Welcome to Lecture 96 of our comprehensive JEE Physics Master Series. In this lecture, we undertake an exhaustive comparative analysis of the two fundamental reversible quasi-static thermodynamic processes: the Isothermal Process ($T = \text{const}$) and the Adiabatic Process ($Q = 0$). We examine their differential slope geometry on $P$-$V$ indicator diagrams, contrast their work-energy behaviors during expansion versus compression, evaluate thermodynamic bulk moduli, and solve graded JEE problems.
Section 1: Comprehensive Concept Capsule
1. Governing Equations & Differential Path Invariants
For an ideal gas of $n$ moles with degrees of freedom $f$ and adiabatic index $\gamma = 1 + \frac{2}{f} = \frac{C_p}{C_v} > 1$:
- Isothermal Path ($T = \text{constant}$):
Governing Equation: $PV = nRT = \text{constant}$
Differential Form: $P\,dV + V\,dP = 0 \implies \left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$
First Law Partition: Since $\Delta T = 0 \implies \Delta U = 0$, all absorbed heat is converted entirely into boundary work:
$$Q_{\text{iso}} = W_{\text{iso}} = nRT \ln\left(\frac{V_f}{V_i}\right) = nRT \ln\left(\frac{P_i}{P_f}\right)$$ - Adiabatic Path ($Q = 0$):
Governing Equation (Poisson’s Relations):
$$PV^\gamma = \text{constant}, \quad TV^{\gamma-1} = \text{constant}, \quad T^\gamma P^{1-\gamma} = \text{constant}$$
Differential Form: $\gamma P V^{\gamma-1}\,dV + V^\gamma\,dP = 0 \implies \left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P}{V}$
First Law Partition: Since $\delta Q = 0$, work is performed exclusively by depleting internal thermal energy:
$$W_{\text{adia}} = -\Delta U = -n C_v (T_f – T_i) = \frac{P_i V_i – P_f V_f}{\gamma – 1} = \frac{nR(T_i – T_f)}{\gamma – 1}$$
2. Slope Comparison on the $P$-$V$ Indicator Diagram
At any identical intersection point $(P_0, V_0)$ on a $P$-$V$ plane:
$$\left(\frac{dP}{dV}\right)_{\text{adia}} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}}$$
Because $\gamma > 1$ for all gases ($\gamma \approx 1.67$ for monoatomic, $1.40$ for rigid diatomic, $1.33$ for polyatomic):
$$\left|\left(\frac{dP}{dV}\right)_{\text{adia}}\right| > \left|\left(\frac{dP}{dV}\right)_{\text{iso}}\right|$$
Core Geometric Law: At any intersection point $(P_0, V_0)$, the adiabatic curve on a $P$-$V$ diagram is strictly steeper than the isothermal curve by a factor of $\gamma$. (On a $V$-$P$ plane where volume is the vertical axis, the isothermal curve is steeper).
3. Expansion vs. Compression Comparison Framework
| Transformation | Constraint | Relative Position on $P$-$V$ | Final Pressure & Temperature | Work Done Comparison |
|---|---|---|---|---|
| Expansion | Same final volume $V_f > V_0$ | Isothermal curve lies above Adiabatic | $P_{\text{iso}} > P_{\text{adia}}$ $T_{\text{iso}} (= T_0) > T_{\text{adia}}$ |
$W_{\text{iso}} > W_{\text{adia}} > 0$ (Larger area under isothermal curve) |
| Expansion | Same final pressure $P_f < P_0$ | Isothermal extends further right | $V_{\text{iso}} > V_{\text{adia}}$ $T_{\text{iso}} > T_{\text{adia}}$ |
$W_{\text{iso}} > W_{\text{adia}} > 0$ |
| Compression | Same final volume $V_f < V_0$ | Adiabatic curve lies above Isothermal | $P_{\text{adia}} > P_{\text{iso}}$ $T_{\text{adia}} > T_{\text{iso}} (= T_0)$ |
$|W_{\text{adia}}| > |W_{\text{iso}}|$ (More work must be performed on gas adiabatically) |
| Compression | Same final pressure $P_f > P_0$ | Adiabatic reaches $P_f$ at larger volume | $V_{\text{adia}} > V_{\text{iso}}$ $T_{\text{adia}} > T_{\text{iso}}$ |
$|W_{\text{adia}}| < |W_{\text{iso}}|$ |
4. Volume Elasticity (Bulk Modulus) & Sound Speed
The volumetric Bulk Modulus is defined as $B = -V \frac{dP}{dV}$:
- Isothermal Bulk Modulus ($B_T$): $B_T = -V \left(-\frac{P}{V}\right) = P$
- Adiabatic Bulk Modulus ($B_S$): $B_S = -V \left(-\gamma\frac{P}{V}\right) = \gamma P$
- Ratio: $\frac{B_S}{B_T} = \gamma > 1$. Sound oscillations in air occur rapidly with zero time for heat exchange, confirming Laplace’s adiabatic formulation $v = \sqrt{B_S/\rho} = \sqrt{\gamma P/\rho}$.
5. Molar Heat Capacities
- Isothermal path: $C_{\text{iso}} = \frac{\delta Q}{n\,dT} = \pm \infty$ (since $dT = 0$).
- Adiabatic path: $C_{\text{adia}} = \frac{\delta Q}{n\,dT} = 0$ (since $\delta Q = 0$).
Section 2: Four Graded Solved Examples
Example 1 (JEE Main): Work Ratio in Dual Isothermal-Adiabatic Expansion
Problem: One mole of a monoatomic ideal gas ($\gamma = 5/3$) at initial state $(P_0, V_0, T_0)$ expands to a volume $V_f = 8V_0$. Find the ratio of work done in isothermal expansion to adiabatic expansion, $\frac{W_{\text{iso}}}{W_{\text{adia}}}$.
Solution:
- For isothermal expansion:
$$W_{\text{iso}} = nRT_0 \ln\left(\frac{8V_0}{V_0}\right) = RT_0 \ln(8) = 3RT_0 \ln 2$$ - For adiabatic expansion:
$$T_f = T_0 \left(\frac{V_0}{8V_0}\right)^{\gamma-1} = T_0 \left(\frac{1}{8}\right)^{2/3} = \frac{T_0}{4}$$
$$W_{\text{adia}} = \frac{nR(T_0 – T_f)}{\gamma – 1} = \frac{R(T_0 – T_0/4)}{5/3 – 1} = \frac{\frac{3}{4}RT_0}{2/3} = \frac{9}{8} RT_0$$ - Taking the ratio:
$$\frac{W_{\text{iso}}}{W_{\text{adia}}} = \frac{3RT_0 \ln 2}{\frac{9}{8}RT_0} = \frac{8 \ln 2}{3} \approx \frac{8 \times 0.693}{3} \approx 1.848$$
Example 2 (JEE Main/Advanced): Dual Compression to Identical Pressure
Problem: An ideal diatomic gas ($\gamma = 1.4$) at $(P_0, V_0)$ is compressed to a final pressure $P_f = 32 P_0$. In Path A, compression is isothermal; in Path B, compression is adiabatic. Determine the ratio of the final volumes $\frac{V_{f,B}}{V_{f,A}}$.
Solution:
- Isothermal compression: $P_0 V_0 = P_f V_{f,A} \implies V_{f,A} = \frac{V_0}{32}$.
- Adiabatic compression: $P_0 V_0^\gamma = P_f V_{f,B}^\gamma \implies V_{f,B} = V_0 \left(\frac{1}{32}\right)^{1/1.4} = V_0 (2^{-5})^{5/7} = V_0 2^{-25/7}$.
- Ratio:
$$\frac{V_{f,B}}{V_{f,A}} = \frac{V_0 2^{-25/7}}{V_0 2^{-5}} = 2^{5 – 25/7} = 2^{10/7} \approx 2.69$$
Takeaway: Because adiabatic compression warms the gas, pressure builds much faster, terminating the compression at a larger volume.
Example 3 (JEE Advanced): Volume Ratio Equivalence in Mixed Carnot Branches
Problem: A cycle comprises isothermal expansion $1 \to 2$ at $T_1$, adiabatic expansion $2 \to 3$ from $T_1$ to $T_2$, isothermal compression $3 \to 4$ at $T_2$, and adiabatic compression $4 \to 1$ from $T_2$ to $T_1$. Prove that $\frac{V_2}{V_1} = \frac{V_3}{V_4}$.
Solution:
- For adiabatic branch $2 \to 3$: $T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1} \implies \frac{T_1}{T_2} = \left(\frac{V_3}{V_2}\right)^{\gamma-1}$.
- For adiabatic branch $4 \to 1$: $T_2 V_4^{\gamma-1} = T_1 V_1^{\gamma-1} \implies \frac{T_1}{T_2} = \left(\frac{V_4}{V_1}\right)^{\gamma-1}$.
- Equating temperature ratios:
$$\left(\frac{V_3}{V_2}\right)^{\gamma-1} = \left(\frac{V_4}{V_1}\right)^{\gamma-1} \implies \frac{V_3}{V_2} = \frac{V_4}{V_1} \iff \frac{V_2}{V_1} = \frac{V_3}{V_4}$$
Example 4 (JEE Advanced): Difference in Net Work Between Intersecting Paths
Problem: An isothermal and an adiabatic curve intersect at $A(P_0, V_0)$. The gas expands from $V_0$ to $2V_0$ along both paths to points $B_{\text{iso}}$ and $B_{\text{adia}}$. Find the area between the two curves in terms of $P_0, V_0$, and $\gamma$.
Solution:
- The difference in work represents the enclosed area:
$$\Delta W = W_{\text{iso}} – W_{\text{adia}} = \int_{V_0}^{2V_0} P_{\text{iso}}\,dV – \int_{V_0}^{2V_0} P_{\text{adia}}\,dV$$ - Integrating:
$$W_{\text{iso}} = P_0 V_0 \ln 2$$
$$W_{\text{adia}} = \frac{P_0 V_0 [1 – 2^{1-\gamma}]}{\gamma – 1}$$ - $$\Delta W = P_0 V_0 \left[ \ln 2 – \frac{1 – 2^{1-\gamma}}{\gamma – 1} \right]$$
For monoatomic gas ($\gamma = 5/3$):
$$\Delta W = P_0 V_0 \left[ 0.693 – \frac{1 – 2^{-2/3}}{2/3} \right] = P_0 V_0 [0.693 – 0.555] = 0.138 P_0 V_0$$
Section 3: 10-Question Practice Worksheet
Single Choice Questions (JEE Main Standard)
Q1. If an ideal gas expands from volume $V_1$ to $V_2$ isothermally doing work $W_1$, and adiabatically doing work $W_2$, then:
(A) $W_1 > W_2$
(B) $W_1 < W_2$
(C) $W_1 = W_2$
(D) $W_1 = \gamma W_2$
Q2. At a common intersection point of an isothermal and an adiabatic curve on a $P$-$V$ diagram, the ratio of the slope of the adiabatic curve to that of the isothermal curve is:
(A) $1/\gamma$
(B) $\gamma$
(C) $\gamma – 1$
(D) $1 – \gamma$
Q3. A sample of gas at $(P_0, V_0)$ is compressed to half its initial volume ($V_0/2$). The work done on the gas is maximum when the compression is:
(A) Isothermal
(B) Isobaric
(C) Adiabatic
(D) Equal in all processes
Q4. The isothermal and adiabatic bulk moduli of a gas at pressure $P$ are $B_T$ and $B_S$ respectively. The ratio $B_S / B_T$ is:
(A) $P$
(B) $1$
(C) $\gamma$
(D) $\gamma P$
Q5. An ideal monoatomic gas ($\gamma = 5/3$) and a diatomic gas ($\gamma = 7/5$) initially at the same temperature and pressure are expanded adiabatically to twice their initial volume. The final pressure will be:
(A) Greater for monoatomic gas
(B) Greater for diatomic gas
(C) Equal for both gases
(D) Independent of volume
Multiple Correct Questions (JEE Advanced Standard)
Q6. An ideal gas initially at $(P_0, V_0)$ is taken through two distinct paths to reach the same final pressure $P_f < P_0$: Path 1 is isothermal and Path 2 is adiabatic. Which of the following statements are correct?
(A) Final volume $V_{\text{iso}} > V_{\text{adia}}$
(B) Final temperature $T_{\text{iso}} > T_{\text{adia}}$
(C) Work done $W_{\text{iso}} > W_{\text{adia}}$
(D) Change in internal energy $\Delta U_{\text{iso}} < \Delta U_{\text{adia}}$
Q7. Which of the following statements regarding $P$-$V$ indicator curves are TRUE?
(A) Two isothermal curves for different temperatures can never intersect each other.
(B) Two adiabatic curves for the same gas can never intersect each other.
(C) An isothermal curve and an adiabatic curve can intersect at only one point.
(D) In a cyclic process consisting of two isotherms and two adiabatics, the net work done is non-zero.
Q8. For a reversible compression from $(P_0, V_0)$ to $(P_f, V_f)$:
(A) If compressed to the same final volume, adiabatic compression results in higher pressure than isothermal compression.
(B) If compressed to the same final volume, work done on the gas is greater in isothermal compression.
(C) If compressed to the same final pressure, final volume in adiabatic compression is greater than in isothermal compression.
(D) If compressed to the same final pressure, work done on the gas is greater in isothermal compression.
Numerical Answer Type Questions
Q9. One mole of an ideal gas ($\gamma = 1.5$) expands adiabatically from $(P_0, V_0, T_0)$ such that its final temperature becomes $0.64 T_0$. If the same gas were expanded isothermally at $T_0$ to the same final volume, the work done in the isothermal process is $k \times RT_0 \ln(1.25)$. Find the integer value of $k$.
Q10. An ideal gas at initial pressure $1.0 \times 10^5\text{ N/m}^2$ has adiabatic bulk modulus $1.4 \times 10^5\text{ N/m}^2$. At what pressure (in $10^5\text{ N/m}^2$) will its isothermal bulk modulus equal its initial adiabatic bulk modulus?
Section 4: Detailed Step-by-Step Solutions & Quick Answer Key
Quick Answer Key
| Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 |
|---|---|---|---|---|---|---|---|---|---|
| A | B | C | C | B | A, B, C | A, B, C, D | A, C, D | 2 | 1.4 |
Step-by-Step Solutions
Sol 1. In expansion from $V_1$ to $V_2$ ($V_2 > V_1$), the isothermal curve lies strictly above the adiabatic curve. Since $W = \int P\,dV$, $W_{\text{iso}} > W_{\text{adia}}$. Correct Option: (A)
Sol 2. From differential analysis: $\left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P}{V}$ and $\left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$. Thus, $\frac{(dP/dV)_{\text{adia}}}{(dP/dV)_{\text{iso}}} = \gamma$. Correct Option: (B)
Sol 3. In compression from $V_0$ to $V_0/2$, the curve that lies highest encloses the greatest area beneath it on the $P$-$V$ plane. For $V < V_0$, $P_{\text{adia}} > P_{\text{iso}} > P_{\text{isobaric}}$. Hence, the magnitude of work done on the gas $|W| = \int P\,dV$ is maximum for adiabatic compression. Correct Option: (C)
Sol 4. $B_S = \gamma P$ and $B_T = P$. Hence, $\frac{B_S}{B_T} = \frac{\gamma P}{P} = \gamma$. Correct Option: (C)
Sol 5. For adiabatic expansion, $P_f = P_0 (V_0/V_f)^\gamma = P_0 (1/2)^\gamma$. For monoatomic gas ($\gamma_1 = 5/3 \approx 1.67$), $P_{f,1} = P_0 / 2^{1.67}$. For diatomic gas ($\gamma_2 = 1.4$), $P_{f,2} = P_0 / 2^{1.4}$. Since $2^{1.67} > 2^{1.4}$, $P_{f,2} > P_{f,1}$. The final pressure is greater for the diatomic gas. Correct Option: (B)
Sol 6.
– For isothermal: $P_0 V_0 = P_f V_{\text{iso}} \implies V_{\text{iso}} = V_0 (P_0/P_f)$.
– For adiabatic: $P_0 V_0^\gamma = P_f V_{\text{adia}}^\gamma \implies V_{\text{adia}} = V_0 (P_0/P_f)^{1/\gamma}$.
Since $P_0/P_f > 1$ and $1/\gamma < 1$, $(P_0/P_f) > (P_0/P_f)^{1/\gamma}$, so $V_{\text{iso}} > V_{\text{adia}}$ (Statement A is True).
– In isothermal expansion, $T_f = T_0$; in adiabatic expansion, gas cools so $T_f < T_0$. Hence $T_{\text{iso}} > T_{\text{adia}}$ (Statement B is True).
– Since the isothermal curve lies above the adiabatic curve throughout expansion, $W_{\text{iso}} > W_{\text{adia}}$ (Statement C is True).
– $\Delta U_{\text{iso}} = 0$, while $\Delta U_{\text{adia}} = -W_{\text{adia}} < 0$. Thus $\Delta U_{\text{iso}} > \Delta U_{\text{adia}}$ (Statement D is False).
Correct Options: (A, B, C)
Sol 7.
– (A) True: Different isotherms correspond to different temperatures $T_1 \neq T_2$. If they intersected, that state would have two temperatures simultaneously.
– (B) True: If two adiabatics intersected, a cycle could be formed with an isotherm that violates Kelvin-Planck’s statement of the Second Law.
– (C) True: Since $(dP/dV)_{\text{adia}} = \gamma (dP/dV)_{\text{iso}}$ and $\gamma > 1$ monotonically, they can intersect at at most one point.
– (D) True: The enclosed area between two isotherms and two adiabatics forms a Carnot cycle where $W_{\text{net}} > 0$.
Correct Options: (A, B, C, D)
Sol 8.
– For compression to same $V_f < V_0$, $P_{\text{adia}} = P_0 (V_0/V_f)^\gamma > P_{\text{iso}} = P_0(V_0/V_f)$ since $\gamma > 1$ (Statement A is True; B is False).
– For compression to same $P_f > P_0$, $V_{\text{adia}} = V_0 (P_0/P_f)^{1/\gamma}$ and $V_{\text{iso}} = V_0 (P_0/P_f)$. Since $P_0/P_f < 1$, $(P_0/P_f)^{1/\gamma} > (P_0/P_f)$, so $V_{\text{adia}} > V_{\text{iso}}$ (Statement C is True).
– Under compression to same final pressure, plotting on $P$-$V$ with $P$ integrated along vertical or evaluating $\int_{V_f}^{V_0} P\,dV$: the isothermal path must compress to a smaller volume, sweeping more area against the pressure axis, yielding $|W_{\text{iso}}| > |W_{\text{adia}}|$ (Statement D is True).
Correct Options: (A, C, D)
Sol 9.
For adiabatic expansion:
$$T_0 V_0^{\gamma-1} = T_f V_f^{\gamma-1} \implies \left(\frac{V_f}{V_0}\right)^{\gamma-1} = \frac{T_0}{T_f} = \frac{1}{0.64} = \frac{25}{16}$$
Given $\gamma = 1.5 \implies \gamma – 1 = 0.5 = 1/2$:
$$\left(\frac{V_f}{V_0}\right)^{1/2} = \frac{5}{4} \implies \frac{V_f}{V_0} = \left(\frac{5}{4}\right)^2 = \left(\frac{1}{0.8}\right)^2 = 1.25^2 = 1.5625$$
For isothermal expansion at $T_0$ to volume $V_f$:
$$W_{\text{iso}} = nRT_0 \ln\left(\frac{V_f}{V_0}\right) = 1 \cdot RT_0 \ln(1.25^2) = 2RT_0 \ln(1.25)$$
Comparing with $k \times RT_0 \ln(1.25)$, we get $k = 2$.
Sol 10.
Adiabatic bulk modulus $B_S = \gamma P_0 = 1.4 \times 10^5\text{ N/m}^2$.
Since $P_0 = 1.0 \times 10^5\text{ N/m}^2$, we have $\gamma = 1.4$.
Isothermal bulk modulus at pressure $P’$ is $B_T’ = P’$.
We require $B_T’ = B_S \implies P’ = 1.4 \times 10^5\text{ N/m}^2$.
Hence, in units of $10^5\text{ N/m}^2$, the pressure is 1.4.