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A system consisting of 1 mol of an ideal gas undergoes a reversible process, $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ (schematically indicated in the figure below). If the temperature at the starting point $\mathrm{A}$ is $300 \mathrm{~K}$ and the work done in the process $\mathrm{B} \rightarrow \mathrm{C}$ is $1 \mathrm{~L} \mathrm{~atm}$, the heat exchanged in the entire process in $\mathrm{L}$ atm is
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The correct answer is:
$1.5$
For process $\mathrm{A} \rightarrow \mathrm{C}, \mathrm{W}=0, \Delta \mathrm{E}=0, \mathrm{q}=0$ For process $\mathrm{A} \rightarrow \mathrm{B}$ $\mathrm{W}=-1(1.5-1)=-0.5$ lit. atm $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ $0=\mathrm{q}+\mathrm{W}$ $\mathrm{q}=-\mathrm{W}=-(-0.5)=0.5$ lit. atm
For process $\mathrm{B} \rightarrow \mathrm{C}, \mathrm{W}=1 \mathrm{~atm}$
$\begin{array}{l}
\Delta \mathrm{E}=\mathrm{q}+\mathrm{W} \\
0=\mathrm{q}-\mathrm{W} \\
-\mathrm{q}=\mathrm{W}=-1 \mathrm{~atm}
\end{array}$
$\therefore$ Total heat exchanged $=1.5$
For process $\mathrm{B} \rightarrow \mathrm{C}, \mathrm{W}=1 \mathrm{~atm}$
$\begin{array}{l}
\Delta \mathrm{E}=\mathrm{q}+\mathrm{W} \\
0=\mathrm{q}-\mathrm{W} \\
-\mathrm{q}=\mathrm{W}=-1 \mathrm{~atm}
\end{array}$
$\therefore$ Total heat exchanged $=1.5$
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