Animated Solution for Physics - Thermodynamics: In the given P-V diagram, a monoatomic gas $\left(\gamma =\frac{5}{3}\right)$ is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given: ${\left(\frac{1}{3}\right)}^{0.6}\simeq 0.5,\ln 2\simeq 0.7$]. Which of the following statement(s) is(are) correct?

Imagine you are tracking the journey of a monoatomic gas as it undergoes a series of thermodynamic transformations. The P-V diagram is our map. We start at state A, where the gas is at a pressure of $100\text{ kPa}$ and occupies a volume of $0.80{\text{ m}}^3$. The gas is then compressed adiabatically to state B, where the pressure spikes to $300\text{ kPa}$.

Before we can calculate any work, we need to find the missing piece of the puzzle: the volume at state B ($V_B$).

For an adiabatic process, the relationship between pressure and volume is governed by the equation $P_A{V}_A^{\gamma }=P_B{V}_B^{\gamma }$. We are given that the gas is monoatomic, so $\gamma =\frac{5}{3}$.

Let's plug in our known values:

Rearranging to solve for $V_B$, we get:

The problem kindly provides the approximation ${\left(\frac{1}{3}\right)}^{0.6}\simeq 0.5$. Since $3/5$ is exactly $0.6$, we can easily compute:

Now, let's calculate the work done during this adiabatic compression. The formula is:

Substituting our values:

The negative sign makes perfect physical sense—the gas was compressed, meaning work was done on the gas. The magnitude of this work is $60\text{ kJ}$, which makes option (C) correct.

Next, the gas expands isothermally from state B to state C. For an isothermal process, the work done is given by:

Since we don't have the temperature, we can use the ideal gas law ($PV=nRT$) to substitute $P_B{V}_B$ for $nRT$:

Looking at the P-V diagram, state C lies on the same vertical dashed line as state A. This means they share the same volume: $V_C=V_A=0.8{\text{ m}}^3$. Let's calculate the work:

Using the given approximation $\ln 2\simeq 0.7$:

The magnitude of the work done in process $B\to C$ is indeed $84\text{ kJ}$, making option (B) correct.

Finally, let's consider the path from C back to A. As we noted earlier, both states lie on the same vertical line, meaning the volume remains constant at $0.8{\text{ m}}^3$.

This is an isochoric process. Because there is no change in volume ($\Delta V=0$), the gas does no work:

This confirms that option (D) is correct.

To wrap things up, let's check the total work done over the entire path $A\to B\to C$:

The magnitude of the total work is $24\text{ kJ}$, not $144\text{ kJ}$. Therefore, option (A) is incorrect.