Welcome to Lecture 100 of our comprehensive JEE Physics Master Series! This milestone lecture serves as the grand synthesis and advanced problem-solving culmination for Unit 8: Thermodynamics. Having mastered thermal equilibrium, the First Law, specific heat capacities, isothermal and adiabatic processes, cyclic paths, the Second Law, and thermal machines, we now tackle the highest-yield, multi-concept problem archetypes tested in JEE Main and JEE Advanced. We dissect non-standard indicator cycles, straight-line $P$-$V$ paths with heat inversion points, generalized polytropic regimes ($P V^m = \text{const}$), and composite thermodynamic systems with moving partitions.
Detailed Concept Card: Master Problem-Solving Toolkit for Thermodynamics
1. First Law Synthesis & State vs. Path Function Rules
Every complex thermodynamic calculation rests upon the rigorous separation of state functions from path-dependent quantities:
- Internal Energy ($U$): An absolute state function. For any ideal gas, internal energy depends solely on temperature: $$\Delta U = n C_v \Delta T = \frac{n R \Delta T}{\gamma – 1} = \frac{\Delta (PV)}{\gamma – 1} = \frac{P_f V_f – P_i V_i}{\gamma – 1}$$ For any closed thermodynamic cycle returning to the initial state, the net change in internal energy is strictly zero: $$\mathbf{\oint dU = \Delta U_{\text{cycle}} = 0}$$
- Work Done ($W$): A path-dependent quantity represented by the area under the process path on an indicator diagram ($W = \int P \, dV$):
- Isochoric ($V = \text{const}$): $W = 0$.
- Isobaric ($P = \text{const}$): $W = P(V_f – V_i) = n R \Delta T$.
- Isothermal ($T = \text{const}$): $W = n R T \ln\left(\frac{V_f}{V_i}\right) = P_i V_i \ln\left(\frac{P_i}{P_f}\right)$.
- Reversible Adiabatic ($P V^\gamma = \text{const}$): $W = \frac{P_i V_i – P_f V_f}{\gamma – 1} = \frac{n R (T_i – T_f)}{\gamma – 1} = -\Delta U$.
- Linear $P$-$V$ Path ($P = -aV + b$): $W = \frac{P_i + P_f}{2} (V_f – V_i)$ (area of a trapezoid).
- Heat Exchanged ($Q$): Determined by the First Law energy balance: $$\mathbf{Q = \Delta U + W}$$ Over a complete cycle: $Q_{\text{net}} = W_{\text{net}}$.
2. Generalized Polytropic Processes ($P V^m = \text{constant}$)
Any process obeying the relation $P V^m = \text{constant}$ (where $m$ is the polytropic exponent) is governed by unified master relations:
| Property | Mathematical Formula | Special Cases |
|---|---|---|
| $T$-$V$ & $P$-$T$ Relations | $T V^{m-1} = \text{const}, \quad T^m P^{1-m} = \text{const}$ | $m=0$ (Isobaric), $m=1$ (Isothermal), $m=\gamma$ (Adiabatic), $m\to\infty$ (Isochoric) |
| Work Done ($m \neq 1$) | $$W = \frac{P_i V_i – P_f V_f}{m – 1} = \frac{n R (T_i – T_f)}{m – 1}$$ | For $m = 1$, $W = nRT \ln(V_f / V_i)$ |
| Molar Heat Capacity ($C$) | $$C = C_v + \frac{R}{1 – m} = \frac{R}{\gamma – 1} + \frac{R}{1 – m}$$ | Negative heat capacity regime: $C < 0$ when $1 < m < \gamma$ |
| Adiabatic Bulk Modulus ($B$) | $$B = -V \frac{dP}{dV} = m P$$ | $B_{\text{iso}} = P$, $B_{\text{adia}} = \gamma P$ |
3. Straight-Line $P$-$V$ Paths: Temperature Extremum & Heat Inversion
A classic JEE Advanced theme involves straight-line indicator paths of the form $P(V) = -aV + b$ connecting state $(P_1, V_1)$ to $(P_2, V_2)$:
- Maximum Temperature ($T_{\text{max}}$):
Since $T(V) = \frac{P(V) V}{n R} = \frac{-a V^2 + b V}{n R}$, the temperature profile is a downward parabola with vertex at: $$V_{T_{\text{max}}} = \frac{b}{2a} = \frac{V_1 + V_2}{2} \quad (\text{if intercepts are symmetric}), \quad P_{T_{\text{max}}} = \frac{b}{2}$$ $$T_{\text{max}} = \frac{b^2}{4 a n R}$$ - Point of Zero Heat Exchange ($dQ = 0$ / Heat Inversion):
Applying the differential First Law: $$dQ = n C_v dT + P dV = \frac{1}{\gamma – 1}(P dV + V dP) + P dV = \frac{\gamma}{\gamma – 1} P dV + \frac{1}{\gamma – 1} V dP = 0$$ Dividing by $dV$ gives the tangency condition where the linear path touches an adiabatic curve: $$\mathbf{\frac{dP}{dV} = -\gamma \frac{P}{V}}$$ Substituting $P = -aV + b$ and $\frac{dP}{dV} = -a$: $$-a = -\gamma \frac{-aV + b}{V} \implies aV = -\gamma a V + \gamma b \implies (\gamma + 1) a V = \gamma b \implies \mathbf{V_{dQ=0} = \frac{\gamma}{\gamma + 1} \frac{b}{a}}$$ For volumes below $V_{dQ=0}$, $dQ > 0$ (heat absorbed); for volumes above $V_{dQ=0}$, $dQ < 0$ (heat rejected).
4. Cyclic Thermal Efficiency Algorithm
To determine the thermal efficiency $\eta$ of any arbitrary cycle without falling into sign traps:
- Compute net cycle work $W_{\text{net}}$ from the enclosed area ($W_{\text{net}} > 0$ for clockwise cycles).
- Identify all individual segments or sub-segments where heat is absorbed ($dQ > 0$).
- Sum the positive heat inputs: $Q_{\text{absorbed}} = \sum_{dQ > 0} Q$.
- Calculate efficiency: $$\mathbf{\eta = \frac{W_{\text{net}}}{Q_{\text{absorbed}}} = \frac{W_{\text{net}}}{W_{\text{net}} + |Q_{\text{rejected}}|}}$$
Four Graded Solved Examples
Example 1 (JEE Main Level: Complete Energy Accounting in a Three-Leg Cycle)
Problem Statement: One mole of an ideal monoatomic gas ($\gamma = 5/3, C_v = 1.5R$) is taken through a cyclic process $A \to B \to C \to A$ on a $P$-$V$ plane:
– Leg $A \to B$: Isochoric heating from $(P_0, V_0)$ to $(2P_0, V_0)$.
– Leg $B \to C$: Linear expansion along a straight line from $(2P_0, V_0)$ to $(P_0, 2V_0)$.
– Leg $C \to A$: Isobaric compression from $(P_0, 2V_0)$ back to initial state $(P_0, V_0)$.
Determine for each leg: the work done $W$, change in internal energy $\Delta U$, and heat exchanged $Q$. Find the net work output $W_{\text{net}}$ and the overall thermal efficiency $\eta$ of the cycle.
Solution:
Step 1: Analyze Leg $A \to B$ (Isochoric Heating, $V = V_0$).
$$W_{AB} = 0$$
$$\Delta U_{AB} = \frac{P_B V_B – P_A V_A}{\gamma – 1} = \frac{2P_0 V_0 – P_0 V_0}{5/3 – 1} = \frac{P_0 V_0}{2/3} = \mathbf{+1.5 \, P_0 V_0}$$
$$Q_{AB} = \Delta U_{AB} + W_{AB} = \mathbf{+1.5 \, P_0 V_0} \quad (\text{Heat absorbed})$$
Step 2: Analyze Leg $B \to C$ (Linear Expansion).
The area under the straight line connecting $(2P_0, V_0)$ and $(P_0, 2V_0)$ is a trapezoid:
$$W_{BC} = \frac{P_B + P_C}{2}(V_C – V_B) = \frac{2P_0 + P_0}{2}(2V_0 – V_0) = \frac{3P_0}{2}(V_0) = \mathbf{+1.5 \, P_0 V_0}$$
Change in internal energy:
$$\Delta U_{BC} = \frac{P_C V_C – P_B V_B}{\gamma – 1} = \frac{P_0 (2V_0) – (2P_0) V_0}{2/3} = \frac{2P_0 V_0 – 2P_0 V_0}{2/3} = \mathbf{0}$$
Since $\Delta U_{BC} = 0$, states $B$ and $C$ lie on the same isotherm ($T_B = T_C = \frac{2P_0 V_0}{R}$).
Heat exchanged along $B \to C$:
$$Q_{BC} = \Delta U_{BC} + W_{BC} = 0 + 1.5 \, P_0 V_0 = \mathbf{+1.5 \, P_0 V_0} \quad (\text{Net heat absorbed})$$
Step 3: Analyze Leg $C \to A$ (Isobaric Compression, $P = P_0$).
$$W_{CA} = P_0 (V_A – V_C) = P_0 (V_0 – 2V_0) = \mathbf{-1.0 \, P_0 V_0}$$
$$\Delta U_{CA} = \frac{P_A V_A – P_C V_C}{\gamma – 1} = \frac{P_0 V_0 – 2P_0 V_0}{2/3} = \frac{-P_0 V_0}{2/3} = \mathbf{-1.5 \, P_0 V_0}$$
$$Q_{CA} = \Delta U_{CA} + W_{CA} = -1.5 \, P_0 V_0 – 1.0 \, P_0 V_0 = \mathbf{-2.5 \, P_0 V_0} \quad (\text{Heat rejected})$$
Step 4: Net Cycle Work and Thermal Efficiency.
Net work done over the cycle:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 0 + 1.5 \, P_0 V_0 – 1.0 \, P_0 V_0 = \mathbf{+0.5 \, P_0 V_0}$$
Notice that this matches the triangular area enclosed: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(V_0)(P_0) = 0.5 \, P_0 V_0$.
Net internal energy change: $\Delta U_{\text{net}} = +1.5 + 0 – 1.5 = 0$ (verified!).
Total heat absorbed ($Q > 0$): $Q_{\text{in}} = Q_{AB} + Q_{BC} = 1.5 \, P_0 V_0 + 1.5 \, P_0 V_0 = 3.0 \, P_0 V_0$.
Thermal efficiency:
$$\mathbf{\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{0.5 \, P_0 V_0}{3.0 \, P_0 V_0} = \frac{1}{6} \approx 16.67\%}$$
Example 2 (JEE Main/Advanced: Polytropic Expansion with Negative Exponent $P = \alpha V^2$)
Problem Statement: One mole of an ideal rigid diatomic gas ($\gamma = 1.4 = 7/5, C_v = 2.5R$) expands from an initial state $(P_0, V_0, T_0)$ to a final volume $2V_0$ following the process equation $P = \alpha V^2$, where $\alpha = P_0 / V_0^2$ is a positive constant. Determine:
1. The polytropic index $m$ of the process.
2. The molar heat capacity $C$ of the gas during this process.
3. The final temperature $T_f$ in terms of $T_0$.
4. The work done $W$, change in internal energy $\Delta U$, and heat absorbed $Q$ in terms of $P_0 V_0$.
Solution:
Step 1: Identify Polytropic Index $m$.
Rearranging $P = \alpha V^2$ into standard polytropic form $P V^m = \text{const}$:
$$P V^{-2} = \alpha \implies \mathbf{m = -2}$$
Step 2: Calculate Molar Heat Capacity $C$.
Using the polytropic heat capacity formula:
$$C = C_v + \frac{R}{1 – m} = \frac{5}{2}R + \frac{R}{1 – (-2)} = \frac{5}{2}R + \frac{R}{3} = \frac{15 + 2}{6}R = \mathbf{\frac{17}{6}R}$$
Step 3: Final State and Temperature.
At $V_f = 2V_0$:
$$P_f = \alpha (2V_0)^2 = 4 \alpha V_0^2 = 4 P_0$$
From the ideal gas equation ($P V = n R T$ with $n = 1$):
$$T_f = \frac{P_f V_f}{R} = \frac{(4P_0)(2V_0)}{R} = 8 \frac{P_0 V_0}{R} = \mathbf{8 T_0}$$
Step 4: Energy Quantities $W, \Delta U$, and $Q$.
Work done:
$$W = \int_{V_0}^{2V_0} P \, dV = \int_{V_0}^{2V_0} \alpha V^2 dV = \frac{\alpha}{3} \left[(2V_0)^3 – V_0^3\right] = \frac{7}{3} \alpha V_0^3 = \mathbf{\frac{7}{3} P_0 V_0 \approx 2.333 \, P_0 V_0}$$
Change in internal energy:
$$\Delta U = n C_v (T_f – T_0) = \frac{5}{2} R (8T_0 – T_0) = \frac{5}{2} R (7T_0) = \frac{35}{2} R T_0 = \mathbf{\frac{35}{2} P_0 V_0 = 17.5 \, P_0 V_0}$$
Heat absorbed by First Law:
$$Q = \Delta U + W = \frac{35}{2} P_0 V_0 + \frac{7}{3} P_0 V_0 = \frac{105 + 14}{6} P_0 V_0 = \mathbf{\frac{119}{6} P_0 V_0 \approx 19.833 \, P_0 V_0}$$
Cross-check using molar heat capacity:
$$Q = n C \Delta T = (1)\left(\frac{17}{6}R\right)(7T_0) = \frac{119}{6} R T_0 = \frac{119}{6} P_0 V_0 \quad (\text{Exact match!})$$
Example 3 (JEE Advanced Level: Maximum Temperature & Heat Inversion along a Linear Path)
Problem Statement: One mole of an ideal monoatomic gas ($\gamma = 5/3, C_v = 1.5R$) expands from state $A(P_0, V_0)$ to state $B(P_0/3, 3V_0)$ along a straight-line path in the $P$-$V$ plane. Determine:
1. The maximum temperature $T_{\text{max}}$ attained by the gas during the expansion and the volume at which it occurs.
2. The volume $V^*$ at which the gas ceases to absorb heat and begins rejecting heat ($dQ = 0$).
3. The total heat absorbed $Q_{\text{absorbed}}$ by the gas during the heat-absorption phase ($V_0 \to V^*$).
Solution:
Step 1: Equation of the Path and Maximum Temperature.
Slope of the linear path:
$$M = \frac{P_B – P_A}{V_B – V_A} = \frac{P_0/3 – P_0}{3V_0 – V_0} = \frac{-2P_0 / 3}{2V_0} = -\frac{P_0}{3V_0}$$
Equation of the line:
$$P(V) = P_0 – \frac{P_0}{3V_0}(V – V_0) = \frac{4}{3}P_0 – \frac{P_0}{3V_0}V$$
Temperature as a function of volume ($n = 1$):
$$T(V) = \frac{P(V) V}{R} = \frac{1}{R}\left(\frac{4}{3}P_0 V – \frac{P_0}{3V_0}V^2\right)$$
To find the maximum temperature, set $\frac{dT}{dV} = 0$:
$$\frac{4}{3}P_0 – \frac{2P_0}{3V_0}V = 0 \implies \mathbf{V_{T_{\text{max}}} = 2V_0}$$
At $V = 2V_0$, pressure is $P = \frac{4}{3}P_0 – \frac{2}{3}P_0 = \frac{2}{3}P_0$.
$$T_{\text{max}} = \frac{(2/3 P_0)(2V_0)}{R} = \mathbf{\frac{4}{3} \frac{P_0 V_0}{R} = \frac{4}{3} T_0 \approx 1.333 \, T_0}$$
Step 2: Heat Inversion Point ($dQ = 0$).
From the differential First Law:
$$dQ = n C_v dT + P dV = \frac{3}{2}(P dV + V dP) + P dV = \frac{5}{2}P dV + \frac{3}{2}V dP = 0$$
Dividing by $dV$:
$$\frac{5}{2}P + \frac{3}{2}V \left(\frac{dP}{dV}\right) = 0$$
Substitute $P = \frac{4}{3}P_0 – \frac{P_0}{3V_0}V$ and $\frac{dP}{dV} = -\frac{P_0}{3V_0}$:
$$\frac{5}{2}\left(\frac{4}{3}P_0 – \frac{P_0}{3V_0}V\right) + \frac{3}{2}V\left(-\frac{P_0}{3V_0}\right) = 0$$
$$\frac{10}{3}P_0 – \frac{5}{6}\frac{P_0}{V_0}V – \frac{1}{2}\frac{P_0}{V_0}V = 0 \implies \frac{10}{3} = \left(\frac{5}{6} + \frac{3}{6}\right)\frac{V}{V_0} = \frac{4}{3}\frac{V}{V_0}$$
$$\mathbf{V^* = \frac{10}{4} V_0 = 2.5 \, V_0}$$
For $V \in [V_0, 2.5 V_0]$, $dQ > 0$ (heat absorbed). For $V \in (2.5 V_0, 3V_0]$, $dQ < 0$ (heat rejected).
Step 3: Total Heat Absorbed ($Q_{\text{absorbed}}$).
At $V^* = 2.5 V_0$:
$$P^* = \frac{4}{3}P_0 – \frac{2.5}{3}P_0 = \frac{1.5}{3}P_0 = 0.5 \, P_0$$
Internal energy change from $V_0$ to $2.5 V_0$:
$$\Delta U = \frac{P^* V^* – P_0 V_0}{\gamma – 1} = \frac{(0.5 P_0)(2.5 V_0) – P_0 V_0}{2/3} = \frac{1.25 P_0 V_0 – P_0 V_0}{2/3} = \frac{0.25}{2/3} P_0 V_0 = \mathbf{0.375 \, P_0 V_0}$$
Work done by the gas from $V_0$ to $2.5 V_0$ (trapezoid rule):
$$W = \frac{P_0 + 0.5 P_0}{2}(2.5 V_0 – V_0) = \frac{1.5 P_0}{2}(1.5 V_0) = 0.75 \times 1.5 \, P_0 V_0 = \mathbf{1.125 \, P_0 V_0}$$
Total heat absorbed during the absorption phase:
$$\mathbf{Q_{\text{absorbed}} = \Delta U + W = 0.375 \, P_0 V_0 + 1.125 \, P_0 V_0 = 1.50 \, P_0 V_0}$$
Example 4 (JEE Advanced Level: Composite Cylinder with Movable Insulating Piston)
Problem Statement: A rigid, thermally insulated horizontal cylinder of total volume $2V_0$ is divided into two equal compartments, $A$ and $B$, by a movable, frictionless, thermally insulating piston. Initially, each compartment contains $n = 1\text{ mole}$ of an ideal monoatomic gas ($\gamma = 5/3, C_v = 1.5R$) at pressure $P_0$ and temperature $T_0$. An electric heater inside compartment $A$ is energized, slowly transferring heat to gas $A$ until the pressure in both compartments rises to $P_f = 32 P_0$. In terms of $P_0, V_0, T_0$, determine:
1. The final volume $V_B$ and final temperature $T_B$ of gas $B$.
2. The final volume $V_A$ and final temperature $T_A$ of gas $A$.
3. The total electrical heat $Q_A$ supplied by the heater.
Solution:
Step 1: Analyze Compartment $B$ (Reversible Adiabatic Compression).
Because the cylinder and piston are thermally insulating, gas $B$ receives zero heat ($Q_B = 0$). Since the piston moves quasi-statically, compartment $B$ undergoes a reversible adiabatic compression from $(P_0, V_0, T_0)$ to pressure $P_f = 32 P_0$:
$$P_0 V_0^\gamma = P_f V_B^\gamma \implies V_B = V_0 \left(\frac{P_0}{P_f}\right)^{1/\gamma} = V_0 \left(\frac{1}{32}\right)^{3/5} = V_0 \left(\frac{1}{2^5}\right)^{3/5} = \frac{V_0}{2^3} = \mathbf{\frac{V_0}{8} = 0.125 \, V_0}$$
Final temperature of gas $B$:
$$T_B = T_0 \left(\frac{P_f V_B}{P_0 V_0}\right) = T_0 \left(32 \times \frac{1}{8}\right) = \mathbf{4 T_0}$$
Work done ON gas $B$:
$$W_{\text{on B}} = \frac{P_f V_B – P_0 V_0}{\gamma – 1} = \frac{32 P_0 (V_0 / 8) – P_0 V_0}{5/3 – 1} = \frac{4 P_0 V_0 – P_0 V_0}{2/3} = \frac{3 P_0 V_0}{2/3} = \mathbf{4.5 \, P_0 V_0}$$
Work done BY gas $A$ on the piston is equal and opposite: $W_A = W_{\text{on B}} = +4.5 \, P_0 V_0$.
Step 2: Analyze Compartment $A$.
Because the total volume of the rigid cylinder is $2V_0$:
$$V_A = 2V_0 – V_B = 2V_0 – \frac{V_0}{8} = \mathbf{\frac{15}{8} V_0 = 1.875 \, V_0}$$
Final temperature of gas $A$:
$$T_A = T_0 \left(\frac{P_f V_A}{P_0 V_0}\right) = T_0 \left(32 \times \frac{15}{8}\right) = 32 \times 1.875 \, T_0 = \mathbf{60 T_0}$$
Step 3: Total Heat Supplied by Heater ($Q_A$).
Change in internal energy of gas $A$:
$$\Delta U_A = n C_v (T_A – T_0) = \frac{3}{2} R (60T_0 – T_0) = \frac{3}{2} R (59T_0) = 88.5 \, R T_0 = \mathbf{88.5 \, P_0 V_0}$$
Applying the First Law to gas $A$:
$$Q_A = \Delta U_A + W_A = 88.5 \, P_0 V_0 + 4.5 \, P_0 V_0 = \mathbf{93.0 \, P_0 V_0}$$
Alternative Global Energy Balance Check:
Considering both compartments $(A + B)$ together as a single composite isolated system with rigid walls ($W_{\text{external}} = 0$):
$$Q_A = \Delta U_A + \Delta U_B = 88.5 \, P_0 V_0 + \frac{3}{2} R (4T_0 – T_0) = 88.5 \, P_0 V_0 + 4.5 \, P_0 V_0 = \mathbf{93.0 \, P_0 V_0} \quad (\text{Perfect match!})$$
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. An ideal monoatomic gas ($\gamma = 5/3$) undergoes a cyclic process represented by a circle of radius $R$ on a $P$-$V$ diagram where pressure is plotted on the vertical axis (with scale $1\text{ unit} = P_0$) and volume is plotted on the horizontal axis (with scale $1\text{ unit} = V_0$). If the radius of the circle is $R = 2$ in these normalized units and the cycle is traversed clockwise, the net work done by the gas in one complete cycle is:
(A) $2\pi P_0 V_0$
(B) $4\pi P_0 V_0$
(C) $8\pi P_0 V_0$
(D) $\pi P_0 V_0$
Q2. One mole of an ideal gas with adiabatic index $\gamma$ undergoes a polytropic process according to $P V^2 = \text{constant}$. The molar heat capacity $C$ of the gas during this process is:
(A) $C_v + R$
(B) $C_v – R$
(C) $C_v + 2R$
(D) $C_v – 2R$
Q3. A sample of an ideal gas expands from $(2P_0, V_0)$ to $(P_0, 2V_0)$ along a straight line in the $P$-$V$ plane. The maximum temperature attained by the gas during this process is:
(A) $\frac{9}{8} \frac{P_0 V_0}{n R}$
(B) $\frac{25}{16} \frac{P_0 V_0}{n R}$
(C) $\frac{9}{4} \frac{P_0 V_0}{n R}$
(D) $2 \frac{P_0 V_0}{n R}$
Q4. An ideal monoatomic gas ($\gamma = 5/3$) expands along the path $P = k V$ (where $k$ is a positive constant) from volume $V_0$ to $2V_0$. The ratio of the heat absorbed to the work done by the gas ($\frac{Q}{W}$) is:
(A) $1$
(B) $2$
(C) $4$
(D) $5/2$
Q5. Two moles of an ideal gas are taken through a reversible cycle consisting of an isothermal expansion at $T_1$, an isochoric cooling to $T_2$, an isothermal compression at $T_2$, and an isochoric heating back to $T_1$. If the volume doubles during the isothermal expansion, the net work done in one cycle is:
(A) $2 R (T_1 – T_2) \ln 2$
(B) $R (T_1 – T_2) \ln 2$
(C) $4 R (T_1 – T_2) \ln 2$
(D) Zero
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. One mole of an ideal monoatomic gas expands along a straight line path from $A(P_0, V_0)$ to $B(P_0/2, 2V_0)$. Which of the following statements is/are correct?
(A) The maximum temperature along the path occurs at $V = 1.5 V_0$.
(B) The maximum temperature reached is $T_{\text{max}} = 1.125 \frac{P_0 V_0}{R}$.
(C) The work done by the gas during the expansion is $0.75 P_0 V_0$.
(D) The gas absorbs heat throughout the entire path from $A$ to $B$ without any heat rejection.
Q7. A horizontal cylinder closed at both ends contains an ideal gas divided into two equal compartments by a movable, frictionless, thermally insulating piston. Initially, the pressure and temperature on both sides are $P_0$ and $T_0$. The gas on the left is heated slowly by an electric coil until its volume increases to $1.5$ times its initial value. Which of the following statements is/are correct?
(A) The process undergone by the gas on the right compartment is reversible adiabatic compression.
(B) The final volume of the gas on the right compartment is $0.5$ times its initial volume.
(C) The final pressure in both compartments is equal.
(D) The temperature of the gas on the right compartment remains equal to $T_0$.
Q8. Which of the following statements is/are correct regarding polytropic processes $P V^m = \text{constant}$ for an ideal gas with adiabatic index $\gamma$?
(A) If $m = \gamma$, the molar heat capacity is zero ($C = 0$).
(B) If $1 < m < \gamma$, the molar heat capacity is negative ($C < 0$).
(C) During an expansion with $1 < m < \gamma$, heat is absorbed by the gas while its temperature decreases.
(D) If $m < 1$, the gas cools during an adiabatic-like compression.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. One mole of an ideal monoatomic gas ($\gamma = 5/3$) initially at temperature $T_0 = 300\text{ K}$ expands from state $A(P_0, V_0)$ to state $B(P_0/3, 3V_0)$ along the straight-line path $P(V) = \frac{4}{3}P_0 – \frac{P_0}{3V_0}V$. Determine the maximum temperature $T_{\text{max}}$ attained by the gas during this path (in $\text{K}$).
Q10. An insulated horizontal cylinder of total volume $20\text{ L}$ is divided into two equal compartments of volume $V_0 = 10\text{ L}$ each by a frictionless insulating piston. Each compartment contains one mole of an ideal monoatomic gas at $P_0 = 100\text{ kPa}$ and $T_0 = 300\text{ K}$. An electric heater slowly heats the gas in compartment $A$ until the pressure in compartment $B$ reaches $P_f = 3200\text{ kPa} = 32 P_0$. Find the total heat energy supplied by the heater to compartment $A$ (in $\text{kJ}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (B) $4\pi P_0 V_0$ | Circular Cycle Area on $P$-$V$ Indicator Diagram |
| Q2 | (B) $C_v – R$ | Molar Heat Capacity for Polytropic $PV^2 = \text{const}$ |
| Q3 | (C) $\frac{9}{4} \frac{P_0 V_0}{n R}$ | Maximum Temperature on Linear Path |
| Q4 | (C) $4$ | Heat-to-Work Ratio for $P \propto V$ Path |
| Q5 | (A) $2 R (T_1 – T_2) \ln 2$ | Stirling-Type Rectangular Cycle Work |
| Q6 | (A), (B), (C) | Straight-Line Path Analysis & Inversion |
| Q7 | (A), (B), (C) | Movable Insulating Piston Dynamics |
| Q8 | (A), (B), (C) | Polytropic Regime & Negative Heat Capacity |
| Q9 | 400 | Maximum Temperature along Straight Line ($4/3 \times 300\text{ K}$) |
| Q10 | 93 | Composite Cylinder Heater Energy ($93 P_0 V_0 = 93\text{ kJ}$) |
Complete Step-by-Step Solutions
Solution 1:
The area enclosed by an ellipse (or circle) on a $P$-$V$ indicator diagram is given by:
$$W = \pi \times a_P \times b_V$$
where $a_P$ is the semi-axis along pressure and $b_V$ is the semi-axis along volume. In normalized units, $a_P = R \cdot P_0 = 2 P_0$ and $b_V = R \cdot V_0 = 2 V_0$.
Therefore:
$$W_{\text{net}} = \pi (2 P_0)(2 V_0) = \mathbf{4\pi P_0 V_0}$$
Since the cycle is traversed clockwise, work done is positive.
Correct Option: (B)
Solution 2:
For a polytropic process $P V^m = \text{constant}$, the molar heat capacity is:
$$C = C_v + \frac{R}{1 – m}$$
Here $m = 2$:
$$C = C_v + \frac{R}{1 – 2} = C_v – R$$
Correct Option: (B)
Solution 3:
Slope of the line passing through $(2P_0, V_0)$ and $(P_0, 2V_0)$:
$$M = \frac{P_0 – 2P_0}{2V_0 – V_0} = -\frac{P_0}{V_0}$$
Equation of the line:
$$P(V) – 2P_0 = -\frac{P_0}{V_0}(V – V_0) \implies P(V) = 3P_0 – \frac{P_0}{V_0}V$$
Temperature profile:
$$T(V) = \frac{P(V) V}{n R} = \frac{3P_0 V – \frac{P_0}{V_0}V^2}{n R}$$
Setting $\frac{dT}{dV} = 0$:
$$3P_0 – 2\frac{P_0}{V_0}V = 0 \implies V = \frac{3}{2}V_0 = 1.5 V_0$$
At this volume, pressure is $P = 3P_0 – 1.5 P_0 = 1.5 P_0 = \frac{3}{2}P_0$.
Maximum temperature:
$$T_{\text{max}} = \frac{(1.5 P_0)(1.5 V_0)}{n R} = \mathbf{\frac{9}{4} \frac{P_0 V_0}{n R} = 2.25 \frac{P_0 V_0}{n R}}$$
Correct Option: (C)
Solution 4:
For $P = k V$, this is a polytropic process $P V^{-1} = \text{const}$ with $m = -1$.
Work done from $V_0$ to $2V_0$:
$$W = \int_{V_0}^{2V_0} k V \, dV = \frac{k}{2}[(2V_0)^2 – V_0^2] = \frac{3}{2} k V_0^2$$
Internal energy change for a monoatomic gas ($\gamma = 5/3, C_v = 1.5R$):
$$\Delta U = \frac{P_f V_f – P_i V_i}{\gamma – 1} = \frac{(2kV_0)(2V_0) – (kV_0)(V_0)}{2/3} = \frac{4kV_0^2 – kV_0^2}{2/3} = \frac{3kV_0^2}{2/3} = \frac{9}{2} k V_0^2$$
Heat absorbed:
$$Q = \Delta U + W = \frac{9}{2} k V_0^2 + \frac{3}{2} k V_0^2 = 6 k V_0^2$$
Taking the ratio:
$$\frac{Q}{W} = \frac{6 k V_0^2}{\frac{3}{2} k V_0^2} = \frac{6}{1.5} = \mathbf{4}$$
Correct Option: (C)
Solution 5:
The cycle consists of:
1. Isothermal expansion at $T_1$ from $V_0$ to $2V_0$: $W_1 = n R T_1 \ln(2V_0 / V_0) = 2 R T_1 \ln 2$.
2. Isochoric cooling from $T_1$ to $T_2$ at $2V_0$: $W_2 = 0$.
3. Isothermal compression at $T_2$ from $2V_0$ to $V_0$: $W_3 = n R T_2 \ln(V_0 / 2V_0) = -2 R T_2 \ln 2$.
4. Isochoric heating from $T_2$ to $T_1$ at $V_0$: $W_4 = 0$.
Net work done:
$$W_{\text{net}} = W_1 + W_2 + W_3 + W_4 = 2 R T_1 \ln 2 + 0 – 2 R T_2 \ln 2 + 0 = \mathbf{2 R (T_1 – T_2) \ln 2}$$
Correct Option: (A)
Solution 6:
Line connecting $A(P_0, V_0)$ and $B(P_0/2, 2V_0)$:
$$P(V) = P_0 – \frac{P_0/2}{V_0}(V – V_0) = 1.5 P_0 – 0.5 \frac{P_0}{V_0}V$$
(A) Correct: $T(V) \propto 1.5 P_0 V – 0.5 \frac{P_0}{V_0} V^2$. Maximum temperature occurs at $V = \frac{1.5}{2(0.5)} V_0 = 1.5 V_0$.
(B) Correct: At $V = 1.5 V_0$, $P = 1.5 P_0 – 0.75 P_0 = 0.75 P_0$. $T_{\text{max}} = \frac{(0.75 P_0)(1.5 V_0)}{R} = 1.125 \frac{P_0 V_0}{R}$.
(C) Correct: $W = \frac{P_0 + 0.5 P_0}{2}(2V_0 – V_0) = 0.75 P_0 V_0$.
(D) Incorrect: At $V > \frac{\gamma}{\gamma + 1} \frac{b}{a} = \frac{5/3}{8/3}\frac{1.5}{0.5} V_0 = \frac{5}{8}(3 V_0) = 1.875 V_0$, $dQ$ becomes negative (heat is rejected between $1.875 V_0$ and $2V_0$).
Correct Options: (A), (B), (C)
Solution 7:
(A) Correct: The cylinder and piston are insulating, and the piston moves slowly, so the right side undergoes quasi-static adiabatic compression.
(B) Correct: Total volume is $2V_0$. Since the left side expands to $1.5 V_0$, the right side occupies $2V_0 – 1.5 V_0 = 0.5 V_0$.
(C) Correct: The piston is movable and frictionless, so mechanical equilibrium ensures equal pressure on both sides at all times.
(D) Incorrect: Adiabatic compression heats the gas on the right ($T_f = T_0 (V_0 / 0.5 V_0)^{\gamma – 1} = T_0 \cdot 2^{2/3} > T_0$).
Correct Options: (A), (B), (C)
Solution 8:
(A) Correct: For $m = \gamma$, $C = C_v – \frac{R}{\gamma – 1} = C_v – C_v = 0$.
(B) Correct: When $1 < m < \gamma$, $C = \frac{(m - \gamma)R}{(\gamma - 1)(m - 1)} < 0$.
(C) Correct: For $C < 0$, expansion cools the gas ($\Delta T < 0$), so $Q = n C \Delta T > 0$ (heat is absorbed).
(D) Incorrect: For $m < 1$, compression ($dV < 0$) causes temperature to change according to $T V^{m-1} = \text{const}$. Since $m - 1 < 0$, $T \propto V^{1 - m}$ which increases as $V$ increases, so compression reduces temperature only if adiabatic, but here $m < 1$ is not an adiabatic path.
Correct Options: (A), (B), (C)
Solution 9:
From Example 3 derivation:
$$T(V) = \frac{1}{R}\left(\frac{4}{3}P_0 V – \frac{P_0}{3V_0}V^2\right)$$
Setting $\frac{dT}{dV} = 0 \implies V = 2V_0$, which gives $P = \frac{2}{3}P_0$.
$$T_{\text{max}} = \frac{(2/3 P_0)(2V_0)}{R} = \frac{4}{3} \frac{P_0 V_0}{R} = \frac{4}{3} T_0$$
Given $T_0 = 300\text{ K}$:
$$T_{\text{max}} = \frac{4}{3} \times 300\text{ K} = \mathbf{400\text{ K}}$$
Answer: 400
Solution 10:
From Example 4 derivation, the total heat supplied to compartment $A$ is:
$$Q_A = 93.0 \, P_0 V_0$$
Given $P_0 = 100\text{ kPa} = 10^5\text{ Pa}$ and $V_0 = 10\text{ L} = 10^{-2}\text{ m}^3$:
$$P_0 V_0 = 10^5\text{ Pa} \times 10^{-2}\text{ m}^3 = 1000\text{ J} = 1\text{ kJ}$$
Therefore:
$$Q_A = 93.0 \times 1\text{ kJ} = \mathbf{93\text{ kJ}}$$
Answer: 93