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An ideal gas is taken from state $1$ to state $2$ through optional path $A, B, C \& D$ as shown in $P-V$ diagram. Let $Q, W$ and $U$ represent the heat supplied, work done $\&$ internal energy of the gas respectively. Then
$Q_B - W_B > Q_C - W_C$
$Q_A - Q_D = W_A - W_D$
$Q_A > Q_B > Q_C > Q_D$
Both $(B)$ and $(C)$
Solution
Work done is equal to the area under the $P-V$ curve.
From the graph, area under path $A$ is the largest whereas under path $c$ is the smallest.
Thus $W_{A}>W_{B}>W_{C}>W_{D}$ $…(1)$
From Ist law of thermodynamics, $Q=\Delta U+W$ $…(2)$
As $\Delta U$ is a state function, thus $\Delta U$ is same for all paths.
From $(1)$ $\&(2), \quad Q_{A}>Q_{B}>Q_{C}>Q_{D}$
Also for path $\mathrm{A}$ and $\mathrm{D}, Q_{A}=\Delta U+W_{A}$ and $Q_{D}=\Delta U+W_{D}$
subtracting these two, $\Longrightarrow Q_{A}-Q_{D}=W_{A}-W_{D}$
Similarly, $Q_{B}-W_{B}=Q_{C}-W_{C}$
