Welcome to Lecture 93 of our comprehensive JEE Physics Master Series. In this lecture, we master the foundational pillar of thermal physics: the First Law of Thermodynamics. We formalize the universal conservation of energy for closed thermodynamic systems, establish rigorous physics sign conventions ($Q = \Delta U + W$), deconstruct the critical distinction between state functions ($U$) and path functions ($Q, W$), formulate Joule’s law of internal energy for ideal gases, and explore comprehensive applications across isochoric, isobaric, isothermal, adiabatic, cyclic, and free expansion processes.
Detailed Concept Card: First Law of Thermodynamics
The First Law of Thermodynamics is the fundamental assertion of the universal principle of conservation of energy applied to thermodynamic systems. It establishes that energy can neither be created nor destroyed; it can only change forms between heat ($Q$), mechanical work ($W$), and microscopic internal energy ($U$).
1. Mathematical Formulation & Sign Conventions
When an infinitesimal amount of heat $\delta Q$ is transferred across the boundary of a thermodynamic system, part of this energy increases the microscopic internal energy of the system by $dU$, and the remainder is delivered to the surroundings as macroscopic mechanical work $\delta W$:
$$\delta Q = dU + \delta W \quad \text{or for a finite process:} \quad Q = \Delta U + W$$
Alternatively written in terms of internal energy change:
$$\Delta U = Q – W$$
Standard Physics Sign Conventions:
- Heat absorbed by the system ($Q > 0$): Thermal energy flows from the surroundings into the system (endothermic transition).
- Heat rejected by the system ($Q < 0$): Thermal energy flows from the system to the surroundings.
- Work done by the system ($W > 0$): The system expands against external restraining pressure ($dV > 0$).
- Work done on the system ($W < 0$): External agents compress the system ($dV < 0$).
- Change in internal energy ($\Delta U$): If temperature rises ($T_f > T_i$), $\Delta U > 0$; if temperature falls ($T_f < T_i$), $\Delta U < 0$.
Critical Trap (Physics vs. Chemistry Conventions): In IUPAC Chemistry, the First Law is expressed as $\Delta U = q + w$, where $w_{\text{chem}}$ is defined as work done on the system ($w_{\text{chem}} = -\int P_{\text{ext}}\, dV$). In Physics (and across all JEE Physics papers), $W$ is defined strictly as work done by the system ($W = \int P\, dV$), making $Q = \Delta U + W$. Never mix the signs!
2. Path Functions vs. State Functions
- Internal Energy ($U$): A pure state function and an exact differential ($dU$). Its value depends strictly on the current equilibrium state coordinates $(P, V, T)$. For any arbitrary closed cyclic process: $$\oint dU = 0 \implies \Delta U_{\text{cycle}} = 0$$
- Heat ($Q$) and Work ($W$): Boundary phenomena and path functions (inexact differentials $\delta Q$ and $\delta W$). They do not exist “inside” a system; they only characterize energy in transit across boundaries. Consequently, $\oint \delta Q = Q_{\text{net}} \neq 0$ and $\oint \delta W = W_{\text{net}} \neq 0$.
- The First Law Synthesis: While $Q$ and $W$ individually depend on the path connecting states $A$ and $B$, their difference $Q – W = \Delta U$ is universally invariant and identical for all imaginable reversible and irreversible paths connecting $A$ and $B$.
3. Internal Energy of an Ideal Gas (Joule’s Law)
For an ideal gas, intermolecular potential energies are zero. Therefore, internal energy is entirely kinetic and depends solely on absolute temperature ($U = U(T)$), regardless of pressure or volume variations:
$$dU = n C_v dT \implies \Delta U = n C_v \Delta T = n \left(\frac{R}{\gamma – 1}\right) (T_2 – T_1)$$
Using the ideal gas equation of state $PV = nRT$, this can be evaluated directly from macroscopic state variables without knowing $T$:
$$\Delta U = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1} = \frac{f}{2} (P_2 V_2 – P_1 V_1)$$
where $f$ is the degrees of freedom ($f = 3$ for monoatomic, $f = 5$ for rigid diatomic, $f = 6$ for non-linear polyatomic).
4. Applications of First Law to Standard Thermodynamic Processes
| Process | Condition | Work Done ($W$) | Change in Internal Energy ($\Delta U$) | Heat Transferred ($Q$) |
|---|---|---|---|---|
| Isochoric | $V = \text{const}$, $dV = 0$ | $W = 0$ | $\Delta U = n C_v \Delta T$ | $Q = \Delta U = n C_v \Delta T$ |
| Isobaric | $P = \text{const}$ | $W = P(V_2 – V_1) = n R \Delta T$ | $\Delta U = n C_v \Delta T$ | $Q = n C_p \Delta T = \Delta U + W$ |
| Isothermal | $T = \text{const}$, $\Delta T = 0$ | $W = n R T \ln\left(\frac{V_2}{V_1}\right)$ | $\Delta U = 0$ | $Q = W = n R T \ln\left(\frac{V_2}{V_1}\right)$ |
| Adiabatic | $Q = 0$ (thermally insulated) | $W = -\Delta U = \frac{P_1 V_1 – P_2 V_2}{\gamma – 1}$ | $\Delta U = -W = n C_v \Delta T$ | $Q = 0$ |
| Cyclic | Initial state = Final state | $W_{\text{net}} = \text{Area of loop on } P\text{-}V$ | $\Delta U_{\text{net}} = 0$ | $Q_{\text{net}} = W_{\text{net}}$ |
| Free Expansion | Expansion into vacuum ($P_{\text{ext}} = 0$) | $W = 0$, $Q = 0$ | $\Delta U = 0 \implies \Delta T = 0$ (ideal) | $Q = 0$ |
5. Energy Partitioning in Isobaric Processes
When an ideal gas is heated at constant pressure, the heat supplied $Q$ divides into internal energy increase $\Delta U$ and expansion work $W$ in fixed stoichiometric ratios determined exclusively by the adiabatic index $\gamma$:
$$\text{Fraction converted to internal energy:} \quad \frac{\Delta U}{Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{1}{\gamma}$$
$$\text{Fraction converted to work:} \quad \frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{C_p – C_v}{C_p} = 1 – \frac{1}{\gamma} = \frac{\gamma – 1}{\gamma}$$
- Monoatomic Gas ($\gamma = 5/3$): $\frac{\Delta U}{Q} = \frac{3}{5} = 60\%$, $\frac{W}{Q} = \frac{2}{5} = 40\%$.
- Diatomic Gas ($\gamma = 7/5$): $\frac{\Delta U}{Q} = \frac{5}{7} \approx 71.4\%$, $\frac{W}{Q} = \frac{2}{7} \approx 28.6\%$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Multi-Path First Law Accounting)
Problem Statement: An ideal gas is taken from state $A$ to state $B$ along three different reversible paths: $1$, $2$, and $3$. Along path $1$, heat absorbed is $Q_1 = 50\text{ J}$ and work done by the gas is $W_1 = 20\text{ J}$. Along path $2$, the work done by the gas is $W_2 = 35\text{ J}$. If the gas is returned from state $B$ to state $A$ along a fourth curved path $4$ where work done on the gas is $15\text{ J}$ ($W_4 = -15\text{ J}$), determine:
- The change in internal energy $\Delta U_{A \to B}$.
- The heat transferred $Q_2$ along path $2$.
- The heat transferred $Q_4$ along the return path $4$.
Solution:
Step 1: Determine $\Delta U_{A \to B}$ from Path 1.
Applying the First Law to path 1:
$$\Delta U_{A \to B} = Q_1 – W_1 = 50\text{ J} – 20\text{ J} = 30\text{ J}$$
Since internal energy is a state function, $\Delta U_{A \to B} = U_B – U_A = +30\text{ J}$ for every path connecting state $A$ to state $B$.
Step 2: Find $Q_2$ along Path 2.
Using the First Law for path 2:
$$Q_2 = \Delta U_{A \to B} + W_2 = 30\text{ J} + 35\text{ J} = +65\text{ J}$$
Thus, $65\text{ J}$ of heat is absorbed by the gas along path 2.
Step 3: Find $Q_4$ for the return path $B \to A$.
For the return process from $B$ to $A$:
$$\Delta U_{B \to A} = U_A – U_B = -\Delta U_{A \to B} = -30\text{ J}$$
Given work done on the gas is $15\text{ J}$, the work done by the gas is $W_4 = -15\text{ J}$. Applying the First Law:
$$Q_4 = \Delta U_{B \to A} + W_4 = -30\text{ J} + (-15\text{ J}) = -45\text{ J}$$
The negative sign indicates that $45\text{ J}$ of heat is rejected by the gas to the surroundings.
Takeaway: Internal energy change reverses sign upon path reversal ($\Delta U_{B \to A} = -\Delta U_{A \to B}$), whereas work done and heat transferred depend on the specific trajectory taken.
Example 2 (Algebraic & Conceptual: Isobaric Heating & Work Fraction)
Problem Statement: A sample of $n = 2\text{ moles}$ of an ideal monoatomic gas ($\gamma = 5/3$) is heated at constant atmospheric pressure $P = 1.01 \times 10^5\text{ Pa}$ from an initial temperature $T_1 = 300\text{ K}$ to a final temperature $T_2 = 500\text{ K}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$, compute:
- The total heat supplied $Q$.
- The work done $W$ by the expanding gas.
- The change in internal energy $\Delta U$.
- Verify that $Q = \Delta U + W$ and verify the theoretical partitioning fractions.
Solution:
Step 1: Compute Molar Heat Capacities.
For a monoatomic ideal gas:
$$C_v = \frac{3}{2}R, \quad C_p = C_v + R = \frac{5}{2}R$$
Temperature increase:
$$\Delta T = T_2 – T_1 = 500\text{ K} – 300\text{ K} = 200\text{ K}$$
Step 2: Calculate Heat Supplied $Q$.
$$Q = n C_p \Delta T = (2) \left(\frac{5}{2} \times 8.314\right) (200) = 5 \times 8.314 \times 200 = 8314\text{ J}$$
Step 3: Calculate Change in Internal Energy $\Delta U$.
$$\Delta U = n C_v \Delta T = (2) \left(\frac{3}{2} \times 8.314\right) (200) = 3 \times 8.314 \times 200 = 4988.4\text{ J}$$
Step 4: Calculate Work Done $W$.
For an isobaric expansion:
$$W = P \Delta V = n R \Delta T = (2)(8.314)(200) = 3325.6\text{ J}$$
Step 5: Verification & Energy Fractions.
$$\Delta U + W = 4988.4\text{ J} + 3325.6\text{ J} = 8314.0\text{ J} = Q \quad \text{(First Law satisfied!)}$$
Fraction into internal energy: $\frac{\Delta U}{Q} = \frac{4988.4}{8314} = 0.60 = 60\% = \frac{1}{\gamma}$.
Fraction into mechanical work: $\frac{W}{Q} = \frac{3325.6}{8314} = 0.40 = 40\% = 1 – \frac{1}{\gamma}$.
Takeaway: In any isobaric heating of an ideal monoatomic gas, exactly $40\%$ of input heat transforms into useful boundary work, and $60\%$ accumulates as thermal internal energy.
Example 3 (JEE Advanced Level: Constrained Piston with Spring & Atmospheric Load)
Problem Statement: A vertical cylinder of cross-sectional area $A = 0.02\text{ m}^2$ fitted with a frictionless, light, movable piston contains $n = 1\text{ mole}$ of an ideal monoatomic gas at atmospheric pressure $P_0 = 10^5\text{ Pa}$ and initial temperature $T_0 = 300\text{ K}$. An ideal linear spring of force constant $k = 2000\text{ N/m}$ is attached to the top of the piston and the ceiling. Initially, the spring is at its relaxed natural length. Heat is slowly supplied to the gas until the piston rises by a distance $h = 0.1\text{ m}$. Determine:
- The final pressure $P_f$ of the gas.
- The total work done $W$ by the expanding gas.
- The change in internal energy $\Delta U$ of the gas.
- The total heat $Q$ supplied to the gas.
Solution:
Step 1: Equilibrium and Final Pressure.
At any displacement $x$ of the piston, the downward forces acting on the piston are atmospheric force $P_0 A$ and spring restoring force $k x$. For quasi-static equilibrium:
$$P(x) A = P_0 A + k x \implies P(x) = P_0 + \frac{k}{A} x$$
At $x = h = 0.1\text{ m}$:
$$P_f = P_0 + \frac{k h}{A} = 10^5 + \frac{(2000)(0.1)}{0.02} = 10^5 + 10^4 = 1.10 \times 10^5\text{ Pa}$$
Step 2: Total Work Done $W$.
The volume change is $dV = A\, dx$. Thus, work done by the gas is:
$$W = \int_0^h P(x) A\, dx = \int_0^h (P_0 A + k x)\, dx = P_0 A h + \frac{1}{2} k h^2$$
Substituting values:
$$P_0 A h = (10^5)(0.02)(0.1) = 200\text{ J}$$
$$\frac{1}{2} k h^2 = \frac{1}{2}(2000)(0.1)^2 = 10\text{ J}$$
$$W = 200\text{ J} + 10\text{ J} = 210\text{ J}$$
Step 3: Change in Internal Energy $\Delta U$.
Initial volume: $V_0 = \frac{n R T_0}{P_0} = \frac{(1)(8.314)(300)}{10^5} = 0.024942\text{ m}^3$.
Volume increase: $\Delta V = A h = (0.02)(0.1) = 0.002\text{ m}^3$.
Final volume: $V_f = V_0 + \Delta V = 0.026942\text{ m}^3$.
For a monoatomic gas ($f = 3$):
$$\Delta U = \frac{f}{2} (P_f V_f – P_0 V_0) = \frac{3}{2} \left(P_f(V_0 + \Delta V) – P_0 V_0\right)$$
$$\Delta U = \frac{3}{2} \left[(P_f – P_0)V_0 + P_f \Delta V\right]$$
Substitute values:
$$(P_f – P_0) V_0 = (10^4)(0.024942) = 249.42\text{ J}$$
$$P_f \Delta V = (1.10 \times 10^5)(0.002) = 220\text{ J}$$
$$\Delta U = \frac{3}{2} (249.42 + 220) = \frac{3}{2} (469.42) = 704.13\text{ J}$$
Step 4: Total Heat Supplied $Q$.
By the First Law:
$$Q = \Delta U + W = 704.13\text{ J} + 210\text{ J} = 914.13\text{ J}$$
Takeaway: Work done by a gas expanding against a spring-loaded piston equals work done against the atmosphere ($P_0 \Delta V$) plus elastic potential energy stored in the spring ($\frac{1}{2} k h^2$).
Example 4 (Non-Standard Polytropic Path: Variable Heat Capacity)
Problem Statement: One mole of a monoatomic ideal gas undergoes a quasi-static process governed by the relationship $P = \alpha V$, where $\alpha$ is a positive dimensional constant. The volume expands from $V_0$ to $2V_0$. Determine:
- The work done $W$ by the gas in terms of $P_0$ and $V_0$ (where $P_0 = \alpha V_0$).
- The change in internal energy $\Delta U$.
- The molar heat capacity $C$ of the gas during this process.
- The total heat absorbed $Q$.
Solution:
Step 1: Calculate Work Done $W$.
$$W = \int_{V_0}^{2V_0} P\, dV = \int_{V_0}^{2V_0} (\alpha V)\, dV = \frac{\alpha}{2} \left[(2V_0)^2 – V_0^2\right] = \frac{\alpha}{2} (3 V_0^2) = \frac{3}{2} \alpha V_0^2$$
Since $P_0 = \alpha V_0$, we have:
$$W = \frac{3}{2} P_0 V_0$$
Step 2: Calculate $\Delta U$.
Initial state: $P_1 = P_0, V_1 = V_0 \implies P_1 V_1 = P_0 V_0$.
Final state: $V_2 = 2V_0, P_2 = \alpha(2V_0) = 2P_0 \implies P_2 V_2 = (2P_0)(2V_0) = 4 P_0 V_0$.
For monoatomic gas ($C_v = \frac{3}{2}R$):
$$\Delta U = \frac{3}{2} (P_2 V_2 – P_1 V_1) = \frac{3}{2} (4 P_0 V_0 – P_0 V_0) = \frac{3}{2} (3 P_0 V_0) = \frac{9}{2} P_0 V_0$$
Step 3: Calculate Total Heat $Q$.
Applying the First Law:
$$Q = \Delta U + W = \frac{9}{2} P_0 V_0 + \frac{3}{2} P_0 V_0 = 6 P_0 V_0$$
Step 4: Determine Molar Heat Capacity $C$.
The process equation $P V^{-1} = \alpha = \text{const}$ matches a polytropic process $P V^m = \text{const}$ with polytropic exponent $m = -1$.
General polytropic molar heat capacity formula:
$$C = C_v + \frac{R}{1 – m} = \frac{3}{2}R + \frac{R}{1 – (-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$$
Let us verify via $Q = n C \Delta T$:
$$\Delta T = T_2 – T_1 = \frac{P_2 V_2 – P_1 V_1}{n R} = \frac{4 P_0 V_0 – P_0 V_0}{R} = \frac{3 P_0 V_0}{R}$$
$$Q = (1)(2R) \left(\frac{3 P_0 V_0}{R}\right) = 6 P_0 V_0 \quad \text{(Perfect match!)}$$
Takeaway: For any polytropic process $P V^m = \text{const}$, the effective molar heat capacity is constant and given by $C = C_v + \frac{R}{1 – m}$. Here with $m = -1$, $C = 2R$.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. An ideal gas undergoes a cyclic process $ABCA$ on a $P$-$V$ diagram. Along path $AB$, $50\text{ J}$ of heat is absorbed and $40\text{ J}$ of work is done by the gas. Path $BC$ involves no heat exchange ($Q_{BC} = 0$) and zero work ($W_{BC} = 0$). Along path $CA$, $70\text{ J}$ of heat is rejected. The work done by the gas along path $CA$ is:
(A) $-60\text{ J}$
(B) $-20\text{ J}$
(C) $+20\text{ J}$
(D) $-10\text{ J}$
Q2. One mole of a rigid diatomic ideal gas ($\gamma = 1.4$) is heated at constant pressure from $27^\circ\text{C}$ to $127^\circ\text{C}$. The ratio of the work done by the gas to the total heat supplied to the gas is:
(A) $2/7$
(B) $5/7$
(C) $2/5$
(D) $3/5$
Q3. A system absorbs $1200\text{ J}$ of heat while performing $400\text{ J}$ of work in moving from state $i$ to state $f$ along path $iaf$. Along path $ibf$, the work done by the system is $200\text{ J}$. If the internal energy of the initial state is $U_i = 100\text{ J}$, the heat absorbed along path $ibf$ and the internal energy of the final state $U_f$ are respectively:
(A) $1000\text{ J}, 900\text{ J}$
(B) $1000\text{ J}, 800\text{ J}$
(C) $600\text{ J}, 900\text{ J}$
(D) $800\text{ J}, 1000\text{ J}$
Q4. An ideal gas is enclosed in an insulated rigid container of volume $V$. A partition divides the container into two equal chambers: one chamber contains the gas at pressure $P_0$ and temperature $T_0$, while the other chamber is completely evacuated. The partition is suddenly removed, and the gas freely expands to fill the entire container. After equilibrium is restored, the final temperature $T_f$ and final pressure $P_f$ are:
(A) $T_f = T_0 / 2, \quad P_f = P_0 / 2$
(B) $T_f = T_0, \quad P_f = P_0 / 2$
(C) $T_f = T_0 / \gamma, \quad P_f = P_0 / 2^\gamma$
(D) $T_f = T_0, \quad P_f = P_0$
Q5. During an adiabatic expansion of $2\text{ moles}$ of a gas, the internal energy decreases by $600\text{ J}$. The work done by the gas and the heat absorbed by the gas are:
(A) $W = +600\text{ J}, \quad Q = 0$
(B) $W = -600\text{ J}, \quad Q = 0$
(C) $W = +600\text{ J}, \quad Q = +600\text{ J}$
(D) $W = 0, \quad Q = -600\text{ J}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. For an arbitrary closed cyclic process undergone by an ideal gas, which of the following statements is/are correct?
(A) The cyclic integral $\oint dU = 0$ for every cyclic process.
(B) The net heat absorbed equals the net work done: $Q_{\text{net}} = W_{\text{net}}$.
(C) The cyclic integral $\oint \delta Q = 0$ for every reversible cyclic process.
(D) If the cycle runs clockwise on a $P$-$V$ diagram, the net work done by the system is positive.
Q7. A sample of $n$ moles of an ideal monoatomic gas undergoes a process described by $P V^2 = \text{constant}$. Which of the following assertions is/are true for this process during an expansion from $V_0$ to $2V_0$?
(A) The temperature of the gas decreases.
(B) The molar heat capacity for this process is $C = \frac{1}{2}R$.
(C) The work done by the gas is positive.
(D) Heat is absorbed by the gas during expansion ($Q > 0$).
Q8. An ideal gas is taken from state $1$ ($P_0, V_0$) to state $2$ ($2P_0, 2V_0$) along a straight-line path on a $P$-$V$ diagram. Which of the following statements is/are correct?
(A) The work done by the gas during this transition is $\frac{3}{2} P_0 V_0$.
(B) If the gas is monoatomic, the change in internal energy is $\frac{9}{2} P_0 V_0$.
(C) The total heat supplied to a monoatomic gas along this path is $6 P_0 V_0$.
(D) The temperature of the gas increases monotonically from state $1$ to state $2$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. In a thermodynamic cycle consisting of four quasi-static processes $1 \to 2$, $2 \to 3$, $3 \to 4$, and $4 \to 1$, the heat exchanges are $Q_{12} = +150\text{ kJ}$, $Q_{23} = -40\text{ kJ}$, $Q_{34} = -80\text{ kJ}$, and $Q_{41} = +20\text{ kJ}$. The work done during the processes are $W_{12} = +60\text{ kJ}$, $W_{23} = -20\text{ kJ}$, and $W_{34} = -30\text{ kJ}$. Find the magnitude of work done $W_{41}$ (in $\text{kJ}$) during the process $4 \to 1$.
Q10. An ideal monoatomic gas ($\gamma = 5/3$) undergoes an isobaric expansion at pressure $P = 2 \times 10^5\text{ Pa}$ from an initial volume $V_1 = 0.01\text{ m}^3$ to a final volume $V_2 = 0.04\text{ m}^3$. Calculate the total heat supplied to the gas $Q$ (in $\text{kJ}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $-60\text{ J}$ | Cyclic First Law & Path Invariance |
| Q2 | (A) $2/7$ | Isobaric Energy Partitioning Ratio |
| Q3 | (A) $1000\text{ J}, 900\text{ J}$ | Multi-Path State Function Analysis |
| Q4 | (B) $T_f = T_0, P_f = P_0 / 2$ | Adiabatic Free Expansion into Vacuum |
| Q5 | (A) $W = +600\text{ J}, Q = 0$ | Adiabatic Work-Internal Energy Relation |
| Q6 | (A), (B), (D) | First Law Principles in Cyclic Processes |
| Q7 | (A), (B), (C) | Polytropic Process $PV^2 = C$ Dynamics |
| Q8 | (A), (B), (C), (D) | Linear $P$-$V$ Path Thermodynamic Properties |
| Q9 | 40 | First Law Cyclic Net Work Balance ($\text{kJ}$) |
| Q10 | 15 | Isobaric Heat Calculation ($15\text{ kJ}$) |
Complete Step-by-Step Solutions
Solution 1:
For any complete cyclic process ($ABCA$), the initial and final states coincide. Because internal energy $U$ is a state function, the net change in internal energy over the entire cycle is strictly zero:
$$\Delta U_{\text{cycle}} = \oint dU = 0$$
Applying the First Law of Thermodynamics to the entire cycle:
$$Q_{\text{net}} = \Delta U_{\text{cycle}} + W_{\text{net}} \implies Q_{\text{net}} = W_{\text{net}}$$
The net heat absorbed over the three segments is:
$$Q_{\text{net}} = Q_{AB} + Q_{BC} + Q_{CA} = 50\text{ J} + 0\text{ J} + (-70\text{ J}) = -20\text{ J}$$
The net work done by the gas over the cycle is:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 40\text{ J} + 0\text{ J} + W_{CA} = 40\text{ J} + W_{CA}$$
Equating $W_{\text{net}} = Q_{\text{net}}$:
$$40\text{ J} + W_{CA} = -20\text{ J} \implies W_{CA} = -20\text{ J} – 40\text{ J} = -60\text{ J}$$
The negative sign confirms that $60\text{ J}$ of work is done on the gas along path $CA$.
Correct Option: (A)
Solution 2:
For an ideal gas undergoing an isobaric process, the heat supplied is $Q = n C_p \Delta T$ and the work done is $W = P \Delta V = n R \Delta T$.
The ratio of work done to heat supplied is:
$$\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$$
For a rigid diatomic gas:
$$C_p = \frac{7}{2}R \implies \frac{W}{Q} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$$
Alternatively, $\frac{W}{Q} = 1 – \frac{1}{\gamma} = 1 – \frac{1}{1.4} = 1 – \frac{5}{7} = \frac{2}{7}$.
Correct Option: (A)
Solution 3:
Along path $iaf$:
$$Q_{iaf} = 1200\text{ J}, \quad W_{iaf} = 400\text{ J}$$
$$\Delta U = U_f – U_i = Q_{iaf} – W_{iaf} = 1200\text{ J} – 400\text{ J} = 800\text{ J}$$
Given $U_i = 100\text{ J}$:
$$U_f = U_i + \Delta U = 100\text{ J} + 800\text{ J} = 900\text{ J}$$
Now for path $ibf$, since internal energy is a state function:
$$\Delta U_{ibf} = \Delta U = 800\text{ J}$$
Given $W_{ibf} = 200\text{ J}$, the heat absorbed is:
$$Q_{ibf} = \Delta U + W_{ibf} = 800\text{ J} + 200\text{ J} = 1000\text{ J}$$
Thus, $Q_{ibf} = 1000\text{ J}$ and $U_f = 900\text{ J}$.
Correct Option: (A)
Solution 4:
In an adiabatic free expansion into vacuum:
1. The container is insulated, so $Q = 0$.
2. The gas expands against zero opposing pressure ($P_{\text{ext}} = 0$), so $W = \int P_{\text{ext}}\, dV = 0$.
3. Applying the First Law:
$$\Delta U = Q – W = 0 – 0 = 0$$
For an ideal gas, $U$ depends solely on temperature: $\Delta U = 0 \implies T_f = T_0$.
Now applying Boyle’s Law ($T = \text{constant}$):
$$P_0 V_0 = P_f V_f \implies P_0 \left(\frac{V}{2}\right) = P_f (V) \implies P_f = \frac{P_0}{2}$$
Therefore, $T_f = T_0$ and $P_f = P_0 / 2$.
Correct Option: (B)
Solution 5:
For an adiabatic process, $Q = 0$.
Applying the First Law:
$$Q = \Delta U + W \implies 0 = \Delta U + W \implies W = -\Delta U$$
Given that the internal energy decreases by $600\text{ J}$, we have $\Delta U = -600\text{ J}$.
Therefore:
$$W = -(-600\text{ J}) = +600\text{ J}$$
Thus, $W = +600\text{ J}$ and $Q = 0$.
Correct Option: (A)
Solution 6:
Let us examine each statement:
(A) Correct: $U$ is a state function, so $\oint dU = 0$ for any closed cycle.
(B) Correct: Since $\Delta U_{\text{cycle}} = 0$, the First Law yields $Q_{\text{net}} = W_{\text{net}}$.
(C) Incorrect: $\oint \delta Q = Q_{\text{net}} = W_{\text{net}} \neq 0$ unless the net work enclosed is zero. (Note that $\oint \frac{\delta Q}{T} = 0$ is the Clausius theorem, not $\oint \delta Q = 0$).
(D) Correct: On a standard $P$-$V$ diagram, tracing clockwise traverses upper expansion paths at higher pressures and lower compression paths at lower pressures, resulting in a positive enclosed area ($W_{\text{net}} > 0$).
Correct Options: (A), (B), (D)
Solution 7:
The process is polytropic with $P V^2 = C \implies m = 2$.
Using $P = \frac{nRT}{V}$, this gives:
$$\left(\frac{nRT}{V}\right) V^2 = C \implies T V = \text{constant} \implies T \propto \frac{1}{V}$$
As $V$ increases, $T$ decreases. Hence (A) is correct.
The molar heat capacity is:
$$C = C_v + \frac{R}{1 – m} = \frac{3}{2}R + \frac{R}{1 – 2} = \frac{3}{2}R – R = \frac{1}{2}R$$
Hence (B) is correct.
For expansion from $V_0$ to $2V_0$, $dV > 0$ and $P > 0$, so $W = \int P\, dV > 0$. Hence (C) is correct.
Now for heat transferred: $Q = n C \Delta T$. Since $T$ decreases, $\Delta T < 0$. With $C = \frac{1}{2}R > 0$, $Q = n \left(\frac{1}{2}R\right) \Delta T < 0$ (heat is released, not absorbed). Hence (D) is incorrect.
Correct Options: (A), (B), (C)
Solution 8:
The straight line connects $(P_0, V_0)$ to $(2P_0, 2V_0)$ on a $P$-$V$ plane.
(A) Work done is the area of the trapezium under the line:
$$W = \frac{P_1 + P_2}{2} (V_2 – V_1) = \frac{P_0 + 2P_0}{2} (2V_0 – V_0) = \frac{3}{2} P_0 V_0 \quad \text{(Correct)}$$
(B) For a monoatomic gas ($C_v = \frac{3}{2}R$):
$$\Delta U = \frac{3}{2} (P_2 V_2 – P_1 V_1) = \frac{3}{2} \left[(2P_0)(2V_0) – P_0 V_0\right] = \frac{3}{2} (3 P_0 V_0) = \frac{9}{2} P_0 V_0 \quad \text{(Correct)}$$
(C) By First Law:
$$Q = \Delta U + W = \frac{9}{2} P_0 V_0 + \frac{3}{2} P_0 V_0 = 6 P_0 V_0 \quad \text{(Correct)}$$
(D) Let the line equation be $P(V) = P_0 + \frac{P_0}{V_0} (V – V_0) = \frac{P_0}{V_0} V$.
Then $T(V) = \frac{P(V) V}{n R} = \frac{P_0}{n R V_0} V^2$.
Since $\frac{dT}{dV} = \frac{2 P_0 V}{n R V_0} > 0$ for all $V \in [V_0, 2V_0]$, temperature strictly and monotonically increases from $T_1 = \frac{P_0 V_0}{nR}$ to $T_2 = 4 T_1$. Hence (D) is correct.
Correct Options: (A), (B), (C), (D)
Solution 9:
For any complete cyclic process, the net change in internal energy is zero ($\Delta U_{\text{cycle}} = 0$).
By the First Law:
$$Q_{\text{net}} = W_{\text{net}}$$
Total heat supplied:
$$Q_{\text{net}} = Q_{12} + Q_{23} + Q_{34} + Q_{41} = 150 – 40 – 80 + 20 = 50\text{ kJ}$$
Total work done:
$$W_{\text{net}} = W_{12} + W_{23} + W_{34} + W_{41} = 60 – 20 – 30 + W_{41} = 10 + W_{41}$$
Equating $W_{\text{net}} = Q_{\text{net}}$:
$$10 + W_{41} = 50 \implies W_{41} = 40\text{ kJ}$$
The magnitude of work done during process $4 \to 1$ is $40\text{ kJ}$.
Answer: 40
Solution 10:
For an isobaric expansion of an ideal gas at constant pressure $P$:
$$W = P(V_2 – V_1) = (2 \times 10^5\text{ Pa})(0.04 – 0.01\text{ m}^3) = (2 \times 10^5)(0.03) = 6000\text{ J} = 6\text{ kJ}$$
For a monoatomic gas ($\gamma = 5/3$), the internal energy change is:
$$\Delta U = \frac{3}{2} P \Delta V = \frac{3}{2} (6000\text{ J}) = 9000\text{ J} = 9\text{ kJ}$$
By the First Law:
$$Q = \Delta U + W = 9\text{ kJ} + 6\text{ kJ} = 15\text{ kJ}$$
Alternatively, $Q = n C_p \Delta T = \frac{5}{2} P \Delta V = \frac{5}{2} (6000\text{ J}) = 15000\text{ J} = 15\text{ kJ}$.
Answer: 15