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Two moles of a certain gas at a temperature $T_{o} = 300 K$ were cooled isochorically so that the pressure of the gas reduced $2$ times. Then as a result of isobaric process, the gas is allowed to expand till its temperature go back to the initial value. The total amount of heat absorbed by gas (in $J$ ) in this process is $X \times 500$ .The value of X is:

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Solution

Consider all the states of the system:
$P_{o}, V_{0}, 300 K$
$\frac{P_{o}}{2}, V_{o}, 150 K$
$\frac{P_{o}}{2}, 2 V_{o}, 300 K$
i.e Work done = $W_{i s o c h o r i c} + W_{i s o b a r i c}$
Work done = $0 + P (V_{2} - V 1)$
$w = \frac{P_{o}}{2} (2 V_{o} - V_{o}) = \frac{P_{0} V_{0}}{2}$
also
Gas equation:
$P V = n R T$
$P_{o} V_{o} = 2 \times 8.314 \times 300 = 5000 J$
i.e. $w = \frac{P_{o} V_{o}}{2} = \frac{5000}{2} = 2500 J$

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