# Work done by an expanding gas, how to solve this particular example?

I just need help with one of these types of questions and I can handle the rest. It is from a practice quiz on Coursera, Physical Chemistry.

How much expansion work is done on the system when exactly 1 mol of solid ammonium chloride, $\ce{NH4Cl}$, decomposes completely to yield gaseous ammonia, $\ce{NH3}$, and hydrogen chloride, $\ce{HCl}$, at a temperature of 1280 K. Treat the expansion as irreversible and the gases formed as perfect.

I'm not sure how to compute the two volumes required to use the formula:

$$W = n R T \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V}$$

Do I assume $1\text{ atm} = 101.325 \text{ kPa}$. I did that and was able to use the ideal gas law: $V = \frac{nRT}{P}$ to get the final volume of the 1 mole of gas, but I'm unsure how to choose the initial volume (0 won't work here as it leads to $-\infty$).

1. Is this the correct approach?
2. If 1., then how do I also get the initial volume.
3. If not 1., then how should I approach it?
• I think that the question is missing what the pressure is but the formula you are using is correct. – Nanoputian Oct 8 '15 at 7:25
• The initial volume is assumed to be 0 as the volume of the solid is negligible to the volume of gas produced. – Nanoputian Oct 8 '15 at 7:26
• That was my initial understanding, too. Then I realized that plugging zero initial volume in the formula produces an infinite result. – Ivan Neretin Oct 8 '15 at 9:00

You have system where 1 mole of solid (0 moles of gas) decomposes into two moles of gas at a constant temperature (given) and likely a constant pressure (not given).

Your formula for work is not correct for this problem. You cannot have $n$ outside of the integral since $n$ is not constant.

Let's write the total differential of the ideal gas equation:

\begin{align} \mathrm{d}(PV)&=\mathrm{d}(nRT) \\ P\mathrm{d}V + V\mathrm{d}P &= nR\mathrm{d}T + RT\mathrm{d}n \end{align}

Since $T$ is constant and $P$ is constant, we know $\mathrm{d}T=0$ and $\mathrm{d}P=0$, so this simplifies to:

$$P\mathrm{d}V = RT\mathrm{d}n$$

Can you do something with this relationship that avoids integrating with respect to (unknown) volumes?

• This is kind of a weird question. Keeping the temperature constant and uniform like that in an irreversible expansion would be problematic. But, if you could somehow do it, it would seem that you would also have to assume that the external pressure is kept constant at, say, 1 atm throughout the process to solve the problem. – Chet Miller Oct 8 '15 at 11:32
• On second thought, maybe this isn't as weird a problem statement as I thought in my previous comment. In this particular problem, it doesn't matter whether the temperature is constant or uniform throughout the irreversible expansion. The only thing that matters is the temperature in the final equilibrium state (1280 K) and the fact that the external pressure is held constant. This is sufficient to determine the amount of work done by the gas on its surroundings. – Chet Miller Oct 8 '15 at 14:18

I am currently taking this course and now understand this question after seeing my initial mistakes. I have a background in chemistry at the graduate level but this is a thermodynamics question that requires some basic knowledge of undergrad stoichiometry.

One mole of ammonium chloride at $1280\ \mathrm{K}$ decomposes completely to yield gaseous ammonia ($\ce{NH3}$) and hydrogen chloride ($\ce{HCl}$). This expansion is irreversible and the gases formed are perfect, which I think is to say they are not in some equilibrium or reversible state with the original solid.

The first law of thermodynamics is what is needed here and the equation needed is in the form of $\Delta U= q_\mathrm{in} + W_\mathrm{on}$. Work on $W_\mathrm{on} = -P \times \Delta V$. This is not the reversible isothermal expansion work problem so you do not need the $W_\mathrm{on}=-nRT \ln (V_\mathrm{f}/V_\mathrm{i})$ equation as I saw in some explanations this was eluded to. Since volume from the ideal gas law equation is $nRT/p$ the equation becomes $W_\mathrm{on} = -p (nRT/p)$.

You also need to know that one mole of the ammonium chloride yields one mole of ammonium and one mole of hydrogen chloride for a total of 2 moles of gas so: $$W_\mathrm{on}= -p\times 2\times 8.314 \times 1280/p=21283.84\ \mathrm{J} = 21.3\ \mathrm{kJ}$$

• I edited your post with MathJax for the chemical and physical formulae. I may have introduced a few errors while doing so because I wasn’t always sure what you meant. Please check if I did it correctly and, if necessary, improve. We have a few introductions to MathJax on meta and the basics can be found in the help center. – Jan Jan 11 '17 at 0:55