Welcome to Lecture 95 of our comprehensive JEE Physics Master Series. In this lecture, we master the physics of the Adiabatic Process. We analyze thermodynamic transformations occurring without thermal exchange ($Q = 0$), derive Poisson’s equations of state ($PV^\gamma = C$, $TV^{\gamma-1} = C$, and $T^\gamma P^{1-\gamma} = C$), formulate the work done from first principles ($W = -\Delta U$), analyze the steep hyperbolic slope on $P$-$V$ indicator diagrams, compute the adiabatic bulk modulus ($B_S = \gamma P$), deconstruct the critical distinction between reversible and irreversible adiabatic pathways, and solve high-yield JEE Main and JEE Advanced problems.
Detailed Concept Card: Adiabatic Process
An adiabatic process is a thermodynamic transformation in which no heat energy enters or leaves the system across its boundary during the entire course of the change:
$$Q = 0 \implies \delta Q = 0$$
1. Essential Physical Conditions for an Adiabatic Process
In classical thermodynamics, two strict operational criteria must be satisfied to ensure an adiabatic transformation:
- Perfectly Insulating Boundary (Adiabatic Walls): The cylinder walls, base, and piston must be fabricated from materials of near-zero thermal conductivity (e.g., thick asbestos, vacuum jackets, or Dewar enclosures) to prevent boundary thermal conduction.
- Extremely Rapid Execution (Sudden Process): The physical transformation must occur so swiftly that virtually zero time is afforded for heat exchange between the system and its surroundings (e.g., the sudden bursting of an inflated tyre, propagation of longitudinal sound compressions in air, or rapid expansion of combustion gases in an engine cylinder).
2. First Law Formulation: Work Done at the Expense of Internal Energy
Applying the First Law of Thermodynamics ($Q = \Delta U + W$) under the adiabatic condition $Q = 0$:
$$0 = \Delta U + W \implies \mathbf{W = -\Delta U \quad \text{or} \quad \Delta U = -W}$$
In differential form:
$$\delta W = -dU = -n C_v dT$$
- Adiabatic Expansion ($W > 0$): The gas performs positive mechanical work on the surroundings entirely at the expense of its internal energy ($\Delta U < 0$). Because $U = n C_v T$ for an ideal gas, a decrease in internal energy causes a mandatory drop in absolute temperature ($T_f < T_i$). Hence, adiabatic expansion causes cooling.
- Adiabatic Compression ($W < 0$): Work is performed on the gas by external agents. This mechanical energy accumulates directly as internal thermal kinetic energy ($\Delta U > 0$), resulting in an increase in absolute temperature ($T_f > T_i$). Hence, adiabatic compression causes heating (the operational principle of diesel engine auto-ignition).
3. Poisson’s Equations of State (Reversible Adiabatic Path)
For an ideal gas undergoing a reversible, quasi-static adiabatic transformation, the state variables obey Poisson’s Laws:
Derivation: From the First Law, $\delta Q = dU + P\, dV = 0$. Using $dU = n C_v dT$ and the ideal gas differential equation $P\, dV + V\, dP = nR\, dT$:
$$n C_v \left(\frac{P\, dV + V\, dP}{nR}\right) + P\, dV = 0 \implies C_v (P\, dV + V\, dP) + R P\, dV = 0$$
Since Mayer’s relation gives $C_v + R = C_p$:
$$C_p P\, dV + C_v V\, dP = 0 \implies \frac{dP}{P} + \left(\frac{C_p}{C_v}\right) \frac{dV}{V} = 0$$
Defining the adiabatic index (ratio of specific heats) $\gamma = \frac{C_p}{C_v}$ and integrating:
$$\ln P + \gamma \ln V = \text{constant} \implies \mathbf{P V^\gamma = \text{constant}}$$
Using the ideal gas law $P = \frac{nRT}{V}$ and $V = \frac{nRT}{P}$, Poisson’s law takes three equivalent mathematical forms:
- Pressure-Volume Relation: $$P_1 V_1^\gamma = P_2 V_2^\gamma = \text{constant}$$
- Temperature-Volume Relation: $$T_1 V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1} = \text{constant}$$
- Temperature-Pressure Relation: $$\frac{T_1^\gamma}{P_1^{\gamma – 1}} = \frac{T_2^\gamma}{P_2^{\gamma – 1}} = \text{constant} \quad \text{or} \quad T_1^\gamma P_1^{1 – \gamma} = T_2^\gamma P_2^{1 – \gamma}$$
4. Adiabatic Index ($\gamma$) by Molecular Atomicity
The value of $\gamma = 1 + \frac{2}{f}$ depends solely on the molecular degrees of freedom $f$:
| Gas Type | Degrees of Freedom ($f$) | Molar $C_v$ | Molar $C_p$ | Adiabatic Index ($\gamma = C_p / C_v$) |
|---|---|---|---|---|
| Monoatomic (He, Ne, Ar) | $f = 3$ (Translational) | $\frac{3}{2}R$ | $\frac{5}{2}R$ | $\gamma = \frac{5}{3} \approx 1.67$ |
| Rigid Diatomic ($O_2, N_2, H_2$) | $f = 5$ (3 Trans + 2 Rot) | $\frac{5}{2}R$ | $\frac{7}{2}R$ | $\gamma = \frac{7}{5} = 1.40$ |
| Non-Linear Polyatomic ($H_2O, NH_3$) | $f = 6$ (3 Trans + 3 Rot) | $3R$ | $4R$ | $\gamma = \frac{4}{3} \approx 1.33$ |
5. Derivation of Work Done in an Adiabatic Process
For a reversible adiabatic transition from initial state $(P_1, V_1, T_1)$ to final state $(P_2, V_2, T_2)$ along $P V^\gamma = K$ (where $K = P_1 V_1^\gamma = P_2 V_2^\gamma$):
$$W = \int_{V_1}^{V_2} P \, dV = K \int_{V_1}^{V_2} V^{-\gamma} \, dV = K \left[ \frac{V^{1 – \gamma}}{1 – \gamma} \right]_{V_1}^{V_2} = \frac{K V_2^{1 – \gamma} – K V_1^{1 – \gamma}}{1 – \gamma}$$
Substituting $K = P_2 V_2^\gamma$ in the first term and $K = P_1 V_1^\gamma$ in the second term:
$$W = \frac{(P_2 V_2^\gamma) V_2^{1 – \gamma} – (P_1 V_1^\gamma) V_1^{1 – \gamma}}{1 – \gamma} = \frac{P_2 V_2 – P_1 V_1}{1 – \gamma}$$
Rearranging the denominator gives the canonical work formula:
$$\mathbf{W = \frac{P_1 V_1 – P_2 V_2}{\gamma – 1}}$$
Using the ideal gas equation $P_1 V_1 = n R T_1$ and $P_2 V_2 = n R T_2$:
$$\mathbf{W = \frac{n R (T_1 – T_2)}{\gamma – 1} = -n C_v (T_2 – T_1) = -\Delta U}$$
- If the gas expands ($T_1 > T_2$), $W > 0$ (positive work done by the gas).
- If the gas is compressed ($T_1 < T_2$), $W < 0$ (negative work done on the gas).
6. Slope of Adiabatic vs. Isothermal Curves on $P$-$V$ Diagram
To determine the slope of an adiabatic curve at state $(P, V)$, differentiate $P V^\gamma = \text{constant}$:
$$V^\gamma \, dP + P (\gamma V^{\gamma – 1}) \, dV = 0 \implies V \, dP = -\gamma P \, dV$$
$$\mathbf{\left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P}{V}}$$
Recall from Lecture 94 that the isothermal slope is $\left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$. Therefore:
$$\mathbf{\left(\frac{dP}{dV}\right)_{\text{adia}} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}}}$$
- Because $\gamma > 1$ for all gases, the adiabatic curve is strictly steeper than the isothermal curve at any common point of intersection on a $P$-$V$ plane.
- Expansion Comparison: When expanding from identical initial state $(P_0, V_0)$ to a given volume $2V_0$, the adiabatic curve plunges beneath the isothermal curve ($P_{\text{adia}} < P_{\text{iso}}$) because cooling depresses pressure below Boyle's law values. Consequently, isothermal expansion produces more work than adiabatic expansion ($W_{\text{iso}} > W_{\text{adia}}$).
- Compression Comparison: When compressing from $(P_0, V_0)$ to $V_0 / 2$, the adiabatic curve climbs above the isothermal curve ($P_{\text{adia}} > P_{\text{iso}}$) because heating boosts pressure. Consequently, adiabatic compression requires more work than isothermal compression ($|W_{\text{adia}}| > |W_{\text{iso}}|$).
7. Adiabatic Bulk Modulus & Molar Heat Capacity
The Adiabatic Bulk Modulus ($B_S$) is:
$$B_S = -V \left(\frac{dP}{dV}\right)_{\text{adia}} = -V \left(-\gamma \frac{P}{V}\right) = \mathbf{\gamma P}$$
Comparing with the isothermal bulk modulus ($B_T = P$):
$$\mathbf{\frac{B_S}{B_T} = \gamma > 1}$$
This is the foundation of Laplace’s correction for the speed of sound in gases ($v = \sqrt{B_S / \rho} = \sqrt{\gamma P / \rho}$), correcting Newton’s isothermal assumption ($v = \sqrt{P / \rho}$).
Molar Heat Capacity: Because $Q = 0$ while temperature changes ($\Delta T \neq 0$):
$$C_{\text{adia}} = \frac{Q}{n \Delta T} = \frac{0}{n \Delta T} = \mathbf{0}$$
Critical Trap (Reversible vs. Irreversible Adiabatic): Poisson’s relations ($P V^\gamma = \text{const}$) hold only for reversible, quasi-static adiabatic processes. For sudden, irreversible expansions against a constant external pressure $P_{\text{ext}}$, $P V^\gamma = \text{const}$ is invalid! To solve irreversible adiabatic problems, apply the First Law directly: $W = P_{\text{ext}} (V_f – V_i) = -\Delta U = -n C_v (T_f – T_i)$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Reversible Compression of Diatomic Gas)
Problem Statement: A sample of $n = 1\text{ mole}$ of an ideal rigid diatomic gas ($\gamma = 1.4$) at initial pressure $P_1 = 1.0 \times 10^5\text{ Pa}$ and initial temperature $T_1 = 300\text{ K}$ is compressed adiabatically and reversibly to $\frac{1}{32}$ of its initial volume ($V_2 = V_1 / 32$). Given universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $2^5 = 32$, calculate:
- The final temperature $T_2$ of the gas.
- The final pressure $P_2$ of the gas.
- The work done $W$ on the gas during the compression.
Solution:
Step 1: Calculate Final Temperature $T_2$.
Using Poisson’s temperature-volume relation $T_1 V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1}$:
$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma – 1} = 300 \times (32)^{1.4 – 1} = 300 \times (32)^{0.4}$$
Since $32 = 2^5$ and $0.4 = 2/5$:
$$(32)^{0.4} = (2^5)^{2/5} = 2^2 = 4$$
$$T_2 = 300 \times 4 = \mathbf{1200\text{ K}}$$
Step 2: Calculate Final Pressure $P_2$.
Using Poisson’s pressure-volume relation $P_1 V_1^\gamma = P_2 V_2^\gamma$:
$$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = (10^5) \times (32)^{1.4} = (10^5) \times (2^5)^{7/5} = (10^5) \times 2^7$$
Since $2^7 = 128$:
$$P_2 = 128 \times 10^5\text{ Pa} = \mathbf{1.28 \times 10^7\text{ Pa}} = 128\text{ bar}$$
Step 3: Calculate Work Done $W$.
Using the adiabatic work formula:
$$W = \frac{n R (T_1 – T_2)}{\gamma – 1} = \frac{(1)(8.314)(300 – 1200)}{1.4 – 1} = \frac{(8.314)(-900)}{0.4} = -18706.5\text{ J} \approx -18.71\text{ kJ}$$
The work done on the gas is:
$$W_{\text{on}} = -W = \mathbf{+18.71\text{ kJ}}$$
Takeaway: Adiabatic compression to $1/32$ volume quadruples absolute temperature ($300\text{ K} \to 1200\text{ K}$) and escalates pressure by a massive factor of $128$ because work done on the gas increases internal thermal kinetic energy without escape.
Example 2 (Algebraic Comparison: Work Done in Adiabatic vs. Isothermal Expansion)
Problem Statement: One mole of a monoatomic ideal gas ($\gamma = 5/3$) at temperature $T_0$ expands from initial volume $V_0$ to final volume $8V_0$. Derive the exact ratio of the work done in this adiabatic expansion to the work done if the gas had instead expanded isothermally at temperature $T_0$ between the same volume limits. Express the result in terms of $\ln 2$.
Solution:
Step 1: Compute Adiabatic Work $W_{\text{adia}}$.
For a monoatomic gas, $\gamma = 5/3 \implies \gamma – 1 = 2/3$.
The final temperature after adiabatic expansion to $8V_0$ is:
$$T_f = T_0 \left(\frac{V_0}{8V_0}\right)^{\gamma – 1} = T_0 \left(\frac{1}{8}\right)^{2/3} = T_0 \left(\left(\frac{1}{2}\right)^3\right)^{2/3} = T_0 \left(\frac{1}{2}\right)^2 = \frac{T_0}{4}$$
The adiabatic work done by the gas is:
$$W_{\text{adia}} = \frac{n R (T_0 – T_f)}{\gamma – 1} = \frac{(1) R \left(T_0 – \frac{T_0}{4}\right)}{2/3} = \frac{3}{2} R \left(\frac{3}{4} T_0\right) = \mathbf{\frac{9}{8} R T_0 = 1.125 \, R T_0}$$
Step 2: Compute Isothermal Work $W_{\text{iso}}$.
At constant temperature $T_0$:
$$W_{\text{iso}} = n R T_0 \ln\left(\frac{8V_0}{V_0}\right) = R T_0 \ln(2^3) = \mathbf{3 R T_0 \ln 2 \approx 3(0.6931) R T_0 = 2.079 \, R T_0}$$
Step 3: Compute the Ratio.
$$\frac{W_{\text{adia}}}{W_{\text{iso}}} = \frac{\frac{9}{8} R T_0}{3 R T_0 \ln 2} = \mathbf{\frac{3}{8 \ln 2}} \approx \frac{3}{8 \times 0.69315} = \frac{3}{5.5452} \approx \mathbf{0.541}$$
Takeaway: Adiabatic expansion produces only about $54.1\%$ of the work produced during isothermal expansion between the same volumes because cooling continuously reduces the driving internal pressure.
Example 3 (JEE Advanced Level: Irreversible Adiabatic Expansion against Constant External Pressure)
Problem Statement: A thermally insulated vertical cylinder fitted with a frictionless, light, insulated piston contains $n = 1\text{ mole}$ of an ideal monoatomic gas ($\gamma = 5/3, C_v = 1.5R$) at initial pressure $P_1 = 8.0 \times 10^5\text{ Pa}$ and temperature $T_1 = 600\text{ K}$. The piston is restrained by stops. The external atmospheric pressure is $P_{\text{ext}} = 1.0 \times 10^5\text{ Pa}$. The stops are suddenly removed, allowing the gas to expand irreversibly and rapidly against the constant external pressure $P_{\text{ext}}$ until mechanical equilibrium is restored ($P_f = P_{\text{ext}}$). Determine:
- The final equilibrium temperature $T_f$ of the gas.
- The work done $W$ by the expanding gas (take $R = 8.314\text{ J/(mol}\cdot\text{K)}$).
- Compare $T_f$ with the temperature that would have resulted if the expansion had been reversible.
Solution:
Step 1: Formulate the First Law for Irreversible Expansion.
Because the cylinder is thermally insulated, $Q = 0$. By the First Law:
$$W = -\Delta U = -n C_v (T_f – T_1) = n C_v (T_1 – T_f) = \frac{3}{2} n R (T_1 – T_f)$$
Because expansion occurs against a constant external pressure $P_{\text{ext}}$, the mechanical work done is:
$$W = P_{\text{ext}} (V_f – V_1)$$
Using the ideal gas equation $V_f = \frac{n R T_f}{P_{\text{ext}}}$ and $V_1 = \frac{n R T_1}{P_1}$:
$$W = P_{\text{ext}} \left( \frac{n R T_f}{P_{\text{ext}}} – \frac{n R T_1}{P_1} \right) = n R T_f – n R T_1 \left(\frac{P_{\text{ext}}}{P_1}\right)$$
Step 2: Equate Work Expressions to Find $T_f$.
$$n R T_f – n R T_1 \left(\frac{P_{\text{ext}}}{P_1}\right) = \frac{3}{2} n R (T_1 – T_f)$$
Dividing both sides by $nR$:
$$T_f – T_1 \left(\frac{P_{\text{ext}}}{P_1}\right) = \frac{3}{2} T_1 – \frac{3}{2} T_f \implies T_f \left(1 + \frac{3}{2}\right) = T_1 \left(\frac{3}{2} + \frac{P_{\text{ext}}}{P_1}\right)$$
$$\frac{5}{2} T_f = T_1 \left(\frac{3}{2} + \frac{P_{\text{ext}}}{P_1}\right)$$
Given $T_1 = 600\text{ K}$ and $\frac{P_{\text{ext}}}{P_1} = \frac{10^5}{8 \times 10^5} = \frac{1}{8} = 0.125$:
$$\frac{5}{2} T_f = 600 \left(1.5 + 0.125\right) = 600 \times 1.625 = 975\text{ K}$$
$$T_f = \frac{2}{5} \times 975 = \mathbf{390\text{ K}}$$
Step 3: Calculate Work Done $W$.
$$W = \frac{3}{2} n R (T_1 – T_f) = \frac{3}{2} (1)(8.314)(600 – 390) = \frac{3}{2} (8.314)(210) = 315 \times 8.314 = \mathbf{2618.9\text{ J} \approx 2.62\text{ kJ}}$$
Step 4: Comparison with Reversible Expansion.
If the expansion had been reversible, Poisson’s relation would apply:
$$T_{f, \text{rev}} = T_1 \left(\frac{P_f}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} = 600 \left(\frac{1}{8}\right)^{\frac{2/3}{5/3}} = 600 \left(\frac{1}{8}\right)^{2/5} = \frac{600}{8^{0.4}} \approx \frac{600}{2.2974} \approx \mathbf{261.2\text{ K}}$$
Notice that $T_{f, \text{irrev}} (390\text{ K}) > T_{f, \text{rev}} (261.2\text{ K})$.
Takeaway: In irreversible adiabatic expansion, the gas does less work against the low external pressure than in a reversible expansion ($W_{\text{irrev}} < W_{\text{rev}}$). Consequently, less internal energy is expended, leaving the final temperature significantly higher than the reversible prediction.
Example 4 (Limiting Cases: Reversible Carnot Cycle Adiabatic Branches)
Problem Statement: An ideal gas undergoes a closed Carnot cycle consisting of alternating isothermal and adiabatic stages: (1 $\to$ 2) isothermal expansion at $T_H$ from $V_1$ to $V_2$, (2 $\to$ 3) adiabatic expansion from $V_2$ to $V_3$ cooling the gas from $T_H$ to $T_C$, (3 $\to$ 4) isothermal compression at $T_C$ from $V_3$ to $V_4$, and (4 $\to$ 1) adiabatic compression from $V_4$ to $V_1$ heating the gas from $T_C$ to $T_H$. Prove using Poisson’s equations that $\frac{V_2}{V_1} = \frac{V_3}{V_4}$, and show that the net work of the two adiabatic branches cancels identically ($W_{2 \to 3} + W_{4 \to 1} = 0$).
Solution:
Step 1: Apply Poisson’s Relation to Adiabatic Branches.
Along the adiabatic expansion branch $2 \to 3$:
$$T_H V_2^{\gamma – 1} = T_C V_3^{\gamma – 1} \implies \frac{T_H}{T_C} = \left(\frac{V_3}{V_2}\right)^{\gamma – 1}$$
Along the adiabatic compression branch $4 \to 1$:
$$T_C V_4^{\gamma – 1} = T_H V_1^{\gamma – 1} \implies \frac{T_H}{T_C} = \left(\frac{V_4}{V_1}\right)^{\gamma – 1}$$
Equating the two temperature ratios:
$$\left(\frac{V_3}{V_2}\right)^{\gamma – 1} = \left(\frac{V_4}{V_1}\right)^{\gamma – 1} \implies \frac{V_3}{V_2} = \frac{V_4}{V_1} \implies \mathbf{\frac{V_2}{V_1} = \frac{V_3}{V_4}}$$
Step 2: Evaluate Work Done Along the Two Adiabatic Branches.
For the adiabatic expansion $2 \to 3$:
$$W_{2 \to 3} = \frac{n R (T_H – T_C)}{\gamma – 1}$$
For the adiabatic compression $4 \to 1$:
$$W_{4 \to 1} = \frac{n R (T_C – T_H)}{\gamma – 1} = -\frac{n R (T_H – T_C)}{\gamma – 1}$$
Summing the work of the two branches:
$$W_{2 \to 3} + W_{4 \to 1} = \frac{n R (T_H – T_C)}{\gamma – 1} – \frac{n R (T_H – T_C)}{\gamma – 1} = \mathbf{0}$$
Takeaway: In any closed cycle bounded between two isotherms, the net mechanical work contributed by the paired adiabatic branches is strictly zero; the net cycle work is contributed entirely by the differential isothermal heat transfers ($W_{\text{net}} = Q_H – Q_C$).
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. During a reversible adiabatic expansion of an ideal gas, the volume increases by a factor of 8 while the absolute temperature drops to half its initial value. The ratio of specific heats $\gamma = C_p / C_v$ for this gas is:
(A) $4/3$
(B) $5/3$
(C) $7/5$
(D) $3/2$
Q2. At a given pressure $P$, the ratio of the adiabatic bulk modulus ($B_S$) to the isothermal bulk modulus ($B_T$) for an ideal gas is:
(A) $1$
(B) $\gamma$
(C) $\frac{1}{\gamma}$
(D) $\gamma – 1$
Q3. One mole of an ideal monoatomic gas ($\gamma = 5/3$) undergoes an adiabatic expansion that lowers its temperature from $500\text{ K}$ to $300\text{ K}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$, the work done by the gas is:
(A) $2494.2\text{ J}$
(B) $1662.8\text{ J}$
(C) $4157.0\text{ J}$
(D) $831.4\text{ J}$
Q4. At the point of intersection of an isothermal curve and an adiabatic curve on a $P$-$V$ indicator diagram, the ratio of the magnitude of the slope of the adiabatic curve to that of the isothermal curve is:
(A) $\frac{1}{\gamma}$
(B) $\gamma$
(C) $\gamma – 1$
(D) $\frac{\gamma}{\gamma – 1}$
Q5. When an inflated bicycle tyre suddenly bursts, the air escaping from the valve feels cold. This observed cooling is primarily a consequence of:
(A) Isothermal expansion against atmospheric pressure
(B) Rapid adiabatic expansion where the gas performs work at the expense of internal energy
(C) Conduction of heat from the expanding gas into the rubber tube
(D) Chemical reaction between escaping air and ambient moisture
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements is/are correct for a reversible adiabatic expansion of an ideal gas?
(A) No heat is exchanged across the system boundary ($Q = 0$).
(B) The mechanical work performed equals the negative change in internal energy ($W = -\Delta U$).
(C) The molar heat capacity of the gas during the transformation is zero ($C = 0$).
(D) The product $P V^\gamma$ remains constant at all intermediate quasi-static states.
Q7. An ideal gas at an initial equilibrium state $(P_0, V_0)$ expands to a final volume $2V_0$ along two alternative reversible paths: Path 1 is isothermal, and Path 2 is adiabatic. Which of the following relationships is/are correct?
(A) The final pressure along the adiabatic path is less than that along the isothermal path ($P_{\text{adia}} < P_{\text{iso}}$).
(B) The work done along the isothermal path is greater than that along the adiabatic path ($W_{\text{iso}} > W_{\text{adia}}$).
(C) The final temperature along the adiabatic path is lower than that along the isothermal path ($T_{\text{adia}} < T_{\text{iso}}$).
(D) The change in internal energy along the adiabatic path is positive ($\Delta U_{\text{adia}} > 0$).
Q8. An insulated container is partitioned into two equal chambers: one chamber holds an ideal gas at temperature $T_0$, while the other is evacuated. When the partition is suddenly removed, the gas undergoes free expansion to fill the whole container. Which of the following statements is/are correct?
(A) The process is adiabatic because no heat is exchanged ($Q = 0$).
(B) The mechanical work done by the gas is zero ($W = 0$) because expansion occurs against zero opposing pressure.
(C) The final temperature of the ideal gas remains $T_0$ ($\Delta T = 0$).
(D) The relation $P_1 V_1^\gamma = P_2 V_2^\gamma$ holds accurately for this process.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. One mole of an ideal monoatomic gas ($\gamma = 5/3$) at an initial temperature $T_1 = 400\text{ K}$ expands reversibly and adiabatically until its volume increases to $8$ times its initial volume ($V_2 = 8V_1$). Calculate the drop in temperature $\Delta T = T_1 – T_2$ (in $\text{K}$).
Q10. Two moles of an ideal diatomic gas ($\gamma = 1.4$) undergo an adiabatic expansion doing $1247.1\text{ J}$ of mechanical work. Given universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$, find the magnitude of the temperature decrease $|\Delta T| = |T_2 – T_1|$ of the gas (in $\text{K}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $4/3$ | Poisson’s $T$-$V$ Relation ($T_1 V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1}$) |
| Q2 | (B) $\gamma$ | Ratio of Bulk Moduli ($B_S / B_T = \gamma$) |
| Q3 | (A) $2494.2\text{ J}$ | Adiabatic Work Done ($W = n R \Delta T / (1 – \gamma)$) |
| Q4 | (B) $\gamma$ | Comparison of $P$-$V$ Indicator Slopes |
| Q5 | (B) Rapid expansion cooling | Physical Manifestation of Adiabatic Expansion |
| Q6 | (A), (B), (C), (D) | Fundamental Properties of Adiabatic Process |
| Q7 | (A), (B), (C) | Isothermal vs. Adiabatic Expansion Profiles |
| Q8 | (A), (B), (C) | Free Expansion vs. Reversible Adiabatic Mechanics |
| Q9 | 300 | Temperature Drop in Monoatomic Expansion ($400\text{ K} \to 100\text{ K}$) |
| Q10 | 30 | Diatomic Temperature Drop from Work ($|\Delta T| = 30\text{ K}$) |
Complete Step-by-Step Solutions
Solution 1:
For a reversible adiabatic process:
$$T_1 V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1}$$
Given that $V_2 = 8V_1$ and $T_2 = T_1 / 2$:
$$T_1 V_1^{\gamma – 1} = \left(\frac{T_1}{2}\right) (8V_1)^{\gamma – 1} \implies 1 = \frac{1}{2} (8)^{\gamma – 1} \implies 8^{\gamma – 1} = 2$$
Expressing in base 2:
$$(2^3)^{\gamma – 1} = 2^1 \implies 3(\gamma – 1) = 1 \implies \gamma – 1 = \frac{1}{3} \implies \gamma = 1 + \frac{1}{3} = \frac{4}{3}$$
Correct Option: (A)
Solution 2:
The adiabatic bulk modulus is defined as:
$$B_S = -V \left(\frac{dP}{dV}\right)_{\text{adia}} = -V \left(-\gamma \frac{P}{V}\right) = \gamma P$$
The isothermal bulk modulus is:
$$B_T = -V \left(\frac{dP}{dV}\right)_{\text{iso}} = -V \left(-\frac{P}{V}\right) = P$$
Taking their ratio:
$$\frac{B_S}{B_T} = \frac{\gamma P}{P} = \gamma$$
Correct Option: (B)
Solution 3:
The work done by an ideal gas in an adiabatic expansion is:
$$W = \frac{n R (T_1 – T_2)}{\gamma – 1}$$
For a monoatomic gas, $\gamma = 5/3 \implies \gamma – 1 = 2/3$.
Substituting $n = 1\text{ mole}, T_1 = 500\text{ K}, T_2 = 300\text{ K}$, and $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
$$W = \frac{(1)(8.314)(500 – 300)}{2/3} = \frac{3}{2} \times (8.314)(200) = 300 \times 8.314 = 2494.2\text{ J}$$
Correct Option: (A)
Solution 4:
Differentiating Poisson’s relation $P V^\gamma = \text{const}$ gives the adiabatic slope:
$$\left(\frac{dP}{dV}\right)_{\text{adia}} = -\gamma \frac{P}{V}$$
Differentiating Boyle’s law $PV = \text{const}$ gives the isothermal slope:
$$\left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}$$
At their intersection point $(P, V)$, the ratio of slope magnitudes is:
$$\frac{\left|\left(\frac{dP}{dV}\right)_{\text{adia}}\right|}{\left|\left(\frac{dP}{dV}\right)_{\text{iso}}\right|} = \frac{\gamma \frac{P}{V}}{\frac{P}{V}} = \gamma$$
Correct Option: (B)
Solution 5:
When a tyre bursts suddenly, the compressed air rushes out into the atmosphere extremely rapidly. Because the duration is too brief for significant thermal exchange with the environment, the process is adiabatic ($Q = 0$). As the gas expands against the surrounding atmosphere, it performs positive work ($W > 0$) at the direct expense of its microscopic internal energy ($\Delta U = -W < 0$). For an ideal gas, a drop in internal energy causes a reduction in temperature ($\Delta T < 0$).
Correct Option: (B)
Solution 6:
Let us examine each statement:
(A) Correct: By definition of an adiabatic boundary, $Q = 0$.
(B) Correct: From First Law, $Q = \Delta U + W \implies 0 = \Delta U + W \implies W = -\Delta U$.
(C) Correct: Molar heat capacity is $C = \frac{Q}{n \Delta T} = \frac{0}{n \Delta T} = 0$.
(D) Correct: For a reversible quasi-static adiabatic transformation, $P V^\gamma = \text{constant}$ holds everywhere.
Correct Options: (A), (B), (C), (D)
Solution 7:
Starting from the same state $(P_0, V_0)$:
– Along the isothermal path: $P_{\text{iso}} = P_0 \left(\frac{V_0}{2V_0}\right) = \frac{P_0}{2}$, and $T_{\text{iso}} = T_0$.
– Along the adiabatic path: $P_{\text{adia}} = P_0 \left(\frac{V_0}{2V_0}\right)^\gamma = \frac{P_0}{2^\gamma}$. Since $\gamma > 1$, $2^\gamma > 2 \implies P_{\text{adia}} < P_{\text{iso}}$. Hence (A) is correct.
– Because the isothermal curve lies entirely above the adiabatic curve over the expansion interval $[V_0, 2V_0]$, the area under the isothermal curve is larger: $W_{\text{iso}} > W_{\text{adia}}$. Hence (B) is correct.
– The adiabatic temperature drops ($T_{\text{adia}} = T_0 / 2^{\gamma – 1} < T_0 = T_{\text{iso}}$). Hence (C) is correct.
– In adiabatic expansion, work is positive, so $\Delta U_{\text{adia}} = -W < 0$. Hence (D) is incorrect.
Correct Options: (A), (B), (C)
Solution 8:
(A) Correct: The container is thermally insulated, so no heat flows in or out ($Q = 0$).
(B) Correct: The gas expands into a vacuum where $P_{\text{ext}} = 0$, so $W = \int P_{\text{ext}}\, dV = 0$.
(C) Correct: By First Law, $\Delta U = Q – W = 0 – 0 = 0$. For an ideal gas, $U$ depends only on temperature, so $\Delta U = 0 \implies \Delta T = 0$ ($T_f = T_0$).
(D) Incorrect: Free expansion is an inherently irreversible, highly non-equilibrium process. Poisson’s relation $P V^\gamma = \text{const}$ is valid strictly for reversible quasi-static paths and fails completely here ($P_1 V_1 = P_2 V_2$ by Boyle’s law, but $P_1 V_1^\gamma \neq P_2 V_2^\gamma$).
Correct Options: (A), (B), (C)
Solution 9:
Using Poisson’s temperature-volume relation:
$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma – 1}$$
Given monoatomic gas $\gamma = 5/3 \implies \gamma – 1 = 2/3$, $T_1 = 400\text{ K}$, and $V_2 = 8V_1$:
$$T_2 = 400 \times \left(\frac{1}{8}\right)^{2/3} = 400 \times \left(\left(\frac{1}{2}\right)^3\right)^{2/3} = 400 \times \left(\frac{1}{2}\right)^2 = 400 \times \frac{1}{4} = 100\text{ K}$$
The drop in temperature is:
$$\Delta T = T_1 – T_2 = 400\text{ K} – 100\text{ K} = \mathbf{300\text{ K}}$$
Answer: 300
Solution 10:
The work done in an adiabatic expansion is:
$$W = \frac{n R (T_1 – T_2)}{\gamma – 1}$$
Given $n = 2\text{ moles}$, $\gamma = 1.4 \implies \gamma – 1 = 0.4$, $R = 8.314\text{ J/(mol}\cdot\text{K)}$, and $W = 1247.1\text{ J}$:
$$1247.1 = \frac{2 \times 8.314 \times (T_1 – T_2)}{0.4} = \frac{16.628}{0.4} (T_1 – T_2) = 41.57 \, (T_1 – T_2)$$
$$T_1 – T_2 = \frac{1247.1}{41.57} = \mathbf{30\text{ K}}$$
The decrease in temperature of the gas is $30\text{ K}$.
Answer: 30