Question

One mole of an ideal monoatomic gas is taken from A to C along the path ABC. The temperature of the gas at A is $T_0$. For the process ABC, which one of the following option(s) is/are correct?

• A. Work done by the gas is $R{T}_0$.
• B. Change in internal energy of the gas is $\frac{11}{2}R{T}_0$.
• C. Heat absorbed by the gas is $\frac{11}{2}R{T}_0$.
• D. Heat absorbed by the gas is $\frac{13}{2}R{T}_0$.

Answer 1

Answer: (A), (C)

The correct options are
A Work done by the gas is $R{T}_0$.
C Heat absorbed by the gas is $\frac{11}{2}R{T}_0$.

Given,
Number of moles of ideal monoatomic gas $\left(n\right)=1$
From the given $P-V$ graph, we can infer that
Process $AB$ is Isobaric and Process $BC$ is Isochoric.

From the data given in the $P-V$ graph, we can find
Work done by the ideal mono atomic gas in process ABC
$W=W_{AB}+W_{BC}=P_0(2V_0-V_0)+0$
$=P_0{V}_0=nR{T}_0=R{T}_0$
Temperature at $C$ is given by
$2P_0\times 2V_0=R{T}_C\Rightarrow T_C=4T_0$

Change in internal energy in process ABC
$\Delta U=n\frac{fR}{2}\Delta T$
$=1\times \frac{3}{2}R\times (4T_0-T_0)=\frac{9}{2}R{T}_0$
Using first law of thermodynamics,
$\Delta Q=\Delta U+W$
$\Rightarrow \Delta Q=\frac{9}{2}R{T}_0+R{T}_0=\frac{11}{2}R{T}_0$
Hence, options (a) and (c) are the correct answers.