If one mole of an ideal gas at (p1, V1) is allowed to expand reversibly and isothermally (A to B), i

Q: Why is the work done in process BC zero?

Work in thermodynamics is defined as the area under the p-V curve, mathematically expressed as $\int p \, dV$. In process BC, the path is a vertical line, meaning the volume is constant ($dV = 0$). Without a change in volume, the gas cannot do any mechanical work.

Q: How did we find the volume at state B?

We used Boyle's law. Since process AB is isothermal, the temperature is constant, which means $p_A V_A = p_B V_B$. We know $p_A = p_1$, $V_A = V_1$, and $p_B = p_1/2$. Solving for $V_B$ gives us $2V_1$.

Q: Why is the adiabatic work formula written as $\frac{p_f V_f - p_i V_i}{1 - \gamma}$?

This formula is derived by integrating $p \, dV$ using the adiabatic condition $pV^\gamma = \text{constant}$. Depending on how you arrange the terms, it can also be written as $\frac{p_i V_i - p_f V_f}{\gamma - 1}$. Both are mathematically identical.

Q: Can we find the net work by calculating the area inside the loop?

Yes, absolutely! The net work done in any cyclic process is exactly equal to the area enclosed by the cycle on a p-V diagram. However, because the curves are logarithmic and polynomial, integrating each path individually is the most straightforward way to find that area.

Q: Why did the sign change in the last step for the adiabatic work?

It was purely an algebraic manipulation to match the given options. We had $\frac{1}{1 - \gamma}$. By factoring out a $-1$ from the denominator, it becomes $-\frac{1}{\gamma - 1}$. This makes the final expression look neater.

Question

If one mole of an ideal gas at $(p_1, V_1)$ is allowed to expand reversibly and isothermally ($A$ to $B$), its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value ($B \rightarrow C$). Then, it is restored to its initial state by a reversible adiabatic compression ($C$ to $A$). The net work done by the gas is equal to

Options:

(A) $RT \left( \ln 2 - \frac{1}{2(\gamma - 1)} \right)$

(B) $-\frac{RT}{2(\gamma - 1)}$

(C) $0$

(D) $RT \ln 2$

Sigma Percentile · Accepted Answer

Thermodynamics is like a beautifully choreographed dance of energy, pressure, and volume. In this problem, we are looking at a cyclic process—a journey where a gas leaves its home, undergoes a series of transformations, and eventually returns exactly to where it started. Our goal? To find out how much net work the gas did during this entire trip.

### Analyzing the Setup

Let's break down the cycle into its three distinct acts. We start at state $A$ with an initial pressure $p_1$ and volume $V_1$.

**Act 1: The Isothermal Expansion ($A 	o B$)**
The gas expands isothermally. This means the temperature remains constant. As it expands, its pressure drops to half its original value, becoming $p_1/2$. Because temperature is constant, Boyle's Law ($p_A V_A = p_B V_B$) dictates that if the pressure is halved, the volume must double. So, the volume at state $B$ is $2V_1$.

**Act 2: The Isochoric Cooling ($B 	o C$)**
Next, the gas cools down while keeping its volume strictly constant at $2V_1$. This is an isochoric process. The pressure drops further to $p_1/4$. Because the volume doesn't change, the gas is trapped in a rigid box during this phase.

**Act 3: The Adiabatic Compression ($C 	o A$)**
Finally, the gas is compressed adiabatically back to its original state $A$. Adiabatic means no heat enters or leaves the system. The gas is squeezed, causing its pressure and temperature to spike until it perfectly reaches $p_1$ and $V_1$ again.

### The Master Equation

The net work done in a cyclic process is simply the algebraic sum of the work done in each individual step:

$$W_{net} = W_{AB} + W_{BC} + W_{CA}$$

Let's calculate them one by one.

### Calculating Work for Each Process

**1. Isothermal Work ($W_{AB}$)**
For an isothermal process, the work done is given by the formula:

$$W_{AB} = p_1 V_1 \ln\left(\frac{V_B}{V_A}ight)$$

Substituting our volumes, we get:

$$W_{AB} = p_1 V_1 \ln\left(\frac{2V_1}{V_1}ight) = p_1 V_1 \ln 2$$

**2. Isochoric Work ($W_{BC}$)**
This one is a freebie! Work is defined as the integral of pressure with respect to volume ($W = \int p \, dV$). Since the volume is constant, $dV = 0$. Therefore, no physical work is done.

$$W_{BC} = 0$$

**3. Adiabatic Work ($W_{CA}$)**
The formula for work done during an adiabatic process is:

$$W_{CA} = \frac{p_f V_f - p_i V_i}{1 - \gamma}$$

Here, our initial state is $C$ and our final state is $A$. Let's plug in the coordinates:

$$W_{CA} = \frac{p_A V_A - p_C V_C}{1 - \gamma} = \frac{p_1 V_1 - \left(\frac{p_1}{4}ight)(2V_1)}{1 - \gamma}$$

Simplifying the numerator:

$$W_{CA} = \frac{p_1 V_1 - \frac{p_1 V_1}{2}}{1 - \gamma} = \frac{p_1 V_1}{2(1 - \gamma)}$$

### Final Calculation

Now, we add them all together to find the net work:

$$W_{net} = p_1 V_1 \ln 2 + 0 + \frac{p_1 V_1}{2(1 - \gamma)}$$

To make this match our options, we use the Ideal Gas Law. For $n = 1$ mole, we know that $p_1 V_1 = RT$. Substituting this into our equation gives:

$$W_{net} = RT \ln 2 + \frac{RT}{2(1 - \gamma)}$$

Finally, we can factor out a negative sign from the denominator of the second term to make it look cleaner:

$$W_{net} = RT \left( \ln 2 - \frac{1}{2(\gamma - 1)} ight)$$

And there we have it! The elegant mathematical conclusion to our thermodynamic journey.

Select Answer:

Visualized Solution

Thermodynamic Cycle

Net Work Done

Work Done: $A \to B$

Work Done: $B \to C$

Work Done: $C \to A$

Total Net Work

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Calculating Work for Each Process

Final Calculation

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