Question

One mole of an ideal monoatomic gas is taken through a cycle $a\to b\to c\to a$ as shown in the figure. Find the difference between maximum and minimum values of internal energy of the gas during the cycle.

• A. $9P_0{V}_0$
• B. $6P_0{V}_0$
• C. $\frac{9}{2}{P}_0{V}_0$
• D. $3P_0{V}_0$

Answer 1

The correct option is C $\frac{9}{2}{P}_0{V}_0$

Let temperature in state ${}^{\prime }{a}^{\prime }$ be $T_0$.
Temperature in state ${}^{\prime }{b}^{\prime }$ will be $2T_0$.
$[\because P_b{V}_b=2P_a{V}_a]$
and temperature at state $C$ is $4T_0$
$[\because P_c{V}_c=4P_a{V}_a]$

$\because$ temperature $T$ is least at ${}^{\prime }{a}^{\prime }$ and maximum at ${}^{\prime }{c}^{\prime }$,
$U_{\text{min}}=U_a$ and $U_{\text{max}}=U_c$
Thus,
$U_{\text{max}}-U_{\text{min}}=U_c-U_a=n{C}_V(T_c-T_a)$
$=1\times \frac{3R}{2}\times (4T_0-T_0)=\frac{9}{2}R{T}_0$
$=\frac{9}{2}{P}_0{V}_0$ $[\because P_0{V}_0=R{T}_0]$
Thus, option (c) is the correct answer.