Question

Following figure shows two processes $A$ and $B$ for a gas. If $\Delta Q_A\text{and}\Delta Q_B$ are the amount of heat absorbed by the system in two cases and $\Delta U_A\text{and}\Delta U_B$ are changes in internal energies respectively, then

• A. $\Delta Q_A<\Delta Q_B,\Delta U_A<\Delta U_B$
• B. $\Delta Q_A=\Delta Q_B,\Delta U_A=\Delta U_B$
• C. $\Delta Q_A>\Delta Q_B,\Delta U_A>\Delta U_B$
• D. $\Delta Q_A>\Delta Q_B,\Delta U_A=\Delta U_B$

Answer 1

The correct option is D $\Delta Q_A>\Delta Q_B,\Delta U_A=\Delta U_B$

Internal energy depends only on initial and final state
So, $\Delta U_A=\Delta U_B$
Also $\Delta Q=\Delta U+\Delta W$
We know that area under $P-V$ graph gives the work done.
So from figure we can say,
$W_A>W_B$
$\Rightarrow \Delta Q_A>\Delta Q_B$
Hence option $\left(C\right)$ is correct.