# Homework Help: Estimating the final pressure of a transformation

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1. Aug 7, 2016

### thaalescosta

1. The problem statement, all variables and given/known data
If the Helmholtz Free Energy remains constant, estimate the final pressure of 1.0mol of an ideal gas in the following transformation: (1.0atm, 300k) → (pfinal, 600k). Given Sgas = R.

2. Relevant equations
A = U - TS
dA = -SdT - pdV

3. The attempt at a solution
If the Helmholtz Free Energy is constant, then ΔA = 0.

So from dA = -SdT - pdV I'd get that S⋅ΔT = -pΔV.

But I don't see how that helps me figuring out the final pressure.

2. Aug 7, 2016

### Staff: Mentor

I think they meant for you to assume that the initial value of S for the gas is R (S0=R). If this is the case then, in terms of pfinal and $C_p$ what is Sfinal? What is $\Delta (TS)$? In terms of $C_p$, what is $\Delta U$? What is $\Delta A$?

Is the gas mono-atomic?

Last edited: Aug 7, 2016
3. Aug 8, 2016

### thaalescosta

It did not specify if the gas is mono-atomic nor did it give me the Cp or Cv values.

It only asks me to find the final pressure, which is 4 atm.

4. Aug 8, 2016

### Staff: Mentor

OK. Then please set up the equations as I suggested in post #2, and let's see where that takes us. I'm confident that this is the correct approach to use.

5. Aug 8, 2016

### Staff: Mentor

My solution to this problem gives 4 atm. if Cp/R=2. This is the same as gamma = 2. It don't think that there are any gases with this ratio, however.

Chet

6. Aug 9, 2016

### thaalescosta

You're right.

My solution was making ΔA = 0 and then setting SdT = - pdV.

Since S = R:
- SdT = (1/V)*nRT⋅dV
- dT = (1/V)*nT⋅dV (and n = 1)
- dT/T = (1/V)dV
- ln(T2/T1) = ln(V2/V1)
- ln(2) = ln(V2/V1)
1/2 = V2/V1
V1 = 2V2

(P1⋅V1)/T1 = (P2⋅V2)/T2
(1 atm ⋅2V2)/300 = (P2 ⋅ V2)/600K
P2 = 4 atm

7. Aug 9, 2016

### Staff: Mentor

I didn't assume that R was constant throughout the change. I assumed that only the initial value of S was equal to R.

For the final entropy I got:
$$S_2=R+C_p\ln(T_2/T_1)-R\ln(P_2/P_1)$$
So, $\Delta (TS)$ was:
$$\Delta (TS)=R(T_2-T_1)+C_pT_2\ln(T_2/T_1)-RT_2\ln(P_2/P_1)$$
For the change in internal energy, I got:$$\Delta U=C_v(T2-T_1)=(C_p-R)(T2-T_1)=C_p(T2-T_1)-R(T2-T_1)$$
So, $$\Delta A=\Delta U-\Delta (TS)=(C_p-2R)(T2-T_1)+C_pT_2\ln(T_2/T_1)-RT_2\ln(P_2/P_1)$$
Setting $\Delta A$ equal to zero and solving for $\ln(P_2/P_1)$, we get:
$$\ln(P_2/P_1)=\frac{C_p}{R}\ln(T_2/T_1)+\left(2-\frac{C_p}{R}\right)\left[1-\frac{T_1}{T_2}\right]$$
Only if $\frac{C_p}{R}=2$ does the second term on the right vanish, and do we obtain $$\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^2=4$$
But, even for a mono-atomic gas, $\frac{C_p}{R}>2$.
Our two analyses are consistent with one another only of $\frac{C_p}{R}=2$. That's the only way that S can be constant throughout the change.

In my opinion, this is a pretty hinky problem.

Last edited: Aug 9, 2016
8. Aug 10, 2016

### Staff: Mentor

Here is a simpler analysis of the situation. If the entropy is constant at a value of R, then the change in free energy is given by:
$$\Delta A=C_v(T_2-T_1)-R(T_2-T_1)=(C_v-R)(T_2-T_1)$$
The only way this can be zero is if $C_v=R$.

The change in entropy for an ideal gas is given by:$$\Delta S=C_p\ln(T_2/T_1)-R\ln(P_2/P_1)$$
But, if $\Delta S = 0$, we have:
$$C_p\ln(T_2/T_1)=R\ln(P_2/P_1)$$
But, if $C_v=R$, then $C_p=2R$
So, $$\ln(P_2/P_1)=2\ln(T_2/T_1)$$
But, for all this to hold together, we must have that $C_v=R$, which is physically unrealistic.