# Thermodynamics and internal energy

I've been working on this problem for quite some time trying to figure out the most efficient way of answering it, So here goes the problem:

There is a container, containing a monoatomic gas with a movable piston on the top. The container and piston are all made of perfectly insulating material. Now the container is divided into two compartments by a movable partition that allow slow transfer of heat among the compartments. The adiabatic exponent of gas is $$\gamma$$.

We are asked to calculate the total work done by the gas till it achieves equilibrium. By equilibrium here I mean that initially after the partition was introduced the gas on each side has different pressure and temperature which with passage of time reached a common temperature and pressure.

Now my approach to the problem is:

We know that in a closed adiabatic container:

$$\DeltaQ = 0$$ as there is no net heat change between the compartments.

So, as per first law of thermodynamics we have, $$\Delta Q = \Delta U + W$$

Now considering the whole container to be the system we can apply the law on the system so,

$$\Delta Q = \Delta U + W$$ reduces to $$0$$ = $$\Delta U + W$$ $$\Rightarrow$$ $$W = - \Delta U$$

Now we can calculate $$\Delta U$$ for both compartments at equilibrium and add them up and the work will be given by $$- \Delta U$$. And we're done.

But something about my above method doesn't seem right. Are all my above considerations correct? If not how can I improve? I've tried reading up Joule-Thomson effect and other related texts on thermodynamics and expansion. I'll be highly obliged by your help. I've tried to fit this question in with the standards of StackExchange.

• "We are asked to calculate the total work done by the gas till it achieves equilibrium." In what way is it not in equilibrium to start with? Jan 18, 2015 at 14:43
• In the beginning the gas on each side of partition has different temperature and pressure and even volume and with passage of time it reaches the equilibrium. Jan 18, 2015 at 14:46

This process is shortly adiabatic so you are right to imply $\Delta Q=0$ Simply we know how to calculate work as : $$W = \int_{v_i}^{v_f} P \, dv$$ Adiabatic condition also satisfy $$PV^{\gamma}=K =constant$$ $$\gamma=\frac{5}{3}$$ for monatomic gas. so work (W) becomes simply $$W = K \int_{v_i}^{v_f} \frac{dV}{V^{\gamma}}$$ Integrating yield $$W=\frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{1-\gamma}$$ and you are done.