JEE Advance - Physics (2014 - Paper 2 Offline - No. 15)
Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be
250 R
200 R
100 R
$$-$$100 R
Explanation
Heat lost by monoatomic gas = Heat gained by diatomic gas
$$2 \times {5 \over 2}R \times (700 - T) = 2 \times {5 \over 2}R(T - 400)$$
$$3500 - 5T = 7T - 2800$$
$$12T = 6300$$ or, $$T = 525K$$
Work done by monoatomic gas:
$$ = {n_1}R\Delta T$$ ($$\because$$ PV = nRT)
$$ = 2 \times R \times (700 - 525)$$ ($$\Delta$$T is negative)
$$ = - 350R$$
Work done by diatomic gas $$ = {n_2}R\Delta T$$
$$ = 2R(525 - 400) = 250R$$. (There is gain in temperature)
The net work done = $$-$$350 + 250 = $$-$$100R and $$\Delta$$T is positive.
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