In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium stat

Question

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 $1 \mathrm{cal}=4.19 \mathrm{~J}$ )\

Professor PC Thomas Classes · Answer

In the first case, since the state if the gas is changed adiabatically from an equilibrium state A to another equilibrium state B, $\mathrm{d} 	heta=0$. Further the work done the system, $\mathrm{dw}=-22.3 \mathrm{~J}$.\ From, $\mathrm{d} 	heta=\mathrm{du}+\mathrm{dW}$\ $\Rightarrow 0=\mathrm{du}-22.3$ or $\mathrm{du}=22.3 \mathrm{~J}$\ In the second case, the gas is again taken in the state to state B when net heat absorbed is 9.35 cal, $\mathrm{du}=22.3 \mathrm{~J}$ du has the same value because internal energy is a state variable.\ As $\mathrm{dQ}=9.35 \mathrm{cal}=9.35 	imes 4.19 \mathrm{~J}=39.2 \mathrm{~J}$\ From $\mathrm{dQ}=\mathrm{du}+\mathrm{dw} \Rightarrow \mathrm{dw}=\mathrm{dQ}-\mathrm{dW}$\ $\Rightarrow 39.2-22.3=16.9 \mathrm{~J}$\