In the given P-V diagram, a monoatomic gas ($\gamma =\frac{5}{3}$) is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given: $3^{-0.6}\approx 0.5$, $ln2\approx 0.7$].

Which of the following statements (s) is (are) correct ?

The correct statements are:

• The magnitude of the work done in the process B → C is 84 kJ.
• The magnitude of the work done in the process A → B is 60 kJ.
• The magnitude of the work done in the process C → A is zero.

Step-by-step Explanation:

1. Extracting data from the P-V diagram:

From the given P-V diagram:

At state A:
Pressure, $P_A=100\text{ kPa}=100\times {10}^3\text{ Pa}$
Volume, $V_A=0.80{\text{m}}^3$
At state B:
Pressure, $P_B=300\text{ kPa}=300\times {10}^3\text{ Pa}$
Monoatomic gas adiabatic exponent: $\gamma =\frac{5}{3}$

2. Analysis of process A → B (Adiabatic Compression):

Since process A → B is adiabatic:
$P_A{V}_A^{\gamma }=P_B{V}_B^{\gamma }$

Taking ratio:
${\left(\frac{V_B}{V_A}\right)}^{\gamma }=\frac{P_A}{P_B}=\frac{100}{300}=3^{-1}$

$\frac{V_B}{V_A}=3^{-1/\gamma }=3^{-3/5}=3^{-0.6}$

Using the given approximation $3^{-0.6}\approx 0.5$:
$V_B=0.5\times V_A=0.5\times 0.80=0.40{\text{m}}^3$

The work done during an adiabatic process is given by:
$W_{A\to B}=\frac{P_A{V}_A-P_B{V}_B}{\gamma -1}$

Substituting the values:
$W_{A\to B}=\frac{(100\times {10}^3\times 0.80)-(300\times {10}^3\times 0.40)}{\frac{5}{3}-1}$

$W_{A\to B}=\frac{80\times {10}^3-120\times {10}^3}{\frac{2}{3}}=\frac{-40\times {10}^3}{\frac{2}{3}}=-60\times {10}^3\text{ J}=-60\text{ kJ}$

Therefore, the magnitude of the work done in process A → B is 60 kJ.

3. Analysis of process B → C (Isothermal Expansion):

Process B → C is an isothermal expansion where the volume expands from $V_B=0.40{\text{m}}^3$ to $V_C=V_A=0.80{\text{m}}^3$.

The work done during an isothermal expansion is given by:
$W_{B\to C}=P_B{V}_B ln\left(\frac{V_C}{V_B}\right)$

Substituting the values:
$W_{B\to C}=(300\times {10}^3)\times 0.40\times ln\left(\frac{0.80}{0.40}\right)$
$W_{B\to C}=120\times {10}^3\times ln\left(2\right)$

Using the given approximation $ln2\approx 0.7$:
$W_{B\to C}=120\times {10}^3\times 0.7=84\times {10}^3\text{ J}=84\text{ kJ}$

Therefore, the magnitude of the work done in process B → C is 84 kJ.

4. Analysis of process C → A (Isochoric Process):

Process C → A takes place at a constant volume $V=0.80{\text{m}}^3$ ($dV=0$).
Work done in an isochoric process is zero:
$W_{C\to A}=\int PdV=0$

Therefore, the magnitude of the work done in process C → A is zero.

Total Work Done:

$W_{\text{total}}=W_{A\to B}+W_{B\to C}+W_{C\to A}=-60\text{ kJ}+84\text{ kJ}+0=24\text{ kJ}$
Hence, the total work done is 24 kJ, making the first option incorrect.