Change in entropy, Gibbs and Helmholtz in an isothermal compression

[T-7]

Homework Statement

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.

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

dU = dQ - dW = dQ - PdV

From the second law,

dS = dQ / T \Rightarrow dQ = TdS

Thus:

dU = TdS - PdV

Transposing to get dS:

dS = \frac{dU}{T} + \frac{P}{T}dV

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

dS = \frac{P}{T}dV

From the ideal gas law,

PV = RT \Rightarrow V = \frac{RT}{P} \Rightarrow dV = -\frac{RT}{P^{2}}dP

Thus:

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

For the Helmholtz free energy, I reason:

dF = -SdT - PdV

isothermal => dT = 0. Thus:

dF = -PdV

Again, using the ideal gas law, rewrite as:

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

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

dG = -SdT + VdP = VdP = \frac{RT}{P}dP

So, in this case, \Delta G = \Delta F

Is this the right approach?

Cheers!

 

[dynamicsolo]

I took a slightly different approach which otherwise parallels yours and got the same result. If we start from

PV = RT , with T constant, and differentiate, we have

P dV + V dP = 0 \Rightarrow P dV = - V dP.

For an isothermal process,

dU = dQ - PdV = 0 \Rightarrow dQ = P dV = -V dP ,

so we can write

dS = dQ / T = -(V dP/T) = -(RT/PT) dP = - R (dP/P) ,

using the ideal gas law. This yields the result you found,

\Delta S = R ln(P_{1}/P_{2}) .

If we start from the definitions for the free energies,

F = U - TS and G = U + PV - TS and differentiate, we have

dF = dU - T dS - S dT and

dG = dU + P dV + V dP - T dS - S dT.

If we strip away all the terms equal to zero for an isothermal process (dU = 0 , dT = 0), these reduce to

dF = - T dS and dG = P dV + V dP - T dS ;

but, as we showed above, P dV + V dP = 0 , so

dF = - T dS = dG .

Thus, applying our result for \Delta S,

\Delta F = \Delta G = RT ln(P_{2}/P_{1}).

[Almost missed that: the signs flip again going from dS to dF or dG, so the order of the integration limits doesn't reverse this time...]

I suppose that, given the definitions of what F and G are (see Wiki, for instance), it perhaps isn't surprising that \Delta F and \Delta G are the same for this process.

 

[T-7]

Thanks very much for the confirmation