Thermodynamics : Please read the following carefully, and show the steps needed

Question

Thermodynamics : Please read the following carefully, and show the steps needed to solve each lettered question. Thanks!

Suppose you have a body that has a constant heat capacity given by Cp and a temperature TB. It is put into contact with a reservoir that has a temperature, TR. Under isobaric conditions, there is equilibrium between both the reservoir and body. (a) Find the change of entropy for each the body and the reservoir. (b) Find an expression for the total entropy change, delta x total. (c) Prove that the total entropy change is positive for either sign of (TR - TB)/TR. For this case, you can consider that

Explanation / Answer

Two equal volume tanks are connected by a valve. Tank A is a well-insulated tank and initially contains one bar pressure of air at 400 Kelvin. Tank B is initially evacuated. The walls of Tank B are conductive with no insulation. Both tank walls are rigid.

The valve is then opened and the air freely expands in to the vacuum. Because Tank B's walls are conductive, the entire system comes to thermal equilibrium with the background temperature of 300 Kevin.

Calculate:
A: The final pressure
B: The change in entropy of the system (assume 1 mole of air is present in the system)
C: The total amount of entropy generated, i.e. the net change in entropy of the universe. (assume 1 mole of air is present in the system)

A:
P1*V1 = n*R*T1

V2 = 2*V1
P2*2*V1 = n*R*T2

n*R = P1*V1/T1
P2 = n*R*T2/(2*V1)
P2 = P1*T2/(2*T1)

B:
dSsystem = n*R*(ln(T2/T1)/(k - 1) + ln(V2/V1))dSsystem = n*R*(ln(T2/T1)/(k - 1) + ln(2))

C:
For part C, consider the situation of the process occuring first adiabatically. How much entropy is generated then?

Then, additionally consider the air once expanded to reduce in temperature as it rejects heat isochorically. Consider both the entropy loss due to the air cooling, and the entropy gain due to the background cooling.

Entropy change, assuming a well-insulated system:
deltaSthrottle = n*R*ln(V2/V1)
deltaSthrottle = n*R*ln(2)

Hint: we know temperature doesn't change because there is no change in internal energy when carried out adiabatically.

Entropy change of system air during cooling:
deltaSair = n*R*ln(T2/T1)/(k - 1)

Heat lost by system, which by conservation of energy is also heat gained by background:
Q = n*R*(T1 - T2)/(k - 1)

Entropy gain by background:
deltaSbg = Q/T2
deltaSbg = n*R*(T1 - T2)/(T2*(k-1))

Total entropy change:
deltaSnet = deltaSthrottle + deltaSair+ deltaSbg

deltaSnet = n*R*(ln(2) + ln(T2/T1)/(k - 1) + (T1 - T2)/(T2*(k-1)))

Summary:
A: P2 = P1*T2/(2*T1)
B: dSsystem = n*R*(ln(T2/T1)/(k - 1) + ln(2))
C: deltaSnet = n*R*(ln(2) + ln(T2/T1)/(k - 1) + (T1 - T2)/(T2*(k-1)))

What is k? k is the adiabatic index of air, the ratio of specific heats. Worth memorizing as 1.4.

Data:
n:=1 mole; k:=1.4; R:=8.314 J/mol-K;
T1:=400 K; T2:=300 K; P1:=100 kPa

Results:
A: P2 = 37.5 kPa
B: deltaSsystem = -0.2167 Joules/Kelvin
C: deltaSnet = +6.712 Joules/Kelvin

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