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An ideal gas goes from state $A$ to state $B$ via three different processes as indicated in the $P-V$ diagram. If $Q_1, Q_2, Q_3$ indicate the heat absorbed by the gas along the three processes and $\Delta U_1, \Delta U_2, \Delta U_3$
indicate the change in internal energy along the three processes respectively, then
${Q_1} < {Q_2} < {Q_3}$ and $\Delta U_1=\Delta U_2=\Delta U_3$
${Q_1} < {Q_2} = {Q_3}$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$
${Q_1} = {Q_2} > {Q_3}$ and $\Delta U_1 > \Delta U_2 > \Delta U_3$
${Q_1} > {Q_2} > {Q_3}$ and $\Delta U_1 = \Delta U_2 = \Delta U_3$
Solution
Change in internal energy is path independent and depends only on the initial and final states.
As the initial and final states in the three processes are same. Therefore,
$\Delta {u_1} = \Delta {u_2} = \Delta {u_3}$
Workdone, $W=Area\,under\,P-V\,graph$
As area under curve $1>$ area under curve $2>$ area under curve $3$
$\therefore {W_1} > {W_2} > {W_3}$
According to first law of thermodynamics,
$Q = W + \Delta u$
$As{W_1} > {W_2} > {W_3}\,and\,\Delta {u_1} = \Delta {u_2} = \Delta {u_3}$
$\therefore {Q_1} > {Q_2} > {Q_3}$
