Question

In the figure, $n$ moles of a monoatomic ideal gas undergo the process $ABC$ as shown in the $P-V$ diagram. The process $AB$ is isothermal and $BC$ is isochoric. The temperature of the gas at $A$ is $T_0.$ Total heat given to the gas during the process $ABC$ is measured to be $Q$.


Heat absorbed by the gas in the process $BC$ is

• A. $3nR{T}_0$
• B. $nR{T}_0$
• C. $2nR{T}_0$
• D. $6nR{T}_0$

Answer 1

The correct option is A $3nR{T}_0$

From the $P-V$ diagram given in the question $AB$ is an isothermal process.
Then, using $P_A{V}_A=P_B{V}_B$, we get
$P\times 2V=P_B\times 6V\Rightarrow P_B=\frac{P}{3}$


Now, $BC$ is an isochoric process then
$\frac{P_B}{T_B}=\frac{P_C}{T_C}$
$\frac{P}{3T_0}=\frac{P}{T_C}\Rightarrow T_C=3T_0$
[$T_B=T_A=T_0$]
Heat absorbed during $BC$ is given by
$Q=\Delta U+\Delta W$
$[\because \Delta W=0]$
$Q=\Delta U$
We know that, $\Delta U=\frac{nf}{2}R\Delta T$
$\Rightarrow Q=\frac{n}{2}\times 3\times R\times (T_C-T_B)$
$Q=n\times \frac{3R}{2}\times 2T_0$
$\Rightarrow Q=3nR{T}_0$
Hence, option (a) is the correct answer.