This is a question from Atkin's Physical Chemistry Book.


Suppose that 3mmol $\mathrm{N_2\,(g)}$ occupies $\mathrm{36~cm^3}$ at $\mathrm{300~K}$ and expands to $\mathrm{60~cm^3}$. Calculate $\Delta G$ for the process.

My Attempt

I decided to split this process up into two separate process (I am not sure if this allowed. Please correct me if this wrong). The first process is an reversible isothermal expansion of the gas to $\mathrm{60~cm^3}$. Then the second step is an adiabatic cooling of the gas. I was planning to calculate $\Delta G$ for each step then add them up to get the answer. For the first step, $\Delta U = 0$, therefore: $$q = -w = n~R~T~\ln\frac{60}{36}$$ $$\Delta S = \frac{q}{T} = n~R~\ln\frac{60}{36}$$

Since pressure is constant, $$\Delta H = q = n~R~T~\ln\frac{60}{36}$$ Therefore: $$\Delta G = \Delta H - T\Delta S = 0$$ I stopped her because I am sure that this is incorrect as Gibbs Free Energy shouldn't be equal to $0$.

Can someone please clarify if the above procedure is correct and perhaps a better way to attempt this question?


2 Answers 2


I'm quite certain about the following, because it's from the related (german) woorkbook to the 4th (german) edition. The beneath numbering is from the linked online version, which should be the 9th english edition.

Your way to the solution is part of subsection 3.9 Properties of the Gibbs energy and subsubsection 3.9 c) The variation of the Gibbs energy with pressure.

I seriously entrust you to read this subsubsection and try to solve it with the given information. If this does not help, the following "hidden" section reveals what you need to solve it.

You need to insert the equation for the molar volume of an ideal gas $V_m = R~T/p$ into (3-57b) $$G_m(p_f) = G_m(p_i)+\int_{p_i}^{p_f}V_m~\mathrm{d}p \tag{3-57.b}$$ which yields one of the two equations that you need: $$G_m(p_f)=G_m(p_i)+R~T\int_{p_i}^{p_f}\frac{\mathrm{d}p}{p}=G_m(p_i)+R~T\ln\frac{p_f}{p_i}\tag{3-59}$$ The other one is the Boyle's law.

Good luck.

  • $\begingroup$ Thanks for that. I missed that section when I was reading the chapter before. That makes the question much easier. However would it still be possible to the question from my method? Also was my result for $\Delta G$ being equal to $0$ correct? $\endgroup$
    – Nanoputian
    Commented Aug 10, 2015 at 10:33
  • $\begingroup$ For your first question ... maybe someone knows that but I am not qualified enough. To your second question ... calculate $\Delta G$ with this "new" way and you will see. $\endgroup$ Commented Aug 10, 2015 at 10:39

For an ideal gas, $\Delta H = 0$ for a constant temperature change between two thermodynamic equilibrium states of a system. So, for your system, $\Delta G=T\Delta S$.


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