temperature is the temperature to which the gas is heated.

The value of final temperature of the gas is,

T2=400K

Write the expression for the change in the temperature of the gas.

ΔT=T2T1

Here,

T2 is the final temperature.

T1 is the initial temperature

Substitute 300K for T1 and 400K for T2 in the above expression.

ΔT=400K100K=100K

The ideal gas equation is,

PV=nRT

Here,

P is the pressure.

V is the volume.

n is the number of moles.

T is the temperature.

Rearrange the above equation for the value of V .

V=nRTP

Substitute V2 for V , P2 for P and T2 for T in the above expression.

V2=nRT2P2

Substitute 1.00×105Pa for P2 2.00mol for n , 8.314J/molK   for R and 400K for T2 in the above expression.

V2=(2.00mol)(8.314J/molK  )(400K)(1.00×105Pa)=66.5L

Thus, the final volume of the gas is 66.5L .

The expression for the change in the internal energy of the gas is,

ΔE=nCvΔT

Substitute 72R for Cv in the above expression.

ΔE=n72RΔT

Substitute 2.00mol for n , 8.314J/molK   for R and 100K for ΔT in the above expression.

ΔE=(2.00mol)72(8.314J/molK  )(100K)=5.82×103J=5.82kJ

Thus the change in the internal energy of the gas is 5.82kJ .

The expression for the energy added to the gas by heat is,

Q=nCpΔT

Substitute 92R for Cp in the above expression.

Q=n92RΔT

Substitute 2.00mol for n , 8.314J/molK   for R and 100K for ΔT in the above expression.

Q=(2.00mol)92(8.314J/molK  )(100K)=7.48×103J×103kJ1J=7.48kJ

Thus the energy added to the gas by heat is 7.48kJ .

The formula for the work done on the gas is,

W=ΔEQ

Substitute 5.82kJ for ΔE and 7.48kJ for Q in the above expression.

W=5.82kJ7.48kJ=1.66kJ

Thus the work done on the gas is 1.66kJ .

Conclusion:

Therefore, for the gas, the final pressure is 1.00×105Pa , final volume is 66.5L , final temperature is 400K , change in the internal energy is 5.82kJ , the energy added to the gas by heat is 7.48kJ and the work done on the gas is 1.66kJ .

(b)

Expert Solution
Check Mark
To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is 133kPa , final volume is 49.9L , final temperature is 400K , change in the internal energy is 5.82kJ , the energy added to the gas by heat is 5.82kJ and the work done on the gas is 0 .

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is 72R , the amount of the gas is 2.00mol , the initial pressure of the gas is 1.00×105Pa and the initial temperature of the gas is 300K . The final temperature of the gas is 400K .

The value of universal gas constant is 8.314J/molK   .

The gas is heated at constant volume.

The gas is heated at constant volume, therefore the final volume of the gas is equal to the initial pressure of the gas.

The formula or ideal gas is,

PV=nRT

Rearrange the above equation for the value of V .

V=nRTP

Substitute V1 for V , P1 for P and T1 for T in the above expression.

V1=nRT1P1

Write the expression for the final volume of the gas.

V2=V1

Substitute nRT1P1 for V1 in the above expression.

V2=nRT1P1

Substitute 1.00×105Pa for P1 2.00mol for n , 8.314J/molK   for R and 300K for T1 in the above expression.

V2=(2.00mol)(8.314J/molK  )(300K)(1.00×105Pa)=49.87L49.9L

Thus, the final volume of the gas is 49.9L .

The final temperature is the temperature to which the gas is heated.

The value of final temperature of the gas is,

T2=400K

Write the expression for the change in the temperature of the gas.

ΔT=T2T1

Here,

T2 is the final temperature.

T1 is the initial temperature

Substitute 300K for T1 and 400K for T2 in the above expression.

ΔT=400K100K=100K

Write the expression for ideal gas.

PV=nRT

Rearrange the above equation for the value of P .

P=nRTV

Substitute V1 for V , P1 for P and T1 for T in the above expression.

P1=nRT1V1 (1)

Substitute V2 for V , P2 for P and T2 for T in the rearranged expression of the ideal gas for value of P

P2=nRT2V2 (2)

Divide equation (2) from equation (1).

P2P1=nRT2V2nRT1V1P2P1=T2T1P2=P1T2T1

Substitute 1.00×105Pa for P1 , 300K for T1 and 400K for T2 in the above expression.

P2=(1.00×105Pa)(400K)300K=133.33kPa133kPa

The expression for the change in the internal energy of the gas is,

ΔE=nCvΔT

Substitute 72R for Cv in the above expression.

ΔE=n72RΔT

Substitute 2.00mol for n , 8.314J/molK   for R and 100K for ΔT in the above expression.

ΔE=(2.00mol)72(8.314J/molK  )(100K)=5.82×103J×103J1kJ=5.82kJ

Thus the change in the internal energy of the gas is 5.82kJ .

The formula for the energy added to the gas by heat for constant volume is,

Q=nCvΔT

Substitute 72R for Cv in the above expression.

Q=n72RΔT

Substitute ΔE for n72RΔT in the above expression.

Q=ΔE

Substitute 5.82kJ for ΔE in the above expression.

Q=5.82kJ

Thus the energy added to the gas by heat is 5.82kJ .

The forrmula for the work done on the gas is,

W=ΔEQ

Substitute 5.82kJ for ΔE and 5.82kJ for Q in the above expression.

W=5.82kJ5.82kJ=0

Thus the work done on the gas is 0 .

Conclusion:

Therefore, for the gas, the final pressure is 133kPa , final volume is 49.9L , final temperature is 400K , change in the internal energy is 5.82kJ , the energy added to the gas by heat is 5.82kJ and the work done on the gas is 0 .

(c)

Expert Solution
Check Mark
To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is 120kPa , final volume is 41.6L , final temperature is 300K , change in the internal energy is 0 , the energy added to the gas by heat is 909J and the work done on the gas is +909J .

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is 72R , the amount of the gas is 2.00mol , the initial pressure of the gas is 1.00×105Pa and the initial temperature of the gas is 300K . The final pressure of the gas is 1.20×105Pa .

The value of universal gas constant is 8.314J/molK   .

The gas is compressed at constant temperature.

The gas is heated at constant volume, therefore the final volume of the gas is equal to the initial pressure of the gas.

The formula for ideal gas is,

PV=nRT

Rearrange the above equation for the value of V .

V=nRTP

Substitute V1 for V , P1 for P and T1 for T in the above expression.

V1=nRT1P1

Substitute 1.00×105Pa for P1 2.00mol for n , 8.314J/molK   for R and 300K for T1 in the above expression.

V1=(2.00mol)(8.314J/molK  )(300K)(1.00×105Pa)=49.87L49.9L

Thus, the initial volume of the gas is 49.9L .

Substitute V2 for V , P2 for P and T2 for T in the above expression.

V2=nRT2P2

Substitute 1.20×105Pa for P2 2.00mol for n , 8.314J/molK   for R and 300K for T2 in the above expression.

V2=(2.00mol)(8.314J/molK  )(300K)(1.20×105Pa)=41.58L41.6L

Thus, the final volume of the gas is 41.6L .

The expression for the final temperature of the gas is,

T2=T1

Substitute 300K for T1 in the above expression.

T2=300K

The final temperature of the gas is 300K .

The expression for the change in the temperature of the gas is,

ΔT=T2T1

Substitute 300K for T1 and 300K for T2 in the above expression.

ΔT=300K300K=0

The final pressure of the gas is the pressure at which the gas is compressed.

The value of final; pressure of the gas is,

P2=1.20×105Pa×1kPa103Pa=120kPa

The expression for the change in the internal energy of the gas is,

ΔE=nCvΔT

Substitute 72R for Cv in the above expression.

ΔE=n72RΔT

Substitute 2.00mol for n , 8.314J/molK   for R and 0 for ΔT in the above expression.

ΔE=(2.00mol)72(8.314J/molK  )(0)=0

Thus the change in the internal energy of the gas is 0 .

The expression for the work done on the gas is,

W=PV1V2dV

Divide and multiply the above equation by V

W=PVV1V2dVV

Substitute nrT for PV in the above expression.

W=nrTV1V2dVV=nrTln(V2V1)

Substitute 49.9L for V1 , 41.6L for V2 and 2.00mol for n , 8.314J/molK   for R and 300K for T in the above expression.

W=(2.00mol)(8.314J/molK  )(300K)ln(41.6L49.9L)=907.5J909J

Thus, the work done on the system is 909J .

The expression for the energy added to the gas by heat for constant volume is,

Q=ΔEW

Substitute 909J for W and 0 for ΔE in the above expression.

Q=0909J=909J

Thus the energy added to the gas by heat is 909J .

Conclusion:

Therefore, for the gas, the final pressure is 120kPa , final volume is 41.6L , final temperature is 300K , change in the internal energy is 0 , the energy added to the gas by heat is 909J and the work done on the gas is +909J .

(d)

Expert Solution
Check Mark
To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is 120kPa , final volume is 43.3L , final temperature is 312K , change in the internal energy is +722J , the energy added to the gas by heat is 0 and the work done on the gas is +722J .

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is 72R , the amount of the gas is 2.00mol , the initial pressure of the gas is 1.00×105Pa and the initial temperature of the gas is 300K . The final pressure of the gas is 1.20×105Pa .

The value of universal gas constant is 8.314J/molK   .

The gas is compressed adiabatically to the final pressure.

The value of the final pressure is,

P2=1.20×105Pa×1kPa103Pa=120kPa

Thus, the value of the final pressure is 120kPa .

The expression for the ratio of the specific heats is,

γ=CPCV

Substitute 72R for CV and 92R for CP in the above expression.

γ=92R72R=97

The expression for an adiabatic process for the initial condition of gas is,

P1V1γ=k (3)

Here,

k is constant value.

The expression for an adiabatic process for the final condition of gas is,

P2V2γ=k (4)

Divide equation (4) by equation (3).

P2V2γP1V1γ=kk

Rearrange the above equation for the value of V2 .

V2=V1(P1P2)1γ

Substitute 49.9L for V1 , 1.00×105Pa for P1 , 1.20×105Pa for P2 and 97 for γ in the above expression.

V2=(49.9L)(1.00×105Pa1.20×105Pa)197=43.3L

Thus, the final value of the volume is 43.3L .

The expression for ideal gas is,

PV=nRT

Rearrange the above equation .

VPT=nR (5)

Substitute V1 for V , P1 for P and T1 for T in the above expression.

P1V1T1=nR (6)

Substitute V2 for V , P2 for P and T2 for T in the equation (5).

P2V2T2=nR (7)

Divide equation (7) by equation (6).

P2V2T2P1V1T1=nRnR

Rearrange the above expression for the value of T2 .

T2=T1(P2V2P1V1)

Substitute 1.00×105Pa for P1 , 1.20×105Pa for P2 49.9L for V1 , 43.3L for V2 and 2.00mol for n , 8.314J/molK   for R and 300K for T1 in the above expression.

T2=(300K)((1.20×105Pa)(43.3L)(1.00×105Pa)(49.9L))=312.38K312K

The expression for the change in the temperature of the gas is,

ΔT=T2T1

Here,

T2 is the final temperature.

T1 is the initial temperature

Substitute 300K for T1 and 312.4K for T2 in the above expression.

ΔT=312.4K100K=12.4K

Thus the change in the temperature is 12K .

The expression for the change in the internal energy of the gas is,

ΔE=nCvΔT

Substitute 72R for Cv in the above expression.

ΔE=n72RΔT

Substitute 2.00mol for n , 8.314J/molK   for R and 12.4K for ΔT in the above expression.

ΔE=(2.00mol)72(8.314J/molK  )(12.4K)=721.7J722J

Thus the change in the internal energy of the gas is 722J .

The adiabatic process is insulated to heat supplied externally.

The expression for the energy added to the gas by heat is,

Q=0

Thus the energy added to the gas by heat is 0 .

The expression for the work done on the gas is,

W=ΔEQ

Substitute 722J for ΔE and 0 for Q in the above expression.

W=722J0=722J

Thus the work done on the gas is 722J .

Conclusion:

Therefore, for the gas, the final pressure is 120kPa , final volume is 43.3L , final temperature is 312K , change in the internal energy is +722J , the energy added to the gas by heat is 0 and the work done on the gas is +722J .

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Chapter 21 Solutions

Physics for Scientists and Engineers, Volume 1

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