Question

The sample of monoatomic gas undergoes a process as represented by$T–V$ graph (if ${\mathrm{P}}_0{\mathrm{V}}_0=\frac{1}{3}{\mathrm{RT}}_0$) then:

$$\begin{gathered} \left(\mathrm{P}\right){\mathrm{W}}_{1->2}=\frac{1}{3}{\mathrm{RT}}_0\ln 2 \\ \left(\mathrm{Q}\right){\mathrm{Q}}_{1->2->3}=\frac{1}{6}{\mathrm{RT}}_0(2\ln (2)+3) \\ \left(\mathrm{R}\right){\mathrm{U}}_{1->2}=0 \\ \left(\mathrm{S}\right){\mathrm{W}}_{1->2->3}=\frac{{\mathrm{RT}}_0}{3}\ln 2 \end{gathered}$$

Which of the following options are correct?

• A. $\mathrm{P},\mathrm{Q}$ are incorrect
• B. $\mathrm{R},\mathrm{S}$ are incorrect
• C. $\mathrm{P},\mathrm{Q},\mathrm{S}$ are incorrect
• D. None of these

Answer 1

The correct option is D

None of these

Step 1: Given Data

$$\begin{gathered} \mathrm{f}=\frac{3}{2}\mathrm{R} \\ \mathrm{n}=\frac{1}{3} \end{gathered}$$

Where, $f$is the ratio of specific heats.

$n$ is the number of moles

$$\begin{gathered} \left(\mathrm{P}\right){\mathrm{W}}_{1->2}=\frac{1}{3}{\mathrm{RT}}_0\ln 2 \\ \left(\mathrm{Q}\right){\mathrm{Q}}_{1->2->3}=\frac{1}{6}{\mathrm{RT}}_0(2\ln (2)+3) \\ \left(\mathrm{R}\right){\mathrm{U}}_{1->2}=0 \\ \left(\mathrm{S}\right){\mathrm{W}}_{1->2->3}=\frac{{\mathrm{RT}}_0}{3}\ln 2 \end{gathered}$$

$R$ is the real gas constant

$U$is the internal energy

$W$ is work done by the system

$Q$ is the heat added to the system

$T_0$ is the temperature of the source of heat.

Step 2: Process: P Work done from process 1 to 2

$$\begin{gathered} {\mathrm{W}}_{1-2}={\mathrm{nRT}}_{\mathrm{o}}\ln 2 \\ {\mathrm{W}}_{1-2}=\frac{1}{3}{\mathrm{RT}}_{\mathrm{o}}\ln 2 \end{gathered}$$

$T_0$ is the temperature of the source of heat.

$R$ is the real gas constant

$W_{1->2}$ is the work done from process 1 to 2.

Step 3: Process: Q

Total heat change from process 1 to 2 to 3

$$\begin{gathered} {\mathrm{Q}}_{1\to 2\to 3}={\mathrm{Q}}_{1\to 2}+{\mathrm{Q}}_{2\to 3} \\ {\mathrm{Q}}_{1\to 2\to 3}=d{\mathrm{W}}_{1\to 2}+{\mathrm{dU}}_{2\to 3} \\ {\mathrm{Q}}_{1\to 2\to 3}=\frac{R{T}_o}{3}\ln 2+n\frac{f}{2}R{T}_o \\ {\mathrm{Q}}_{1\to 2\to 3}=\frac{R{T}_o}{3}\ln 2+\frac{1}{3}\times \frac{3}{2}R{T}_o \\ {\mathrm{Q}}_{1\to 2\to 3}=\frac{1}{3}R{T}_o\ln 2+\frac{1}{2}R{T}_o \\ {\mathrm{Q}}_{1\to 2\to 3}=R{T}_o\left(\frac{1}{3}\ln 2+\frac{1}{2}\right) \end{gathered}$$

Step 4: Process: R

The process from1 to 2 is an isothermal process and therefore the change in internal energy is zero.

Step 5: Process: S

${\mathrm{W}}_{1-2-3}=\left(\frac{1}{3}\right){\mathrm{RT}}_{\mathrm{o}}\ln 2$

Therefore, the conditions for the processes P, R, and S provided in the question are correct and process Q is incorrect.

Therefore, the correct option is (D).