One mole of a monatomic ideal gas is taken along two cyclic processes E - F - G - E and E - F -

Q: How do we know which curve is adiabatic and which is isothermal?

On a $p-V$ diagram, the slope of an adiabatic curve is $\gamma$ times the slope of an isothermal curve at the same point. Since $\gamma > 1$, the adiabatic curve is always steeper. Thus, the inner curve $F \rightarrow H$ is adiabatic.

Q: Why is $\gamma = 5/3$?

The problem specifies a monatomic ideal gas. For monatomic gases, the degrees of freedom $f = 3$. The ratio of specific heats $\gamma = 1 + 2/f = 1 + 2/3 = 5/3$.

Q: Why do we take the magnitude of work done?

The problem explicitly asks to match the *magnitudes* of the work done. Compressions like $G \rightarrow E$ and $G \rightarrow H$ have negative work done, but we only consider their absolute values.

Q: How did we calculate the volume at $H$?

Using the adiabatic equation $p_F V_F^\gamma = p_H V_H^\gamma$. Substituting the knowns: $32p_0 V_0^{5/3} = p_0 V_H^{5/3}$. Canceling $p_0$ and taking the $3/5$ power on both sides gives $V_H = (32)^{3/5} V_0 = 8 V_0$.

Q: What is the formula for work done in an isothermal process?

The work done is $W = nRT \ln(V_f/V_i)$. Since $nRT = p_i V_i$, we can also write it as $W = p_i V_i \ln(V_f/V_i)$.

Question

One mole of a monatomic ideal gas is taken along two cyclic processes $E ightarrow F ightarrow G ightarrow E$ and $E ightarrow F ightarrow H ightarrow E$ as shown in the $p-V$ diagram.

The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

$\begin{array}{clcl}
\hline
& 	ext{List I} & & 	ext{List II} \
\hline
	ext{P.} & G ightarrow E & 1. & 160 p_0 V_0 \ln 2 \
	ext{Q.} & G ightarrow H & 2. & 36 p_0 V_0 \
	ext{R.} & F ightarrow H & 3. & 24 p_0 V_0 \
	ext{S.} & F ightarrow G & 4. & 31 p_0 V_0 \
\hline
\end{array}$

Options:

(A) P-4, Q-3, R-2, S-1

(B) P-4, Q-3, R-1, S-2

(C) P-3, Q-1, R-2, S-4

(D) P-1, Q-3, R-2, S-4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's carefully analyze the given $p-V$ diagram. We are presented with two cyclic processes starting from state $E$. The path $E ightarrow F$ is clearly isochoric as the volume remains constant at $V_0$. From $F$, the gas expands along two different curves to reach the isobaric line $p = p_0$.

We know from thermodynamics that an adiabatic curve is always steeper than an isothermal curve on a $p-V$ diagram. Therefore, the inner curve $F ightarrow H$ must be the adiabatic process, and the outer curve $F ightarrow G$ must be the isothermal process.

### Finding the Coordinates

To calculate the work done for each path, we first need the exact pressure and volume at every state. From the graph, state $E$ is at $(V_0, p_0)$ and state $F$ is at $(V_0, 32p_0)$.

Let's find the volume at state $G$. The path $F ightarrow G$ is an isothermal process, which means the product of pressure and volume remains constant. So, $p_F V_F = p_G V_G$. Substituting the known values, we get $(32p_0)(V_0) = (p_0)(V_G)$. Solving this, the pressure $p_0$ cancels out, and we get the volume at $G$ as $V_G = 32V_0$.

Now for state $H$. The path $F ightarrow H$ is an adiabatic process. The governing equation is $pV^\gamma = 	ext{constant}$. For a monatomic gas, the ratio of specific heats $\gamma$ is $5/3$. So, $p_F V_F^{5/3} = p_H V_H^{5/3}$.

Plugging in the values, we get $(32p_0)(V_0)^{5/3} = (p_0)(V_H)^{5/3}$. The number $32$ can be written as $2^5$. Taking the $3/5$ power on both sides, we find that the volume at $H$ is exactly $V_H = (2^5)^{3/5} V_0 = 8V_0$.

### Calculating the Work Done

Now we are ready to match the lists. Let's start with path **P**, which is $G ightarrow E$. This is an isobaric compression at pressure $p_0$. The work done is $p_0$ times the change in volume, which is $V_0 - 32V_0$. This gives $-31p_0 V_0$. The magnitude is **$31p_0 V_0$**, which matches with option **4**.

Next is path **Q**, from $G ightarrow H$. This is also an isobaric compression. The work done is $p_0$ times the final volume $8V_0$ minus the initial volume $32V_0$. This equals $-24p_0 V_0$. Its magnitude is **$24p_0 V_0$**, matching with option **3**.

Path **R** is the adiabatic expansion from $F ightarrow H$. The work done is given by $\frac{p_i V_i - p_f V_f}{\gamma - 1}$. Substituting the values, we get $\frac{32p_0 V_0 - 8p_0 V_0}{5/3 - 1}$. This simplifies to $\frac{24p_0 V_0}{2/3}$, which is **$36p_0 V_0$**. This matches option **2**.

Finally, path **S** is the isothermal expansion from $F ightarrow G$. The work done is $nRT \ln(V_f/V_i)$, which is also equal to $p_i V_i \ln(V_f/V_i)$. This gives $32p_0 V_0 \ln(32V_0 / V_0)$. Since $32$ is $2^5$, the $5$ comes out, giving **$160p_0 V_0 \ln 2$**. This matches option **1**.

### Final Conclusion

So, the correct matching is **P-4, Q-3, R-2, S-1**. This corresponds perfectly to option (a).

Select Answer:

Visualized Solution

$p - V$ Diagram Analysis

Initial States

State G (Isothermal)

Volume at G

State H (Adiabatic)

Volume at H

Path P: $G \to E$

Path Q: $G \to H$

Path R: $F \to H$

Path S: $F \to G$

Final Conclusion

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

Finding the Coordinates

Calculating the Work Done

Final Conclusion

Similar Questions

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 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]

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

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.

The figure shows the $p - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then,

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 ?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p−V Diagram Analysis

Initial States

State G (Isothermal)

Volume at G

State H (Adiabatic)

Volume at H

Path P: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G→E

Path Q: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G→H

Path R: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F→H

Path S: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F→G

Final Conclusion

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

Finding the Coordinates

Calculating the Work Done

Final Conclusion

Similar Questions

List-I

List-II

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]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The figure shows the p−V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

$p - V$ Diagram Analysis

Path P: $G \to E$

Path Q: $G \to H$

Path R: $F \to H$

Path S: $F \to G$

The figure shows the $p - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then,