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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:

$\Delta$$Q = 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.

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  • $\begingroup$ "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? $\endgroup$
    – Time4Tea
    Jan 18, 2015 at 14:43
  • $\begingroup$ 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. $\endgroup$
    – ritvik1512
    Jan 18, 2015 at 14:46

1 Answer 1

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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.

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  • $\begingroup$ This seems okay, but how can I comment upon the volumes as that is not stated? $\endgroup$
    – ritvik1512
    Jan 18, 2015 at 16:11

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