One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the P-V diag

Q: Why is the work done in process 'c' negative?

In process 'c', the volume of the gas is decreasing (from $4V_0$ to $V_0$). Work done by the gas is $P\Delta V$. Since $\Delta V$ is negative, the work done is negative, meaning work is done *on* the gas.

Q: How did we get $P_i V_i$ instead of $nRT$ in the isothermal work formula?

From the ideal gas law, $PV = nRT$. Since temperature $T$ is constant in an isothermal process, the product $PV$ is also constant. Therefore, we can substitute $nRT$ with the initial pressure and volume $P_i V_i$ (or any $PV$ point on that isotherm).

Q: Why is the work done in an isochoric process zero?

Work done by a gas is defined as the integral of pressure with respect to volume ($\int P dV$). In an isochoric process, the volume is constant, so $dV = 0$. Consequently, the integral evaluates to zero.

Q: How do we know the pressure at the end of the isothermal process 'b'?

For an isothermal process, $P_1 V_1 = P_2 V_2$. At the start of 'b', $P_1 = 4P_0$ and $V_1 = 2V_0$. At the end, $V_2 = 4V_0$. Solving $(4P_0)(2V_0) = P_2(4V_0)$ gives $P_2 = 2P_0$, which perfectly matches the graph.

Q: Can we just find the area enclosed by the cycle directly?

Yes, the net work done in a cyclic process is the area enclosed by the loop on a P-V diagram. However, because the cycles involve curved isothermal paths, it's mathematically easier and less error-prone to calculate the work for each segment and sum them up algebraically.

Question

One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the $P-V$ diagrams below. In cycle I, processes a, b, c and d are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes a$'$, b$'$, c$'$ and d$'$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_{	ext{I}}$ and that during cycle II is $W_{	ext{II}}$. The ratio $W_{	ext{I}}/W_{	ext{II}}$ is _______.

Sigma Percentile · Accepted Answer

This is a classic and beautifully constructed thermodynamics problem from JEE Advanced. It tests your fundamental understanding of calculating work done from $P-V$ diagrams for various thermodynamic processes. Let's embark on this journey step-by-step.

### Analyzing the Setup
We are given two distinct cyclic processes, Cycle I and Cycle II, operating on one mole of an ideal gas. The total work done in a cycle is the algebraic sum of the work done in each of its constituent processes.

To find the work done, we rely on the geometric interpretation: the work done by a gas is the area under the curve on a $P-V$ diagram.

For an **isobaric** (constant pressure) process, the work is simply the area of a rectangle: 
$$W_{	ext{isobaric}} = P \Delta V$$

For an **isothermal** (constant temperature) process, the curve is a hyperbola ($PV = 	ext{constant}$), and the work is found by integration:
$$W_{	ext{isothermal}} = nRT \ln\left(\frac{V_f}{V_i}ight) = P_i V_i \ln\left(\frac{V_f}{V_i}ight)$$

For an **isochoric** (constant volume) process, there is no area under the curve, so:
$$W_{	ext{isochoric}} = 0$$

### The Master Equation for Cycle I
Let's break down Cycle I into its four processes:

1. **Process a (Isobaric Expansion):** The gas expands from $V_0$ to $2V_0$ at a constant pressure of $4P_0$.
$$W_a = 4P_0(2V_0 - V_0) = 4P_0 V_0$$

2. **Process b (Isothermal Expansion):** The gas expands from $2V_0$ to $4V_0$. The initial state is $(2V_0, 4P_0)$.
$$W_b = (4P_0)(2V_0) \ln\left(\frac{4V_0}{2V_0}ight) = 8P_0 V_0 \ln 2$$

3. **Process c (Isobaric Compression):** The gas is compressed from $4V_0$ back to $V_0$ at a constant pressure of $2P_0$. Notice the negative sign because volume decreases!
$$W_c = 2P_0(V_0 - 4V_0) = -6P_0 V_0$$

4. **Process d (Isochoric Heating):** Volume is constant at $V_0$.
$$W_d = 0$$

Summing these up gives the total work for Cycle I:
$$W_{	ext{I}} = 4P_0 V_0 + 8P_0 V_0 \ln 2 - 6P_0 V_0$$
$$W_{	ext{I}} = 8P_0 V_0 \ln 2 - 2P_0 V_0 = 2P_0 V_0 (4 \ln 2 - 1)$$

### The Master Equation for Cycle II
Now, let's perform the same analysis for Cycle II:

1. **Process a$'$ (Isothermal Expansion):** The gas expands from $V_0$ to $2V_0$. The initial state is $(V_0, 4P_0)$.
$$W_{a'} = (4P_0)(V_0) \ln\left(\frac{2V_0}{V_0}ight) = 4P_0 V_0 \ln 2$$

2. **Process b$'$ (Isochoric Cooling):** Volume is constant at $2V_0$.
$$W_{b'} = 0$$

3. **Process c$'$ (Isobaric Compression):** The gas is compressed from $2V_0$ to $V_0$ at a constant pressure of $P_0$.
$$W_{c'} = P_0(V_0 - 2V_0) = -P_0 V_0$$

4. **Process d$'$ (Isochoric Heating):** Volume is constant at $V_0$.
$$W_{d'} = 0$$

Summing these up gives the total work for Cycle II:
$$W_{	ext{II}} = 4P_0 V_0 \ln 2 - P_0 V_0 = P_0 V_0 (4 \ln 2 - 1)$$

### Final Calculation
The problem asks for the ratio of the total work done in Cycle I to that in Cycle II. Let's divide our two master equations:

$$\frac{W_{	ext{I}}}{W_{	ext{II}}} = \frac{2P_0 V_0 (4 \ln 2 - 1)}{P_0 V_0 (4 \ln 2 - 1)}$$

Notice the beautiful symmetry! The entire term $(4 \ln 2 - 1)$ and the constants $P_0 V_0$ cancel out perfectly, leaving us with a clean integer.

**Final Answer: 2**

Enter Numerical Value:

Visualized Solution

$P - V Diagrams$

$Work Done Formulas$

$Setup for Cycle I$

$Total Work for Cycle I$

$Setup for Cycle II$

$Total Work for Cycle II$

$Ratio of Work Done$

$The Way Forward$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup We are given two distinct cyclic processes, Cycle I and Cycle II, operating on one mole of an ideal gas

The Master Equation for Cycle I

The Master Equation for Cycle II

Final Calculation The problem asks for the ratio of the total work done in Cycle I to that in Cycle II

Similar Questions

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $p_{0}$ . Choose the correct option(s) from the following.

One mole of a monatomic ideal gas undergoes the cyclic process J $\to$ K $\to$ L $\to$ M $\to$ J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

List-I

List-II

One mole of a monoatomic ideal gas undergoes a cyclic process as shown in the figure (where, $V$ is the volume and $T$ is the temperature). Which of the statements below is (are) true ?