Final Volume of Ideal Gas After Equilibrium

[KFC]

Homework Statement

Two ideal gas (n mole A and m mole B) was separated by a piston (impermeable and diathermal) the whole setup is confine in a adiabatic walls so no heat exchange with outside. Let the piston move, at equilibrium, find the final volume. Assume the final temperature of both gas is T and the total volume is V.

2. The attempt at a solution
First of all, both gas satisfy ideal gas state equation

$$\frac{P_AV_A}{T_A} = nR, \qquad \qquad \frac{P_BV_B}{T_B} = mR$$

Since we know the final temperature, and at equilibrium, the pressure is the same on both compartment (otherwise, the piston will move), so assuming the final volume of A is $$V_A$$, then the final volume of B will be $$V-V_B$$, we conclude that

$$\frac{P}{T} = nRV_A = \frac{P}{T} = mR(V-V_A)$$

We can solve for $$V_A$$ and $$V_B$$, right?

 

[CompuChip]

Your setup looks correct, except for a mathematical error...
If $\frac{P V}{T} = n R$, then $\frac{P}{T} = \cdots$ ?

By the way, once you have found the answer, it'll be nice trying to explain its physical meaning and saying something about whether you could have foreseen the outcome.

 

[KFC]

Your setup looks correct, except for a mathematical error...
If $\frac{P V}{T} = n R$, then $\frac{P}{T} = \cdots$ ?

By the way, once you have found the answer, it'll be nice trying to explain its physical meaning and saying something about whether you could have foreseen the outcome.

Oh, careless. Thanks