(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Determine the change in the entropy, Helmholtz free energy, and Gibbs free energy, when a mole of ideal gas is compressed from 1atm to 100atm at 20C.

3. The attempt at a solution

I am not entirely convinced by my attempt below -- can anyone spot something wrong? (I thought it odd that the entropy change should be temperature-independent...)

From the first law

[tex]dU = dQ - dW = dQ - PdV[/tex]

From the second law,

[tex]dS = dQ / T \Rightarrow dQ = TdS[/tex]

Thus:

[tex]dU = TdS - PdV[/tex]

Transposing to get dS:

[tex]dS = \frac{dU}{T} + \frac{P}{T}dV[/tex]

But the compression takes place at 20C => isothermal => dT = 0 => dU = 0. Thus:

[tex]dS = \frac{P}{T}dV[/tex]

From the ideal gas law,

[tex]PV = RT \Rightarrow V = \frac{RT}{P} \Rightarrow dV = -\frac{RT}{P^{2}}dP[/tex]

Thus:

[tex]dS = -R\frac{dP}{P} \Rightarrow \Delta S = Rln(P_{1}/P_{2})[/tex]

For the Helmholtz free energy, I reason:

[tex]dF = -SdT - PdV[/tex]

isothermal => dT = 0. Thus:

[tex]dF = -PdV[/tex]

Again, using the ideal gas law, rewrite as:

[tex]dF = RT\frac{dP}{P} \Rightarrow \Delta S = RTln(P_{2}/P_{1})[/tex]

And finally, for the Gibbs Free energy, I reason:

[tex]dG = -SdT + VdP = VdP = \frac{RT}{P}dP[/tex]

So, in this case, [tex]\Delta G = \Delta F[/tex]

Is this the right approach?

Cheers!

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# Homework Help: Change in entropy, Gibbs and Helmholtz in an isothermal compression

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