Question

One mole of an ideal gas is taken from state $A$ to state $B$ by three different processes, $\left(a\right)ACB\left(b\right)ADB\left(c\right)AEB$ as shown in the $P-V$ diagram. The heat absorbed by the gas is:

• A. greater in process $\left(b\right)$ then in $\left(a\right)$
• B. the least in process $\left(b\right)$
• C. the same in $\left(a\right)$ and $\left(c\right)$
• D. less in $\left(c\right)$ then in $\left(b\right)$

Answer 1

The correct option is A greater in process $\left(b\right)$ then in $\left(a\right)$

Tn the process $ADB$,

Volume change $=0=$ Work done

$\therefore$ Heat absorbed $=\Delta U$ [Change in internal energy]

But, in case of $ACB$

$dQ=dU-dW$

$\therefore {dQ}_{ADB}>{dQ}_{ACB}$