Welcome to Lecture 94 of our comprehensive JEE Physics Master Series. In this lecture, we explore the physics of the Isothermal Process. We analyze thermodynamic transformations occurring at strictly constant temperature, derive the quasi-static work done formula from first principles, deconstruct the hyperbolic $P$-$V$ indicator diagram, compute the isothermal bulk modulus, establish the exact conversion between boundary work and reservoir heat transfer ($Q = W$), and resolve high-yield JEE Main and JEE Advanced problems.
Detailed Concept Card: Isothermal Process
An isothermal process is a thermodynamic state transformation in which the temperature of the system remains strictly constant throughout the entire duration of the change:
$$T = \text{constant} \implies dT = 0, \quad \Delta T = 0$$
1. Equation of State: Boyle’s Law
For an ideal gas obeying the equation of state $PV = nRT$, if the temperature $T$ and the quantity of matter $n$ remain invariant, the product of pressure and volume is constant:
$$PV = \text{constant} \implies P_1 V_1 = P_2 V_2$$
This is the classical formulation of Boyle’s Law. If the gas expands ($V_2 > V_1$), its pressure decreases proportionally ($P_2 < P_1$). Conversely, if the gas is compressed ($V_2 < V_1$), its pressure increases.
2. Essential Physical Conditions for an Isothermal Process
To realize an isothermal process in practice, two boundary conditions must be satisfied simultaneously:
- Perfectly Conducting Walls (Diathermic Boundary): The cylinder base and walls enclosing the gas must have infinite thermal conductivity so that heat transfers instantaneously between the gas and the external thermal bath without establishing an internal temperature gradient.
- Infinitely Slow Motion (Quasi-Static Execution): The piston must be displaced with negligible acceleration ($v \to 0$). If the gas expands, it tends to cool slightly; an infinitely slow displacement guarantees that sufficient time is available for thermal energy to flow in from the heat bath to replenish the thermal deficit, maintaining thermal equilibrium ($T_{\text{gas}} = T_{\text{bath}}$) at every intermediate state.
3. Internal Energy and the First Law of Thermodynamics
According to Joule’s Law, the internal energy $U$ of an ideal gas depends solely on its absolute temperature ($U = U(T)$), because intermolecular potential energy is identically zero. Therefore, for any isothermal process undergone by an ideal gas:
$$\Delta U = n C_v \Delta T = n C_v (0) = 0$$
Applying the First Law of Thermodynamics ($Q = \Delta U + W$):
$$Q = 0 + W \implies Q = W$$
- Isothermal Expansion ($V_2 > V_1$): The gas performs positive mechanical work on the surroundings ($W > 0$). An equivalent quantity of thermal energy is absorbed from the reservoir ($Q > 0$). The input heat is converted entirely into boundary displacement work without increasing the microscopic kinetic energy of the molecules.
- Isothermal Compression ($V_2 < V_1$): External agents perform work on the gas ($W < 0$). To prevent the temperature from rising, an equal quantity of heat is rejected into the reservoir ($Q < 0$).
Crucial Real-Gas Trap: For a real gas (e.g., Van der Waals gas), internal energy depends on both temperature and volume due to intermolecular attractive forces: $U = U(T, V)$. During isothermal expansion of a real gas, molecules move farther apart against mutual attraction, increasing potential energy. Hence, for a real gas, $\Delta U > 0$, meaning $Q > W$. Do not apply $\Delta U = 0$ to real gases!
4. Mathematical Derivation of Work Done
Consider $n$ moles of an ideal gas maintained at constant temperature $T$ expanding quasi-statically from initial state $(P_1, V_1)$ to final state $(P_2, V_2)$. At any intermediate volume $V$, the pressure is given by $P = \frac{nRT}{V}$.
The total work done by the gas is:
$$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \int_{V_1}^{V_2} \frac{dV}{V}$$
Integrating gives the fundamental work equation:
$$\mathbf{W = nRT \ln\left(\frac{V_2}{V_1}\right) = 2.303 \, nRT \log_{10}\left(\frac{V_2}{V_1}\right)}$$
Using Boyle’s Law ($P_1 V_1 = P_2 V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$), work done can be expressed in terms of pressures:
$$\mathbf{W = nRT \ln\left(\frac{P_1}{P_2}\right) = 2.303 \, nRT \log_{10}\left(\frac{P_1}{P_2}\right)}$$
In terms of initial state coordinates $(P_1, V_1)$ since $nRT = P_1 V_1 = P_2 V_2$:
$$W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) = P_1 V_1 \ln\left(\frac{P_1}{P_2}\right)$$
5. Indicator Diagram & Slope of Isothermal Curves
On a $P$-$V$ indicator diagram, the curve of an isothermal process is an equilateral (rectangular) hyperbola asymptotic to both the pressure and volume axes.
To determine the slope at any state point $(P, V)$, differentiate $PV = \text{constant}$:
$$d(PV) = P \, dV + V \, dP = 0 \implies V \, dP = -P \, dV$$
$$\mathbf{\left(\frac{dP}{dV}\right)_{T} = -\frac{P}{V}}$$
- The slope is negative everywhere, indicating that pressure decreases monotonically as volume increases.
- The magnitude of the slope $|dP/dV| = P/V = \frac{nRT}{V^2}$ is directly proportional to temperature $T$ and inversely proportional to $V^2$.
- Family of Isotherms: For a given gas, isotherms corresponding to higher temperatures ($T_2 > T_1$) lie entirely above and to the right of isotherms corresponding to lower temperatures. Two isotherms never intersect.
6. Isothermal Bulk Modulus & Molar Heat Capacity
The Bulk Modulus of Elasticity ($B$) measures a fluid’s resistance to uniform compression and is defined as $B = -V \frac{dP}{dV}$.
Substituting the isothermal slope $\left(\frac{dP}{dV}\right)_T = -\frac{P}{V}$:
$$\mathbf{B_T = -V \left(-\frac{P}{V}\right) = P}$$
Universal Law: The isothermal bulk modulus of an ideal gas is numerically equal to its instantaneous pressure $P$.
Molar Heat Capacity ($C$): By definition, $C = \frac{Q}{n \Delta T}$. In an isothermal process, heat is exchanged while $\Delta T = 0$:
$$C_{\text{iso}} = \frac{Q}{n (0)} = \pm \infty$$
The molar heat capacity of an ideal gas in an isothermal process is infinite.
Four Graded Solved Examples
Example 1 (JEE Main Level: Work and Heat in Isothermal Expansion)
Problem Statement: Two moles of an ideal gas are maintained at constant temperature $T = 300\text{ K}$ and expand reversibly from an initial volume $V_1 = 5.0\text{ L}$ to a final volume $V_2 = 20.0\text{ L}$. Take universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $\ln 2 = 0.693$. Determine:
- The change in internal energy $\Delta U$ of the gas.
- The work done $W$ by the expanding gas.
- The total heat energy $Q$ absorbed from the thermal reservoir.
Solution:
Step 1: Calculate $\Delta U$.
Because the process is isothermal and the gas is ideal, internal energy depends solely on temperature:
$$\Delta U = n C_v \Delta T = 2 \times C_v \times 0 = 0\text{ J}$$
Step 2: Calculate Work Done $W$.
The volume expansion ratio is:
$$\frac{V_2}{V_1} = \frac{20.0\text{ L}}{5.0\text{ L}} = 4 = 2^2$$
$$\ln\left(\frac{V_2}{V_1}\right) = \ln(2^2) = 2 \ln 2 = 2 \times 0.693 = 1.386$$
Applying the isothermal work formula:
$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = (2)(8.314)(300)(1.386)$$
$$nRT = 2 \times 8.314 \times 300 = 4988.4\text{ J}$$
$$W = 4988.4 \times 1.386 = 6913.9\text{ J} \approx 6.91\text{ kJ}$$
Step 3: Calculate Heat Absorbed $Q$.
From the First Law of Thermodynamics:
$$Q = \Delta U + W = 0 + 6913.9\text{ J} = +6913.9\text{ J} \approx +6.91\text{ kJ}$$
The positive sign confirms that $6.91\text{ kJ}$ of thermal energy is absorbed from the reservoir.
Takeaway: In reversible isothermal expansion of an ideal gas, $100\%$ of the heat drawn from the reservoir is converted into external mechanical work, keeping internal energy invariant.
Example 2 (Algebraic Comparison: Isothermal vs. Isobaric Compression)
Problem Statement: One mole of an ideal gas at state $(P_0, V_0)$ is compressed to half its initial volume ($V_f = V_0 / 2$) along two distinct quasi-static paths: (Path A) an isothermal path, and (Path B) an isobaric path. Derive analytical expressions for the work done on the gas in each process, and determine which path requires greater compressive work.
Solution:
Step 1: Work in Isothermal Compression (Path A).
The work done by the gas is:
$$W_A = nRT_0 \ln\left(\frac{V_f}{V_0}\right) = P_0 V_0 \ln\left(\frac{V_0 / 2}{V_0}\right) = P_0 V_0 \ln\left(\frac{1}{2}\right) = -P_0 V_0 \ln 2$$
The work done on the gas is the magnitude:
$$W_{\text{on, A}} = -W_A = P_0 V_0 \ln 2 \approx 0.693 \, P_0 V_0$$
Step 2: Work in Isobaric Compression (Path B).
Pressure remains constant at $P_0$ during the compression:
$$W_B = P_0 (V_f – V_0) = P_0 \left(\frac{V_0}{2} – V_0\right) = -\frac{1}{2} P_0 V_0$$
The work done on the gas is:
$$W_{\text{on, B}} = -W_B = 0.500 \, P_0 V_0$$
Step 3: Comparison and Physical Rationale.
Comparing the two magnitudes:
$$W_{\text{on, A}} = 0.693 \, P_0 V_0 > W_{\text{on, B}} = 0.500 \, P_0 V_0$$
$$\frac{W_{\text{on, A}}}{W_{\text{on, B}}} = \frac{\ln 2}{0.5} = 2 \ln 2 \approx 1.386$$
During isothermal compression, pressure continuously climbs from $P_0$ to $2P_0$ as volume decreases ($P = P_0 V_0 / V$). In contrast, during isobaric compression, pressure remains fixed at $P_0$. Because the isothermal $P$-$V$ curve lies entirely above the isobaric horizontal line on a compression diagram, the area under the isothermal curve is larger.
Takeaway: Isothermal compression requires approximately $38.6\%$ more work than isobaric compression between the same volume limits because the resisting pressure escalates continuously.
Example 3 (JEE Advanced Level: Diathermic Partition Equilibrium & Work Exchange)
Problem Statement: A rigid, thermally insulated horizontal cylinder of total volume $2V_0$ is divided into two chambers $A$ and $B$ by a frictionless, movable diathermic (thermally conducting) piston. Initially, the piston is held fixed such that each chamber has volume $V_0$. Chamber $A$ contains an ideal gas at pressure $P_0$, and Chamber $B$ contains an ideal gas at pressure $2P_0$. The entire cylinder is kept in a large constant-temperature water bath at temperature $T_0$. The piston is released and allowed to move quasi-statically until mechanical equilibrium is reached. Determine:
- The final equilibrium pressure $P_f$ and final volumes $V_A$ and $V_B$.
- The work done $W_A$ by gas in chamber $A$ and $W_B$ by gas in chamber $B$.
- The net heat $Q_{\text{net}}$ absorbed by the system from the water bath.
Solution:
Step 1: Determine Equilibrium State Variables.
Because the dividing piston is diathermic and the system is in contact with the water bath at $T_0$, both chambers remain at temperature $T_0$ throughout the process.
Initial number of moles:
$$n_A = \frac{P_0 V_0}{R T_0}, \quad n_B = \frac{(2P_0) V_0}{R T_0} = \frac{2 P_0 V_0}{R T_0}$$
Total moles: $n_{\text{total}} = n_A + n_B = \frac{3 P_0 V_0}{R T_0}$.
Total volume is fixed: $V_A + V_B = 2V_0$.
At mechanical equilibrium, pressures on both sides of the piston must equalize: $P_A = P_B = P_f$.
$$P_f = \frac{n_{\text{total}} R T_0}{V_{\text{total}}} = \frac{3 P_0 V_0}{2 V_0} = \frac{3}{2} P_0 = 1.5 P_0$$
The final volumes are:
$$V_A = \frac{n_A R T_0}{P_f} = \frac{P_0 V_0}{\frac{3}{2} P_0} = \frac{2}{3} V_0$$
$$V_B = \frac{n_B R T_0}{P_f} = \frac{2 P_0 V_0}{\frac{3}{2} P_0} = \frac{4}{3} V_0$$
Check: $V_A + V_B = \frac{2}{3} V_0 + \frac{4}{3} V_0 = 2V_0$.
Step 2: Calculate Work Done by Each Gas.
Gas $A$ undergoes isothermal compression from $V_0$ to $\frac{2}{3} V_0$:
$$W_A = n_A R T_0 \ln\left(\frac{V_A}{V_0}\right) = P_0 V_0 \ln\left(\frac{2}{3}\right) = -P_0 V_0 \ln\left(\frac{3}{2}\right) < 0$$
Gas $B$ undergoes isothermal expansion from $V_0$ to $\frac{4}{3} V_0$:
$$W_B = n_B R T_0 \ln\left(\frac{V_B}{V_0}\right) = 2 P_0 V_0 \ln\left(\frac{4}{3}\right) > 0$$
Step 3: Calculate Net Heat Absorbed $Q_{\text{net}}$.
Since temperature is constant everywhere, $\Delta U_{\text{total}} = \Delta U_A + \Delta U_B = 0$.
By the First Law, the net heat absorbed from the water bath is equal to the net boundary work done:
$$Q_{\text{net}} = W_{\text{net}} = W_A + W_B = P_0 V_0 \ln\left(\frac{2}{3}\right) + 2 P_0 V_0 \ln\left(\frac{4}{3}\right)$$
Using logarithmic properties:
$$Q_{\text{net}} = P_0 V_0 \left[ \ln\left(\frac{2}{3}\right) + \ln\left(\left(\frac{4}{3}\right)^2\right) \right] = P_0 V_0 \ln\left(\frac{2}{3} \times \frac{16}{9}\right) = P_0 V_0 \ln\left(\frac{32}{27}\right)$$
Since $\frac{32}{27} \approx 1.185 > 1$, $\ln(32/27) \approx +0.170 > 0$.
Thus, $Q_{\text{net}} = +0.170 \, P_0 V_0 > 0$ (heat is drawn from the bath).
Takeaway: Even though the overall volume of the container is rigid ($V_{\text{total}} = \text{const}$), internal redistributive expansion produces a positive net work output on the piston, drawing heat from the isothermal bath.
Example 4 (Reversibility & Limiting Cases: Multi-Step vs. Reversible Expansion)
Problem Statement: One mole of an ideal gas at temperature $T$ expands isothermally from initial state $(P_1, V_1)$ to final state $(P_2 = P_1 / 4, V_2 = 4V_1)$ via three alternative mechanisms:
- Single-Step Irreversible: Piston is suddenly released against constant external pressure $P_{\text{ext}} = P_2 = P_1 / 4$.
- Two-Step Irreversible: Gas first expands against $P_{\text{ext1}} = P_1 / 2$ until volume reaches $2V_1$, then against $P_{\text{ext2}} = P_1 / 4$ until volume reaches $4V_1$.
- Reversible Isothermal: Gas expands infinitely slowly under continuous quasi-static equilibrium.
Compute the work done in each case and verify the Maximum Work Theorem.
Solution:
Case 1 (Single-Step Irreversible):
$$W_1 = P_{\text{ext}} (V_2 – V_1) = \left(\frac{P_1}{4}\right) (4V_1 – V_1) = \frac{3}{4} P_1 V_1 = 0.75 \, P_1 V_1$$
Case 2 (Two-Step Irreversible):
Step a: $W_{2a} = P_{\text{ext1}} (V_{\text{mid}} – V_1) = \left(\frac{P_1}{2}\right) (2V_1 – V_1) = \frac{1}{2} P_1 V_1 = 0.50 \, P_1 V_1$
Step b: $W_{2b} = P_{\text{ext2}} (V_2 – V_{\text{mid}}) = \left(\frac{P_1}{4}\right) (4V_1 – 2V_1) = \left(\frac{P_1}{4}\right) (2V_1) = \frac{1}{2} P_1 V_1 = 0.50 \, P_1 V_1$
Total two-step work:
$$W_2 = W_{2a} + W_{2b} = 0.50 \, P_1 V_1 + 0.50 \, P_1 V_1 = 1.00 \, P_1 V_1$$
Case 3 (Reversible Quasi-Static):
$$W_3 = nRT \ln\left(\frac{V_2}{V_1}\right) = P_1 V_1 \ln(4) = 2 P_1 V_1 \ln 2 \approx 2(0.693) P_1 V_1 = 1.386 \, P_1 V_1$$
Comparison: $$W_1 (0.75 \, P_1 V_1) < W_2 (1.00 \, P_1 V_1) < W_3 (1.386 \, P_1 V_1)$$ As the number of intermediate stages $N \to \infty$, the stepwise expansion approaches the continuous integral area under the curve.
Takeaway: Reversible isothermal expansion delivers the theoretical maximum possible work between any two given equilibrium states ($W_{\text{rev}} = W_{\max}$). Any finite irreversible expansion produces less work due to uncompensated unrestrained expansion.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. An ideal gas at initial pressure $P_0$ and initial volume $V_0$ expands isothermally until its volume increases to $3V_0$. The work done by the gas during this expansion is:
(A) $P_0 V_0 \ln 3$
(B) $3 P_0 V_0$
(C) $\frac{1}{3} P_0 V_0 \ln 3$
(D) $P_0 V_0 (\ln 3 – 1)$
Q2. The isothermal bulk modulus of elasticity of an ideal gas at an absolute pressure of $P = 1.5 \times 10^5\text{ Pa}$ is:
(A) $1.5 \times 10^5\text{ Pa}$
(B) $2.1 \times 10^5\text{ Pa}$
(C) $2.5 \times 10^5\text{ Pa}$
(D) $0$
Q3. One mole of an ideal gas undergoes a reversible isothermal expansion at temperature $T = 300\text{ K}$ until its volume doubles. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $\ln 2 \approx 0.693$, the change in internal energy $\Delta U$ and heat absorbed $Q$ are respectively:
(A) $\Delta U = 0, \quad Q \approx 1728.5\text{ J}$
(B) $\Delta U \approx 1728.5\text{ J}, \quad Q = 0$
(C) $\Delta U \approx 1728.5\text{ J}, \quad Q \approx 1728.5\text{ J}$
(D) $\Delta U = 0, \quad Q = 0$
Q4. On a standard $P$-$V$ indicator diagram, the slope $\frac{dP}{dV}$ of an isothermal process for an ideal gas at state point $(P, V)$ is:
(A) $-\frac{P}{V}$
(B) $-\gamma \frac{P}{V}$
(C) $+\frac{P}{V}$
(D) $-\frac{V}{P}$
Q5. An ideal gas expands from volume $V_1$ to $V_2$ at constant temperature $T_0$. In Case I, the expansion is reversible against a slowly moving piston. In Case II, the expansion is a free expansion into an insulated evacuated chamber. Which of the following statements correctly compares the work done and heat exchanged?
(A) $W_{\text{I}} = nRT_0 \ln(V_2/V_1), \, Q_{\text{I}} = W_{\text{I}}; \quad W_{\text{II}} = 0, \, Q_{\text{II}} = 0$
(B) $W_{\text{I}} = W_{\text{II}} = nRT_0 \ln(V_2/V_1), \, Q_{\text{I}} = Q_{\text{II}}$
(C) $W_{\text{I}} = nRT_0 \ln(V_2/V_1), \, Q_{\text{I}} = 0; \quad W_{\text{II}} = 0, \, Q_{\text{II}} = nRT_0 \ln(V_2/V_1)$
(D) $W_{\text{I}} = 0, \, Q_{\text{I}} = 0; \quad W_{\text{II}} = nRT_0 \ln(V_2/V_1), \, Q_{\text{II}} = W_{\text{II}}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements is/are correct for an ideal gas undergoing a reversible isothermal process?
(A) The internal energy of the gas remains strictly invariant throughout the process ($\Delta U = 0$).
(B) The total heat exchanged with the surroundings equals the mechanical work done ($Q = W$).
(C) The molar heat capacity of the gas during this process is infinite ($C = \infty$).
(D) The product of pressure and volume remains constant at all intermediate points ($PV = \text{constant}$).
Q7. Two isotherms for a fixed mass of an ideal gas are plotted on a $P$-$V$ indicator diagram at temperatures $T_1$ and $T_2$, where $T_2 > T_1$. Which of the following statements is/are correct?
(A) The isotherm corresponding to $T_2$ lies entirely above and to the right of the isotherm corresponding to $T_1$.
(B) At any given volume $V$, the magnitude of the slope $\left|\frac{dP}{dV}\right|$ of the $T_2$ isotherm is greater than that of the $T_1$ isotherm.
(C) For an identical volume expansion from $V_i$ to $V_f$, the work done along the $T_2$ isotherm exceeds the work done along the $T_1$ isotherm.
(D) The two isotherms will intersect at sufficiently high pressures.
Q8. An ideal gas at initial pressure $P_0$ and volume $V_0$ is compressed isothermally to volume $V_0 / 2$. Which of the following assertions is/are correct?
(A) The final pressure of the gas is $2P_0$.
(B) The work done by the gas is negative and given by $W = -nRT \ln 2$.
(C) The heat transferred is negative ($Q < 0$), meaning heat is rejected to the surroundings.
(D) The root-mean-square speed $v_{\text{rms}}$ of the gas molecules increases during the compression.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. During an isothermal expansion of an ideal gas at temperature $T$, the work done by the gas is $W_1 = 800\text{ J}$ when the volume doubles from $V_0$ to $2V_0$. If the gas is further expanded isothermally at the same temperature from $2V_0$ to $8V_0$, find the additional work done by the gas (in $\text{J}$).
Q10. A sample of $n$ moles of an ideal gas undergoes a three-step thermodynamic cycle $1 \to 2 \to 3 \to 1$:
– $1 \to 2$: Reversible isothermal expansion at temperature $T_0$ from volume $V_0$ to $2V_0$.
– $2 \to 3$: Isobaric compression at pressure $P_2$ from volume $2V_0$ back to volume $V_0$.
– $3 \to 1$: Isochoric heating at volume $V_0$ back to initial temperature $T_0$.
Given $nRT_0 = 2000\text{ J}$ and taking $\ln 2 = 0.693$, calculate the net work done by the gas over one complete cycle (in $\text{J}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $P_0 V_0 \ln 3$ | Isothermal Work Formula ($V_2 / V_1 = 3$) |
| Q2 | (A) $1.5 \times 10^5\text{ Pa}$ | Isothermal Bulk Modulus ($B_T = P$) |
| Q3 | (A) $\Delta U = 0, \, Q \approx 1728.5\text{ J}$ | First Law in Isothermal Expansion ($Q = W$) |
| Q4 | (A) $-\frac{P}{V}$ | Slope of Isotherm on $P$-$V$ Diagram |
| Q5 | (A) $W_{\text{I}} = nRT_0 \ln(V_2/V_1), \, Q_{\text{I}} = W_{\text{I}}; \, W_{\text{II}} = 0, \, Q_{\text{II}} = 0$ | Reversible vs. Free Isothermal Expansion |
| Q6 | (A), (B), (C), (D) | Comprehensive Properties of Isothermal Process |
| Q7 | (A), (B), (C) | $P$-$V$ Isotherm Family & Slope Comparison |
| Q8 | (A), (B), (C) | Isothermal Compression Dynamics |
| Q9 | 1600 | Logarithmic Work Scaling ($2V_0 \to 8V_0$) |
| Q10 | 386 | Cycle Net Work ($W_{12} + W_{23} + W_{31}$) |
Complete Step-by-Step Solutions
Solution 1:
The work done by an ideal gas in an isothermal expansion from $V_1 = V_0$ to $V_2 = 3V_0$ is:
$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = P_0 V_0 \ln\left(\frac{3V_0}{V_0}\right) = P_0 V_0 \ln 3$$
Correct Option: (A)
Solution 2:
The bulk modulus of elasticity is defined as $B = -V \frac{dP}{dV}$.
Differentiating Boyle’s Law ($PV = \text{const}$) yields:
$$P \, dV + V \, dP = 0 \implies \left(\frac{dP}{dV}\right)_T = -\frac{P}{V}$$
Substituting this slope:
$$B_T = -V \left(-\frac{P}{V}\right) = P$$
Given $P = 1.5 \times 10^5\text{ Pa}$, we have:
$$B_T = 1.5 \times 10^5\text{ Pa}$$
Correct Option: (A)
Solution 3:
For an ideal gas, internal energy depends solely on temperature ($U = U(T)$). Since temperature is constant, $\Delta U = 0$.
By the First Law of Thermodynamics:
$$Q = \Delta U + W = 0 + W = W$$
The work done during isothermal expansion where volume doubles is:
$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = (1)(8.314)(300) \ln 2 = (2494.2)(0.693) \approx 1728.5\text{ J}$$
Therefore, $\Delta U = 0$ and $Q \approx 1728.5\text{ J}$.
Correct Option: (A)
Solution 4:
For an isothermal process of an ideal gas, $PV = \text{constant}$.
Taking differentials on both sides:
$$d(PV) = P \, dV + V \, dP = 0 \implies V \, dP = -P \, dV \implies \frac{dP}{dV} = -\frac{P}{V}$$
The slope is negative and equal to $-\frac{P}{V}$.
Correct Option: (A)
Solution 5:
In Case I (reversible quasi-static expansion):
The gas expands against external pressure that balances internal pressure at every instant:
$$W_{\text{I}} = \int_{V_1}^{V_2} P \, dV = nRT_0 \ln\left(\frac{V_2}{V_1}\right)$$
Since $\Delta U = 0$, by the First Law, $Q_{\text{I}} = W_{\text{I}} = nRT_0 \ln(V_2/V_1)$.
In Case II (free expansion into vacuum):
The gas expands against zero opposing pressure ($P_{\text{ext}} = 0$). Hence:
$$W_{\text{II}} = \int P_{\text{ext}} \, dV = 0$$
The container is thermally insulated, so $Q_{\text{II}} = 0$. By First Law, $\Delta U = Q – W = 0$, so $T_f = T_0$.
Therefore, $W_{\text{I}} = nRT_0 \ln(V_2/V_1), Q_{\text{I}} = W_{\text{I}}$, while $W_{\text{II}} = 0, Q_{\text{II}} = 0$.
Correct Option: (A)
Solution 6:
Let us examine each statement:
(A) Correct: For an ideal gas, $U = n C_v T$. Since $T = \text{const}$, $\Delta U = 0$.
(B) Correct: By First Law, $Q = \Delta U + W = 0 + W = W$.
(C) Correct: Molar heat capacity is $C = \frac{Q}{n \Delta T} = \frac{Q}{0} = \infty$.
(D) Correct: For an ideal gas at constant $T$, $PV = nRT = \text{constant}$.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Since $P = \frac{nRT}{V}$, for any volume $V$, higher temperature $T_2 > T_1$ yields higher pressure $P(T_2) > P(T_1)$. Thus, the $T_2$ curve lies entirely above and to the right of $T_1$.
(B) Correct: The slope magnitude is $\left|\frac{dP}{dV}\right| = \frac{P}{V} = \frac{nRT}{V^2}$. At a given $V$, $\left|\frac{dP}{dV}\right| \propto T$. Therefore, the $T_2$ isotherm has a steeper slope magnitude.
(C) Correct: For an identical expansion from $V_i$ to $V_f$, $W = nRT \ln(V_f/V_i) \propto T$. Since $T_2 > T_1$, $W(T_2) > W(T_1)$.
(D) Incorrect: For two isotherms at different temperatures, $P_1 V = n R T_1 \neq n R T_2 = P_2 V$ for any non-zero finite $V$. They never intersect.
Correct Options: (A), (B), (C)
Solution 8:
(A) Correct: By Boyle’s Law, $P_f = P_0 \left(\frac{V_0}{V_0 / 2}\right) = 2P_0$.
(B) Correct: Work done by the gas is $W = nRT \ln\left(\frac{V_0 / 2}{V_0}\right) = nRT \ln(1/2) = -nRT \ln 2 < 0$.
(C) Correct: By First Law, $Q = \Delta U + W = 0 + (-nRT \ln 2) = -nRT \ln 2 < 0$. Heat is released to the surroundings.
(D) Incorrect: Molecular root-mean-square speed is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$. Because temperature is strictly constant in an isothermal process, $v_{\text{rms}}$ remains unchanged.
Correct Options: (A), (B), (C)
Solution 9:
During isothermal expansion from $V_0$ to $2V_0$:
$$W_1 = nRT \ln\left(\frac{2V_0}{V_0}\right) = nRT \ln 2 = 800\text{ J}$$
During the subsequent isothermal expansion from $2V_0$ to $8V_0$:
$$W_2 = nRT \ln\left(\frac{8V_0}{2V_0}\right) = nRT \ln 4 = nRT \ln(2^2) = 2 \, nRT \ln 2$$
Substituting $nRT \ln 2 = 800\text{ J}$:
$$W_2 = 2 \times 800\text{ J} = 1600\text{ J}$$
The additional work done is $1600\text{ J}$.
Answer: 1600
Solution 10:
Let us calculate the work done in each of the three steps:
1. Step $1 \to 2$ (Isothermal Expansion at $T_0$ from $V_0$ to $2V_0$):
$$W_{12} = nRT_0 \ln\left(\frac{2V_0}{V_0}\right) = nRT_0 \ln 2 = (2000)(0.693) = 1386\text{ J}$$
At state 1: $P_1 = \frac{nRT_0}{V_0}$. At state 2: $P_2 = \frac{nRT_0}{2V_0} = \frac{P_1}{2}$.
2. Step $2 \to 3$ (Isobaric Compression at $P_2 = \frac{nRT_0}{2V_0}$ from $2V_0$ to $V_0$):
$$W_{23} = P_2 (V_0 – 2V_0) = -P_2 V_0 = -\left(\frac{nRT_0}{2V_0}\right) V_0 = -\frac{1}{2} nRT_0 = -\frac{1}{2}(2000) = -1000\text{ J}$$
3. Step $3 \to 1$ (Isochoric Heating at volume $V_0$ back to $T_0$):
Because volume is constant ($dV = 0$):
$$W_{31} = 0\text{ J}$$
Total work done over the complete cycle:
$$W_{\text{net}} = W_{12} + W_{23} + W_{31} = 1386\text{ J} + (-1000\text{ J}) + 0\text{ J} = 386\text{ J}$$
Answer: 386