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Q12.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)

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In the first case, the process is adiabatic i.e. \Delta Q=0

22.3 J work is done on the system i.e. \Delta w=-22.3\ J

\\\Delta Q=\Delta u+\Delta w\\ 0=\Delta u-22.3\\ \Delta u=22.3\ J

Since in the latter process as well the initial and final states are the same as those in the former process \Delta u will remain the same for the latter case.

In the latter case net heat absorbed by the system is 9.35 cal

\\\Delta Q=9.35\times 4.2\\ \Delta Q=39.3\ J

\\\Delta w=\Delta Q-\Delta u\\ \Delta w=39.3-22.3\\ \Delta w=17.0\ J

The network done by the system in the latter case is 17.0 J

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