Welcome to Lecture 96 of our comprehensive JEE Physics Master Series. In this lecture, we undertake a systematic, side-by-side Comparison of Isothermal and Adiabatic Processes. We contrast their equations of state, indicator diagram slopes, bulk moduli, and work outputs across both expansion and compression regimes, resolve the classic JEE Advanced traps involving identical initial states versus identical final states, unify all standard thermodynamic paths under the general polytropic framework ($P V^m = \text{constant}$), and master multi-step cyclic problems.
Detailed Concept Card: Isothermal vs. Adiabatic Processes
1. Comprehensive Side-by-Side Comparison Matrix
| Thermodynamic Feature | Isothermal Process | Adiabatic Process |
|---|---|---|
| Defining Constraint | $T = \text{constant}$ ($dT = 0, \Delta T = 0$) | $Q = 0$ ($\delta Q = 0$, no heat exchange) |
| Boundary & Speed | Diathermic (conducting) walls; infinitely slow (quasi-static) execution | Adiabatic (insulating) walls; extremely rapid (sudden) execution |
| Equation of State (Reversible) | Boyle’s Law: $P V = \text{constant}$ | Poisson’s Law: $P V^\gamma = \text{constant}$, $T V^{\gamma – 1} = \text{constant}$, $T^\gamma P^{1 – \gamma} = \text{constant}$ |
| Internal Energy Change ($\Delta U$) | $\Delta U = n C_v \Delta T = 0$ (for ideal gas) | $\Delta U = n C_v (T_2 – T_1) = -W \neq 0$ |
| First Law Formulation | $Q = W$ (heat absorbed equals work done) | $W = -\Delta U$ (work done at expense of internal energy) |
| Work Done ($W$) | $W = nRT \ln\left(\frac{V_2}{V_1}\right) = P_1 V_1 \ln\left(\frac{P_1}{P_2}\right)$ | $W = \frac{P_1 V_1 – P_2 V_2}{\gamma – 1} = \frac{nR(T_1 – T_2)}{\gamma – 1}$ |
| Slope on $P$-$V$ Diagram | $\left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$ | $\left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P}{V} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}}$ |
| Bulk Modulus ($B = -V \frac{dP}{dV}$) | $B_T = P$ | $B_S = \gamma P \implies \frac{B_S}{B_T} = \gamma$ |
| Molar Heat Capacity ($C$) | $C_{\text{iso}} = \frac{Q}{n \Delta T} = \pm \infty$ | $C_{\text{adia}} = \frac{0}{n \Delta T} = 0$ |
2. Indicator Diagram Geometry & Slope Analysis
At any common point of intersection $(P_0, V_0)$ on a $P$-$V$ indicator diagram, the slopes of the isothermal and adiabatic curves are related by:
$$\mathbf{\left(\frac{dP}{dV}\right)_{\text{adia}} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}} = -\gamma \frac{P_0}{V_0}}$$
Because the adiabatic index $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f} > 1$ for every gas, the adiabatic curve is always $\gamma$ times steeper than the isothermal curve at their point of intersection. This fundamental geometric property governs how the two processes behave in expansion versus compression:
Case A: Expansion from the Same Initial State $(P_1, V_1, T_1)$ to Final Volume $V_2$ ($V_2 > V_1$)
- Because the adiabatic curve is steeper, as volume increases from $V_1$ to $V_2$, the adiabatic curve falls below the isothermal curve.
- Final Pressure Ordering: $P_{\text{isobaric}} (= P_1) > P_{\text{iso}} > P_{\text{adia}}$.
- Final Temperature Ordering: $T_{\text{isobaric}} > T_{\text{iso}} (= T_1) > T_{\text{adia}}$ (since adiabatic expansion cools the gas).
- Work Done Ordering (Area under $P$-$V$ curve): $$\mathbf{W_{\text{isobaric}} > W_{\text{iso}} > W_{\text{adia}} > W_{\text{isochoric}} (= 0)}$$
- Effect of Gas Atomicity in Adiabatic Expansion: Since $\gamma_{\text{mono}} (5/3 \approx 1.67) > \gamma_{\text{dia}} (7/5 = 1.40) > \gamma_{\text{poly}} (4/3 \approx 1.33)$, the monoatomic adiabatic curve is the steepest and drops the lowest during expansion: $$P_{\text{poly}} > P_{\text{dia}} > P_{\text{mono}} \quad \text{and} \quad W_{\text{poly}} > W_{\text{dia}} > W_{\text{mono}}$$
Case B: Compression from the Same Initial State $(P_1, V_1, T_1)$ to Final Volume $V_2$ ($V_2 < V_1$)
- Moving leftward from $V_1$ to a smaller volume $V_2$, the steeper adiabatic curve climbs above the isothermal curve.
- Final Pressure Ordering: $P_{\text{adia}} > P_{\text{iso}} > P_{\text{isobaric}} (= P_1)$.
- Final Temperature Ordering: $T_{\text{adia}} > T_{\text{iso}} (= T_1) > T_{\text{isobaric}}$ (since adiabatic compression heats the gas).
- Compressive Work Magnitude Ordering ($|W|$): $$\mathbf{|W_{\text{adia}}| > |W_{\text{iso}}| > |W_{\text{isobaric}}| > |W_{\text{isochoric}}| (= 0)}$$
- Effect of Gas Atomicity in Adiabatic Compression: For a fixed volume compression ratio, the monoatomic gas ($\gamma = 5/3$) rises most steeply, reaching the highest final pressure and requiring the greatest compressive work: $$|W_{\text{mono}}| > |W_{\text{dia}}| > |W_{\text{poly}}|$$
Classic JEE Advanced Trap (Same Initial State vs. Same Final State):
Always check whether the processes share the same initial state or converge to the same final state!
Suppose a gas expands from initial volume $V_1$ to reach the same final state $(P_2, V_2)$ via an isothermal path and an adiabatic path. Because the two curves now intersect at $(P_2, V_2)$ and the adiabatic curve is steeper, tracing backward to $V_1$ places the adiabatic curve above the isothermal curve for all $V \in [V_1, V_2]$! Consequently, for expansion from $V_1$ to a common final state $(P_2, V_2)$, $P_{1,\text{adia}} > P_{1,\text{iso}}$ and $\mathbf{W_{\text{adia}} > W_{\text{iso}}}$—the exact opposite of expansion from a common initial state!
3. Polytropic Unification of Thermodynamic Processes ($P V^m = \text{constant}$)
Every standard thermodynamic process of an ideal gas is a special case of the generalized polytropic process governed by:
$$P V^m = \text{constant}$$
where $m$ is the polytropic exponent. Differentiating gives the general slope $\frac{dP}{dV} = -m \frac{P}{V}$, work done $W = \frac{P_1 V_1 – P_2 V_2}{m – 1} = \frac{nR(T_1 – T_2)}{m – 1}$ (for $m \neq 1$), and molar heat capacity:
$$\mathbf{C = C_v + \frac{R}{1 – m} = \frac{R}{\gamma – 1} + \frac{R}{1 – m} = \frac{(m – \gamma) R}{(\gamma – 1)(m – 1)}}$$
- Isobaric Process ($m = 0$): $P V^0 = \text{const} \implies P = \text{const}$, slope $= 0$, $C = C_v + R = C_p$.
- Isothermal Process ($m = 1$): $P V^1 = \text{const}$, slope $= -\frac{P}{V}$, $C \to \pm \infty$.
- Adiabatic Process ($m = \gamma$): $P V^\gamma = \text{const}$, slope $= -\gamma \frac{P}{V}$, $C = 0$.
- Isochoric Process ($m \to \infty$): $P^{1/m} V = \text{const} \implies V = \text{const}$, slope $= -\infty$, $C = C_v$.
- Negative Molar Heat Capacity Region ($1 < m < \gamma$): For any process curve lying strictly between the isothermal curve ($m = 1$) and the adiabatic curve ($m = \gamma$), the molar heat capacity is negative ($C < 0$)! During expansion along such a path, heat is absorbed ($Q > 0$), yet the gas performs even more work than the heat supplied ($W > Q$), causing its internal energy and temperature to drop ($\Delta T < 0$).
Four Graded Solved Examples
Example 1 (JEE Main Level: Expansion Comparison from Common Initial State)
Problem Statement: Two identical cylinders $A$ and $B$ each contain $n = 1\text{ mole}$ of an ideal rigid diatomic gas ($\gamma = 1.4$) at initial temperature $T_0 = 300\text{ K}$ and initial pressure $P_0$. The gas in cylinder $A$ expands isothermally until its volume doubles ($V_f = 2V_0$), while the gas in cylinder $B$ expands reversibly and adiabatically until its volume also doubles ($V_f = 2V_0$). Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$, $\ln 2 = 0.693$, and $2^{0.4} = 1.32$, determine:
- The ratio of final pressures $\frac{P_A}{P_B}$ and the ratio of final temperatures $\frac{T_A}{T_B}$.
- The work done $W_A$ and $W_B$ by the gases in the two cylinders.
Solution:
Step 1: Final Pressures and Temperatures.
For cylinder $A$ (isothermal expansion to $2V_0$):
$$T_A = T_0 = 300\text{ K}, \quad P_A = P_0 \left(\frac{V_0}{2V_0}\right) = \frac{P_0}{2} = 0.50 \, P_0$$
For cylinder $B$ (adiabatic expansion to $2V_0$):
$$T_B = T_0 \left(\frac{V_0}{2V_0}\right)^{\gamma – 1} = \frac{T_0}{2^{0.4}} = \frac{300}{1.32} \approx 227.27\text{ K}$$
$$P_B = P_0 \left(\frac{V_0}{2V_0}\right)^\gamma = \frac{P_0}{2^{1.4}} = \frac{P_0}{2 \times 2^{0.4}} = \frac{P_0}{2.64} \approx 0.379 \, P_0$$
Taking the ratios:
$$\frac{P_A}{P_B} = \frac{P_0 / 2}{P_0 / 2^{1.4}} = 2^{0.4} = \mathbf{1.32}, \quad \frac{T_A}{T_B} = \frac{T_0}{T_0 / 2^{0.4}} = 2^{0.4} = \mathbf{1.32}$$
Step 2: Calculate Work Done $W_A$ and $W_B$.
For isothermal expansion in cylinder $A$:
$$W_A = n R T_0 \ln\left(\frac{2V_0}{V_0}\right) = (1)(8.314)(300)(0.693) = \mathbf{1728.5\text{ J}}$$
For adiabatic expansion in cylinder $B$:
$$W_B = \frac{n R (T_0 – T_B)}{\gamma – 1} = \frac{(1)(8.314)(300 – 227.27)}{1.4 – 1} = \frac{8.314 \times 72.73}{0.4} = \mathbf{1511.6\text{ J}}$$
Takeaway: Starting from the same initial state and expanding to the same final volume, isothermal expansion maintains higher pressure ($P_A = 1.32 P_B$), higher temperature ($T_A = 1.32 T_B$), and delivers greater mechanical work ($W_A > W_B$).
Example 2 (Conceptual & Graphical: Atomicity Ranking on $P$-$V$ Diagrams)
Problem Statement: Three identical cylinders each contain $1\text{ mole}$ of an ideal gas at the same initial state $(P_0, V_0, T_0)$: Cylinder 1 contains Helium (monoatomic, $\gamma_1 = 5/3$), Cylinder 2 contains Nitrogen (diatomic, $\gamma_2 = 7/5$), and Cylinder 3 contains Argon (monoatomic). Cylinders 1 and 2 undergo reversible adiabatic expansion to volume $8V_0$, while Cylinder 3 undergoes reversible isothermal expansion to $8V_0$. Rank the three curves from top to bottom on a $P$-$V$ diagram, find the ratio of initial slopes $\left|\frac{\text{Slope}_1}{\text{Slope}_2}\right|$ at $(P_0, V_0)$, and calculate the exact work done in Cylinder 1 in terms of $P_0 V_0$.
Solution:
Step 1: Rank the Polytropic Exponents and Curves.
The three processes follow $P V^m = \text{constant}$ with exponents:
– Cylinder 3 (Isothermal): $m_3 = 1.00$
– Cylinder 2 (Diatomic Adiabatic): $m_2 = \gamma_2 = 7/5 = 1.40$
– Cylinder 1 (Monoatomic Adiabatic): $m_1 = \gamma_1 = 5/3 \approx 1.67$
Since slope at $(P_0, V_0)$ is $-m \frac{P_0}{V_0}$ and $m_1 > m_2 > m_3$, Curve 1 is the steepest and drops the fastest during expansion. Thus, from top to bottom on the expansion diagram:
$$\textbf{Top: Curve 3 (Isothermal)} \longrightarrow \textbf{Middle: Curve 2 (Diatomic Adiabatic)} \longrightarrow \textbf{Bottom: Curve 1 (Monoatomic Adiabatic)}$$
Step 2: Ratio of Adiabatic Slopes at $(P_0, V_0)$.
$$\left|\frac{\text{Slope}_1}{\text{Slope}_2}\right| = \frac{\gamma_1 (P_0 / V_0)}{\gamma_2 (P_0 / V_0)} = \frac{5/3}{7/5} = \mathbf{\frac{25}{21} \approx 1.19}$$
Step 3: Exact Work Done in Cylinder 1 (Monoatomic Adiabatic to $8V_0$).
Final pressure in Cylinder 1:
$$P_{1f} = P_0 \left(\frac{V_0}{8V_0}\right)^{5/3} = P_0 \left(\frac{1}{2^3}\right)^{5/3} = \frac{P_0}{2^5} = \frac{P_0}{32}$$
Work done:
$$W_1 = \frac{P_0 V_0 – P_{1f} V_{1f}}{\gamma_1 – 1} = \frac{P_0 V_0 – \left(\frac{P_0}{32}\right)(8V_0)}{\frac{5}{3} – 1} = \frac{P_0 V_0 \left(1 – \frac{1}{4}\right)}{\frac{2}{3}} = \frac{3}{4} \times \frac{3}{2} P_0 V_0 = \mathbf{\frac{9}{8} P_0 V_0 = 1.125 \, P_0 V_0}$$
Takeaway: A gas with fewer degrees of freedom has a larger $\gamma$, making its adiabatic curve steeper; during expansion it cools faster and performs less boundary work.
Example 3 (JEE Advanced Level: Compression to Identical Final Pressure $P_f = 32 P_0$)
Problem Statement: An ideal monoatomic gas ($\gamma = 5/3$) initially at state $(P_0, V_0, T_0)$ is compressed in two separate experiments until its pressure increases by a factor of $32$ ($P_f = 32 P_0$): in Experiment 1, the compression is reversible isothermal; in Experiment 2, the compression is reversible adiabatic. Calculate the final volume, final temperature, and magnitude of work done on the gas in both experiments in terms of $P_0, V_0, T_0$ (take $\ln 2 = 0.693$).
Solution:
Step 1: Experiment 1 (Isothermal Compression to $P_f = 32 P_0$).
By Boyle’s Law ($P_f V_{f,\text{iso}} = P_0 V_0$):
$$V_{f,\text{iso}} = \frac{P_0 V_0}{32 P_0} = \mathbf{\frac{V_0}{32}}, \quad T_{f,\text{iso}} = \mathbf{T_0}$$
Magnitude of work done on the gas:
$$|W_{\text{iso}}| = P_0 V_0 \ln\left(\frac{P_f}{P_0}\right) = P_0 V_0 \ln(32) = 5 P_0 V_0 \ln 2 = 5(0.693) P_0 V_0 = \mathbf{3.465 \, P_0 V_0}$$
Step 2: Experiment 2 (Adiabatic Compression to $P_f = 32 P_0$).
By Poisson’s Law ($P_f V_{f,\text{adia}}^\gamma = P_0 V_0^\gamma$) with $\gamma = 5/3$:
$$\left(\frac{V_0}{V_{f,\text{adia}}}\right)^{5/3} = \frac{P_f}{P_0} = 32 = 2^5 \implies \frac{V_0}{V_{f,\text{adia}}} = (2^5)^{3/5} = 2^3 = 8$$
$$V_{f,\text{adia}} = \mathbf{\frac{V_0}{8}}$$
Final temperature from ideal gas law:
$$T_{f,\text{adia}} = T_0 \left(\frac{P_f V_{f,\text{adia}}}{P_0 V_0}\right) = T_0 \left(\frac{32 P_0 \cdot (V_0 / 8)}{P_0 V_0}\right) = \mathbf{4 T_0}$$
Magnitude of work done on the gas:
$$|W_{\text{adia}}| = \frac{P_f V_{f,\text{adia}} – P_0 V_0}{\gamma – 1} = \frac{32 P_0 (V_0 / 8) – P_0 V_0}{\frac{5}{3} – 1} = \frac{4 P_0 V_0 – P_0 V_0}{2/3} = 3 \times \frac{3}{2} P_0 V_0 = \mathbf{4.50 \, P_0 V_0}$$
Takeaway: To reach $32 P_0$, the adiabatic compression requires compressing the volume by a factor of only $8$ (because temperature quadruples to $4T_0$), whereas isothermal compression requires compressing the volume by a factor of $32$.
Example 4 (JEE Advanced Level: Three-Step Isothermal–Isochoric–Adiabatic Cycle)
Problem Statement: One mole of an ideal monoatomic gas ($\gamma = 5/3$) initially at state $A(P_0, V_0, T_0)$ undergoes a closed three-step reversible cycle $A \to B \to C \to A$:
– Process $A \to B$: Isothermal expansion at temperature $T_0$ from volume $V_0$ to $8V_0$.
– Process $B \to C$: Isochoric cooling at constant volume $8V_0$ until the gas reaches state $C$, which lies on the reversible adiabatic curve passing through state $A$.
– Process $C \to A$: Reversible adiabatic compression from volume $8V_0$ back to initial state $A(P_0, V_0, T_0)$.
Determine the work done and heat exchanged in each step, the net cycle work $W_{\text{net}}$, and the thermal efficiency $\eta$ of the cycle (take $\ln 2 = 0.693$).
Solution:
Step 1: Analyze Process $A \to B$ (Isothermal Expansion, $V_0 \to 8V_0$ at $T_0$).
$$\Delta U_{AB} = 0$$
$$W_{AB} = Q_{AB} = R T_0 \ln\left(\frac{8V_0}{V_0}\right) = 3 R T_0 \ln 2 \approx 3(0.693) R T_0 = \mathbf{2.079 \, R T_0 > 0}$$
Since $Q_{AB} > 0$, heat is absorbed during $A \to B$.
Step 2: Find Coordinates of State $C$ and Analyze Process $B \to C$ (Isochoric Cooling at $8V_0$).
Because $C \to A$ is a reversible adiabatic path connecting $(V_C = 8V_0, T_C)$ to $(V_A = V_0, T_0)$ with $\gamma = 5/3$:
$$T_C (8V_0)^{5/3 – 1} = T_0 V_0^{2/3} \implies T_C (8)^{2/3} = T_0 \implies 4 T_C = T_0 \implies \mathbf{T_C = \frac{T_0}{4}}$$
In isochoric cooling $B \to C$ ($V = 8V_0, dV = 0$):
$$W_{BC} = \mathbf{0}$$
$$Q_{BC} = \Delta U_{BC} = n C_v (T_C – T_B) = \frac{3}{2} R \left(\frac{T_0}{4} – T_0\right) = \mathbf{-\frac{9}{8} R T_0 = -1.125 \, R T_0 < 0}$$
Heat is rejected during $B \to C$.
Step 3: Analyze Process $C \to A$ (Adiabatic Compression, $8V_0 \to V_0$).
$$Q_{CA} = \mathbf{0}$$
$$\Delta U_{CA} = n C_v (T_0 – T_C) = +\frac{9}{8} R T_0 \implies W_{CA} = -\Delta U_{CA} = \mathbf{-\frac{9}{8} R T_0 = -1.125 \, R T_0}$$
Step 4: Net Cycle Work and Thermal Efficiency $\eta$.
Net work done over the cycle:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 3 R T_0 \ln 2 + 0 – \frac{9}{8} R T_0 = \left(3 \ln 2 – \frac{9}{8}\right) R T_0 \approx \mathbf{0.954 \, R T_0}$$
Notice that $Q_{\text{net}} = Q_{AB} + Q_{BC} + Q_{CA} = \left(3 \ln 2 – \frac{9}{8}\right) R T_0 = W_{\text{net}}$, verifying First Law closure.
Heat is absorbed only during $A \to B$ ($Q_{\text{in}} = Q_{AB} = 3 R T_0 \ln 2$). Thus, thermal efficiency is:
$$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{3 R T_0 \ln 2 – \frac{9}{8} R T_0}{3 R T_0 \ln 2} = 1 – \frac{3}{8 \ln 2} \approx 1 – \frac{3}{5.544} = 1 – 0.541 = \mathbf{0.459 \; (45.9\%)}$$
Takeaway: In any cycle combining isothermal, isochoric, and adiabatic steps, use Poisson’s relation on the adiabatic branch first to lock in the corner temperatures before computing $\Delta U$ and $W$.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. An ideal gas initially at state $(P_0, V_0)$ expands to a final volume $3V_0$ along three different reversible paths: Path 1 is isobaric, Path 2 is isothermal, and Path 3 is adiabatic. If $W_1, W_2$, and $W_3$ denote the work done by the gas along Paths 1, 2, and 3 respectively, then:
(A) $W_1 > W_2 > W_3$
(B) $W_3 > W_2 > W_1$
(C) $W_2 > W_1 > W_3$
(D) $W_1 = W_2 = W_3$
Q2. An isothermal curve and a reversible adiabatic curve for an ideal monoatomic gas intersect at state point $(P_0, V_0)$ on a $P$-$V$ diagram. The ratio of the slope of the adiabatic curve to the slope of the isothermal curve at $(P_0, V_0)$ is:
(A) $7/5$
(B) $5/3$
(C) $3/5$
(D) $4/3$
Q3. Two identical samples of an ideal gas initially at the same temperature $T_0$, pressure $P_0$, and volume $V_0$ are compressed to half their initial volume ($V_0 / 2$): Sample 1 isothermally and Sample 2 adiabatically. Which of the following is true regarding their final pressures ($P_1, P_2$) and work done on the gases ($|W_1|, |W_2|$)?
(A) $P_2 > P_1$ and $|W_2| > |W_1|$
(B) $P_1 > P_2$ and $|W_1| > |W_2|$
(C) $P_2 > P_1$ and $|W_1| > |W_2|$
(D) $P_1 = P_2$ and $|W_2| > |W_1|$
Q4. One mole of an ideal gas with adiabatic index $\gamma$ undergoes a polytropic process $P V^m = \text{constant}$, where the process curve on a $P$-$V$ diagram lies strictly between the isothermal curve and the adiabatic curve ($1 < m < \gamma$). During this process, the molar heat capacity $C$ of the gas is:
(A) Zero ($C = 0$)
(B) Positive and greater than $C_p$
(C) Strictly negative ($C < 0$)
(D) Positive and between $0$ and $C_v$
Q5. Equal moles of Helium (monoatomic, $\gamma = 5/3$) and Hydrogen (diatomic, $\gamma = 7/5$) initially at the same state $(P_0, V_0)$ expand reversibly and adiabatically to the same final volume $2V_0$. The ratio of the final pressure of Helium to that of Hydrogen ($P_{\text{He}} / P_{\text{H}_2}$) is:
(A) $2^{-4/15}$
(B) $2^{+4/15}$
(C) $2^{-2/5}$
(D) $1$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. An ideal gas expands from an initial volume $V_1$ to reach the exact same final equilibrium state $(P_2, V_2)$ via two distinct reversible paths: Path A is isothermal, and Path B is adiabatic. Which of the following statements is/are correct?
(A) The initial pressure of the gas in Path B must be greater than in Path A ($P_{1,\text{B}} > P_{1,\text{A}}$).
(B) The initial temperature of the gas in Path B must be greater than in Path A ($T_{1,\text{B}} > T_{1,\text{A}}$).
(C) The work done by the gas along the adiabatic Path B is greater than along the isothermal Path A ($W_{\text{B}} > W_{\text{A}}$).
(D) At the common final state $(P_2, V_2)$, the adiabatic curve is steeper than the isothermal curve by a factor of $\gamma$.
Q7. Which of the following correctly matches the polytropic exponent $m$ in $P V^m = \text{constant}$ with the corresponding thermodynamic property for an ideal gas?
(A) For $m = 0$, the process is isobaric, the $P$-$V$ slope is zero, and molar heat capacity is $C = C_p$.
(B) For $m = 1$, the process is isothermal, bulk modulus is $B = P$, and molar heat capacity is $C = \pm \infty$.
(C) For $m = \gamma$, the process is reversible adiabatic, bulk modulus is $B = \gamma P$, and molar heat capacity is $C = 0$.
(D) In the limit $m \to \infty$, the process is isochoric, work done is zero, and molar heat capacity is $C = C_v$.
Q8. Starting from the same initial state $(P_0, V_0, T_0)$, a sample of an ideal gas is compressed to volume $V_0 / 3$ along an isothermal path (Process I) and a reversible adiabatic path (Process II). Which of the following statements is/are correct?
(A) In Process I, heat is rejected to the surroundings ($Q_{\text{I}} < 0$), whereas in Process II, no heat is exchanged ($Q_{\text{II}} = 0$).
(B) In Process I, internal energy remains constant ($\Delta U_{\text{I}} = 0$), whereas in Process II, internal energy increases ($\Delta U_{\text{II}} > 0$).
(C) The magnitude of work done on the gas is greater in Process II than in Process I ($|W_{\text{II}}| > |W_{\text{I}}|$).
(D) The final pressure in Process I is $3P_0$, whereas in Process II it is $3^\gamma P_0 > 3P_0$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. On a $P$-$V$ indicator diagram, an isothermal curve and a reversible adiabatic curve for an ideal rigid diatomic gas ($\gamma = 1.4$) intersect at state point $(P_0, V_0)$. If the slope of the isothermal curve at this intersection point is $-25\text{ kPa/m}^3$, find the magnitude of the slope of the adiabatic curve at the same point (in $\text{kPa/m}^3$).
Q10. One mole of an ideal monoatomic gas ($\gamma = 5/3$) at an initial state where $n R T_0 = 8000\text{ J}$ expands from volume $V_0$ to $8V_0$ in two separate reversible experiments: first isothermally (performing work $W_{\text{iso}}$) and second adiabatically (performing work $W_{\text{adia}}$). Taking $\ln 2 = 0.693$, calculate the excess work $\Delta W = W_{\text{iso}} – W_{\text{adia}}$ (in $\text{J}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $W_1 > W_2 > W_3$ | Expansion Work Ordering from Common Initial State |
| Q2 | (B) $5/3$ | Slope Ratio $\gamma$ for Monoatomic Gas |
| Q3 | (A) $P_2 > P_1$ and $|W_2| > |W_1|$ | Isothermal vs. Adiabatic Compression Comparison |
| Q4 | (C) Strictly negative ($C < 0$) | Polytropic Molar Heat Capacity for $1 < m < \gamma$ |
| Q5 | (A) $2^{-4/15}$ | Atomicity Effect on Adiabatic Pressure Drop |
| Q6 | (A), (B), (C), (D) | Expansion Converging to Identical Final State $(P_2, V_2)$ |
| Q7 | (A), (B), (C), (D) | Polytropic Unification ($m = 0, 1, \gamma, \infty$) |
| Q8 | (A), (B), (C), (D) | Thermodynamic Variables in Compression to $V_0 / 3$ |
| Q9 | 35 | Adiabatic Slope Magnitude ($1.4 \times 25\text{ kPa/m}^3$) |
| Q10 | 7632 | Work Difference $W_{\text{iso}} – W_{\text{adia}}$ ($16632 – 9000\text{ J}$) |
Complete Step-by-Step Solutions
Solution 1:
Starting from the same state $(P_0, V_0)$ and expanding to $3V_0$:
– Path 1 (Isobaric): Pressure remains constant at $P_0$.
– Path 2 (Isothermal): Pressure decreases as $P = P_0 (V_0 / V)$.
– Path 3 (Adiabatic): Pressure decreases more steeply as $P = P_0 (V_0 / V)^\gamma$.
Because the isobaric curve is highest, the isothermal curve is in the middle, and the adiabatic curve is lowest over the interval $[V_0, 3V_0]$, the areas under the curves satisfy:
$$W_1 > W_2 > W_3$$
Correct Option: (A)
Solution 2:
At the point of intersection $(P_0, V_0)$:
$$\left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P_0}{V_0}, \quad \left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P_0}{V_0}$$
For an ideal monoatomic gas ($f = 3$), $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$. Thus, the ratio of slopes is $\frac{5}{3}$.
Correct Option: (B)
Solution 3:
When compressed from $(P_0, V_0)$ to $V_0 / 2$:
– Sample 1 (Isothermal): $P_1 = 2P_0$.
– Sample 2 (Adiabatic): $P_2 = 2^\gamma P_0$. Since $\gamma > 1$, $P_2 > P_1$.
Because the adiabatic compression curve lies strictly above the isothermal compression curve for all $V \in [V_0/2, V_0]$, the area under the adiabatic curve is larger, meaning greater compressive work is required: $|W_2| > |W_1|$.
Correct Option: (A)
Solution 4:
For a polytropic process $P V^m = \text{constant}$, the molar heat capacity is:
$$C = \frac{R}{\gamma – 1} + \frac{R}{1 – m} = \frac{(m – \gamma) R}{(\gamma – 1)(m – 1)}$$
When $1 < m < \gamma$, the numerator $(m - \gamma) < 0$, while the denominator $(\gamma - 1)(m - 1) > 0$. Therefore, $C < 0$ (strictly negative). Physically, along any curve between the isotherm and the adiabat, the gas expands enough that work done exceeds heat absorbed, causing temperature to fall despite heat input.
Correct Option: (C)
Solution 5:
For reversible adiabatic expansion from $(P_0, V_0)$ to $2V_0$:
$$P_{\text{He}} = P_0 \left(\frac{1}{2}\right)^{\gamma_{\text{He}}} = P_0 \, 2^{-5/3}$$
$$P_{\text{H}_2} = P_0 \left(\frac{1}{2}\right)^{\gamma_{\text{H}_2}} = P_0 \, 2^{-7/5}$$
Taking the ratio:
$$\frac{P_{\text{He}}}{P_{\text{H}_2}} = 2^{-5/3 – (-7/5)} = 2^{\frac{-25 + 21}{15}} = 2^{-4/15}$$
Correct Option: (A)
Solution 6:
Both paths reach the same final state $(P_2, V_2)$ starting from volume $V_1 < V_2$:
(A) Correct: In Path A (isothermal), $P_{1,\text{A}} = P_2 (V_2 / V_1)$. In Path B (adiabatic), $P_{1,\text{B}} = P_2 (V_2 / V_1)^\gamma$. Since $V_2 / V_1 > 1$ and $\gamma > 1$, $P_{1,\text{B}} > P_{1,\text{A}}$.
(B) Correct: Since both start at volume $V_1$ and $P_{1,\text{B}} > P_{1,\text{A}}$, by $T = PV/(nR)$, $T_{1,\text{B}} > T_{1,\text{A}}$ (which also makes sense because Path B cools down to $T_2$ during expansion whereas Path A stays at $T_2$).
(C) Correct: Tracing backward from $(P_2, V_2)$ to $V_1$, the steeper adiabatic curve lies entirely above the isothermal curve, enclosing a larger area under the $P$-$V$ graph: $W_{\text{B}} > W_{\text{A}}$.
(D) Correct: At their intersection point $(P_2, V_2)$, $\left|(dP/dV)_{\text{B}}\right| = \gamma \left|(dP/dV)_{\text{A}}\right|$.
Correct Options: (A), (B), (C), (D)
Solution 7:
Using $P V^m = \text{const}$, $B = -V (dP/dV) = m P$, and $C = C_v + \frac{R}{1 – m}$:
(A) Correct: $m = 0 \implies P = \text{const}$ (isobaric), slope $= 0$, $C = C_v + R = C_p$.
(B) Correct: $m = 1 \implies PV = \text{const}$ (isothermal), $B = P$, $C \to \pm \infty$.
(C) Correct: $m = \gamma \implies P V^\gamma = \text{const}$ (adiabatic), $B = \gamma P$, $C = C_v – \frac{R}{\gamma – 1} = 0$.
(D) Correct: $m \to \infty \implies V = \text{const}$ (isochoric), $W = 0$, $C = C_v + 0 = C_v$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: Isothermal compression requires rejecting heat ($Q_{\text{I}} = W_{\text{I}} < 0$) to keep $T$ constant; adiabatic compression has $Q_{\text{II}} = 0$.
(B) Correct: $\Delta T_{\text{I}} = 0 \implies \Delta U_{\text{I}} = 0$, whereas adiabatic compression does positive work on the gas, raising internal energy ($\Delta U_{\text{II}} = -W_{\text{II}} > 0$).
(C) Correct: Starting from $(P_0, V_0)$, the adiabatic compression curve lies above the isothermal curve, so $|W_{\text{II}}| > |W_{\text{I}}|$.
(D) Correct: $P_{f,\text{I}} = 3P_0$ by Boyle’s Law, and $P_{f,\text{II}} = 3^\gamma P_0 > 3P_0$ by Poisson’s Law.
Correct Options: (A), (B), (C), (D)
Solution 9:
At the point of intersection $(P_0, V_0)$:
$$\left(\frac{dP}{dV}\right)_{\text{adia}} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}}$$
Given $\gamma = 1.4$ and $\left(\frac{dP}{dV}\right)_{\text{iso}} = -25\text{ kPa/m}^3$:
$$\left(\frac{dP}{dV}\right)_{\text{adia}} = 1.4 \times (-25\text{ kPa/m}^3) = -35\text{ kPa/m}^3$$
The magnitude of the slope of the adiabatic curve is $35\text{ kPa/m}^3$.
Answer: 35
Solution 10:
1. Isothermal Expansion Work ($V_0 \to 8V_0$):
$$W_{\text{iso}} = n R T_0 \ln\left(\frac{8V_0}{V_0}\right) = 3 \, n R T_0 \ln 2 = 3 \times 8000 \times 0.693 = 24000 \times 0.693 = 16632\text{ J}$$
2. Adiabatic Expansion Work ($V_0 \to 8V_0$ with $\gamma = 5/3$):
Final temperature:
$$T_f = T_0 \left(\frac{V_0}{8V_0}\right)^{5/3 – 1} = T_0 \left(\frac{1}{8}\right)^{2/3} = \frac{T_0}{4}$$
Work done:
$$W_{\text{adia}} = \frac{n R (T_0 – T_f)}{\gamma – 1} = \frac{n R T_0 (1 – 1/4)}{2/3} = \frac{9}{8} n R T_0 = \frac{9}{8} \times 8000 = 9000\text{ J}$$
3. Excess Work ($\Delta W$):
$$\Delta W = W_{\text{iso}} – W_{\text{adia}} = 16632\text{ J} – 9000\text{ J} = \mathbf{7632\text{ J}}$$
Answer: 7632