Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 15, Problem 91P

(a)

To determine

The heat flow in or out of the gas during each step.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The heat flow in or out of the gas during each step are 36.5kJ_, 36.3kJ_ and 21.8kJ_.

Explanation of Solution

Write the expression for the heat for the isothermal expansion.

Q1=ΔU1Winput (I)

Here, Q1 is the heat for the first process of isothermal expansion of the gas, ΔU1 is the change in the internal energy for the process and Winput is the input work done.

Write the expression for the heat for the isobaric compression.

Q2=ΔU2Winput (II)

Here, Q2 is the heat for the second process of isobaric compression of the gas, ΔU2 is the change in the internal energy for the process and Winput is the input work done.

Write the expression for the heat for the isochoric process.

Q3=ΔU3Winput (III)

Here, Q3 is the heat for the third isochoric process of the gas, ΔU3 is the change in the internal energy for the process and Winput is the input work done.

Write the expression for contraction temperature.

T2f=T2i(V2fV2i) (IV)

Here, T2f is the final temperature, T2i is the initial temperature, V2f is the final volume, V2i is the initial volume.

Conclusion:

Substitute 0 for ΔU1 and 36.5kJ for Winput in equation (I).

Q1=0(36.5kJ)=36.5kJ

Substitute 650.0K for T2i and 19.50 for V2fV2i in equation (IV).

T2f=650.0K(19.50)=68.4K

Substitute 32nRΔT for ΔU2 and nRΔT for Winput in equation (II).

Q2=32nRΔT(nRΔT)=52nRΔT

Here, n is the number of moles, R is the universal gas constant and ΔT is the change in temperature.

Substitute 3.00mol for n, 8.314J/molK for R and 68.4K650K for ΔT in the above equation.

Q2=52(3.00mol)(8.314J/molK)(68.4K650K)=36.3kJ

Substitute 32nRΔT for ΔU3 and 0 for Winput in equation (III).

Q3=32nRΔT0=32nRΔT

Substitute 3.00mol for n, 8.314J/molK for R and 650K68.4K for ΔT in the above equation.

Q3=32(3.00mol)(8.314J/molK)(650K68.4K)=21.8kJ

Therefore, the heat flow in or out of the gas during each step are 36.5kJ_, 36.3kJ_ and 21.8kJ_.

(b)

To determine

The entropy change of the gas during the isothermal step.

(b)

Expert Solution
Check Mark

Answer to Problem 91P

The entropy change of the gas during the isothermal step is +56.2J/K_.

Explanation of Solution

Write the expression to find the change in entropy.

ΔS=Q1T

Here, Q1 is the heat in the isothermal step and T is the temperature.

Conclusion:

Substitute 36.5kJ for Q1 and 650.0K for T in the above equation.

ΔS=36.5kJ650.0K=+56.2J/K

Therefore, the entropy change of the gas during the isothermal step is +56.2J/K_.

(c)

To determine

The entropy change of the gas for the complete cycle.

(c)

Expert Solution
Check Mark

Answer to Problem 91P

The entropy change of the gas for the complete cycle is ΔSgas=0_.

Explanation of Solution

The entropy is a state variable, it depends on the initial and final values of the variable only. Since the process completes a cycle, the change in entropy of the gas is zero. But to validate the second law of thermodynamics, that the entropy of a system always increases, the entropy of the environment increases, so the system as a whole has increase in entropy.

Therefore, the entropy change of the gas for the complete cycle is ΔSgas=0_.

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Chapter 15 Solutions

Physics

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