# Change in entropy, Gibbs and Helmholtz in an isothermal compression

1. Nov 3, 2007

### T-7

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

$$dU = dQ - dW = dQ - PdV$$

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!

Last edited: Nov 3, 2007
2. Nov 4, 2007

### 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.

Last edited: Nov 4, 2007
3. Nov 5, 2007

### T-7

Thanks very much for the confirmation