Welcome to Lecture 92 of our comprehensive JEE Physics Master Series. In this lecture, we explore the core triad of thermal dynamics: Heat, Work, and Internal Energy. We examine the mathematical and physical distinction between path-dependent boundary transfers and state-dependent functions, formulate quasi-static work as the area under $P$-$V$ curves, deconstruct Joule’s law of internal energy, and bridge these three quantities to establish the foundational conservation principles of thermodynamics.
1. Detailed Concept Card
1.1 Thermodynamic Work ($W$)
In thermodynamics, mechanical work represents energy transferred between a system and its surroundings through macroscopic forces acting across a boundary displacement. Consider a gas enclosed in a cylinder fitted with a frictionless, movable piston of cross-sectional area $A$ under internal pressure $P$:
- When the piston moves outward by an infinitesimal displacement $dx$, the force exerted by the gas is $F = P A$.
- The infinitesimal work done by the gas is: $$\delta W = F \, dx = (P A) \, dx = P \, dV$$
- For a finite, quasi-static process transitioning from initial volume $V_1$ to final volume $V_2$: $$\mathbf{W = \int_{V_1}^{V_2} P \, dV}$$
- Geometric Interpretation on $P$-$V$ Diagram: The work done during any quasi-static process is numerically equal to the area under the $P$-$V$ curve projected onto the volume axis.
- Sign Convention in Physics (Standard NCERT / JEE):
- $W > 0$ (Positive): Work done BY the system on surroundings during expansion ($V_2 > V_1, dV > 0$).
- $W < 0$ (Negative): Work done ON the system by surroundings during compression ($V_2 < V_1, dV < 0$).
- Work is a Path Function: The value of $W$ depends not only on the initial and final equilibrium states, but intimately on the specific continuous trajectory (path) taken. Hence, work is an inexact differential, denoted by $\delta W$ or $dW$.
- Cyclic Processes: For a closed cycle returning to its initial state ($\oint dV = 0$):
$$W_{\text{net}} = \oint P \, dV = \text{Area enclosed by the closed loop on the } P\text{-}V \text{ diagram}$$
- Clockwise Cycle: Expansion occurs at higher pressures than compression $\implies \mathbf{W_{\text{net}} > 0}$ (Heat Engine).
- Counter-Clockwise Cycle: Compression occurs at higher pressures than expansion $\implies \mathbf{W_{\text{net}} < 0}$ (Refrigerator / Heat Pump).
1.2 Thermodynamic Heat ($Q$)
Heat is energy in transit across the boundary of a thermodynamic system solely by virtue of a temperature difference between the system and its environment:
- Like work, heat is a boundary phenomenon (it appears only at the boundary during a process) and a Path Function.
- Heat cannot be stored: a body never contains heat. It contains internal energy.
- Infinitesimal heat transfer is an inexact differential ($\delta Q$).
- Sign Convention:
- $Q > 0$ (Positive): Heat absorbed or supplied to the system.
- $Q < 0$ (Negative): Heat rejected or released from the system.
1.3 Internal Energy ($U$) & Joule’s Law
The Internal Energy $U$ of a system is the grand sum of all microscopic forms of energy possessed by its constituent molecules:
$$U = E_k + E_p$$
- $E_k$: Microscopic kinetic energy of molecular motions (translational, rotational, and vibrational).
- $E_p$: Microscopic potential energy arising from intermolecular conservative forces.
- State Function Character: Unlike $Q$ and $W$, internal energy $U$ is a State Function. Its value is uniquely determined by the instantaneous thermodynamic state of the system ($P, V, T$).
- The differential $dU$ is an exact differential. The net change $\Delta U = U_2 – U_1$ depends only on the initial and final states, completely independent of the intermediate path: $$\oint dU = 0$$
Joule’s Law for an Ideal Gas:
In an ideal gas, molecules are point masses with negligible volume and zero intermolecular attractive or repulsive forces ($E_p = 0$). Consequently, the internal energy of an ideal gas is purely kinetic and depends exclusively on absolute temperature $T$:
$$\mathbf{U = f(T) \implies \left(\frac{\partial U}{\partial V}\right)_T = 0, \quad \left(\frac{\partial U}{\partial P}\right)_T = 0}$$
For $n$ moles of an ideal gas having $f$ degrees of freedom:
$$U = n C_v T = \frac{f}{2} n R T$$
For any arbitrary process connecting state $1$ ($P_1, V_1, T_1$) and state $2$ ($P_2, V_2, T_2$):
$$\mathbf{\Delta U = n C_v \Delta T = \frac{f}{2} n R \Delta T = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}}$$
Universal Invariant Rule: $\Delta U = n C_v \Delta T = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}$ holds universally for any thermodynamic path (isobaric, isochoric, isothermal, adiabatic, or non-linear polytropic) because internal energy is a state property.
1.4 The Fundamental First Law Relation ($\Delta Q = \Delta U + W$)
Although heat added ($Q$) and work done ($W$) vary drastically from path to path between two equilibrium states $1$ and $2$, experimental observation (Joule’s paddle-wheel experiments) demonstrates that their algebraic difference is strictly path-independent:
$$\mathbf{Q – W = \Delta U = U_2 – U_1 = \text{invariant}}$$
In differential form, the First Law of Thermodynamics is stated as:
$$\mathbf{dQ = dU + dW = dU + P \, dV}$$
- Physical Interpretation: When heat $dQ$ is supplied to a system, a portion goes toward raising the microscopic internal energy ($dU$), while the remainder is expended as macroscopic mechanical work ($dW = P dV$) done against external constraints.
- Exact vs Inexact Differentials: The sum of two path-dependent inexact differentials equals an exact state differential: $\delta Q – \delta W = dU$.
1.5 Work Done Across Standard Thermodynamic Paths
| Process | Constraint | Work Done ($W = \int P dV$) | Internal Energy ($\Delta U$) | Heat Transfer ($Q$) |
|---|---|---|---|---|
| Isochoric | $V = \text{constant} \implies dV = 0$ | $W = 0$ | $\Delta U = n C_v \Delta T$ | $Q = \Delta U = n C_v \Delta T$ |
| Isobaric | $P = \text{constant}$ | $W = P(V_2 – V_1) = n R \Delta T$ | $\Delta U = n C_v \Delta T$ | $Q = n C_p \Delta T$ |
| Isothermal | $T = \text{constant} \implies \Delta T = 0$ | $W = n R T \ln\left(\frac{V_2}{V_1}\right) = n R T \ln\left(\frac{P_1}{P_2}\right)$ | $\Delta U = 0$ | $Q = W = n R T \ln\left(\frac{V_2}{V_1}\right)$ |
| Adiabatic | $Q = 0$ (Insulated boundary) | $W = -\Delta U = \frac{P_1 V_1 – P_2 V_2}{\gamma – 1}$ | $\Delta U = -W = n C_v \Delta T$ | $Q = 0$ |
1.6 Critical Conceptual Insights & JEE Traps
- Trap 1: Physics vs Chemistry Sign Convention:
- Physics (Clausius convention): $\Delta Q = \Delta U + W$, where $W = \int P dV$ is work done BY the system. Expansion is positive work.
- Chemistry (IUPAC convention): $\Delta U = q + w$, where $w = -\int P_{\text{ext}} dV$ is work done ON the system. Compression is positive work.
- Both formulations express the identical physical law of conservation of energy.
- Trap 2: Area on $V$-$P$ Axes (Axes Inversion): If a problem plots $V$ on the vertical axis and $P$ on the horizontal axis, the work done is the area projected onto the vertical (volume) axis, NOT the horizontal axis!
- Trap 3: Free Expansion into Vacuum (Joule Expansion): When an ideal gas expands into an evacuated, thermally insulated chamber ($P_{\text{ext}} = 0$): $$W = 0 \quad (\text{no opposing force}), \quad Q = 0 \quad (\text{adiabatic})$$ $$\Delta U = Q – W = 0 \implies \mathbf{T_f = T_i}$$ Free expansion of an ideal gas is strictly isothermal, but it is highly irreversible (entropy increases).
- Trap 4: Cyclic Process Heat vs Work: In any closed cyclic process, $\Delta U_{\text{cycle}} = 0$, which mandates: $$\mathbf{Q_{\text{net}} = W_{\text{net}}}$$ The net heat absorbed over the entire cycle is identically equal to the net mechanical work performed.
2. Four Graded Solved Examples
Example 1: Multi-Path State Transitions on a $P$-$V$ Diagram (JEE Main Level)
Problem Statement: An ideal gas is taken from initial state $A$ ($P_1 = 1.0 \times 10^5\text{ Pa}, V_1 = 2.0 \times 10^{-3}\text{ m}^3$) to final state $B$ ($P_2 = 3.0 \times 10^5\text{ Pa}, V_2 = 6.0 \times 10^{-3}\text{ m}^3$) along three different paths on a $P$-$V$ diagram:
- Path 1 ($ACB$): Isobaric expansion at $P_1$ from $V_1$ to $V_2$, followed by isochoric heating at $V_2$ from $P_1$ to $P_2$. Along this path, the heat absorbed is $Q_{ACB} = 1100\text{ J}$.
- Path 2 ($ADB$): Isochoric pressurization at $V_1$ from $P_1$ to $P_2$, followed by isobaric expansion at $P_2$ from $V_1$ to $V_2$.
- Path 3 ($AB$): Direct straight-line path connecting states $A$ and $B$.
Determine:
- (a) The change in internal energy $\Delta U = U_B – U_A$.
- (b) The work done $W_{ADB}$ and heat absorbed $Q_{ADB}$ along Path 2.
- (c) The work done $W_{AB}$ and heat absorbed $Q_{AB}$ along the direct linear Path 3.
- (d) If the gas returns from state $B$ to state $A$ along a curved path where work done on the gas is $500\text{ J}$ ($W_{BA} = -500\text{ J}$), determine the heat exchange $Q_{BA}$.
Step-by-Step Solution:
Step 1: Calculate work along Path 1 ($ACB$) and determine $\Delta U$:
$$W_{ACB} = W_{AC} + W_{CB} = P_1(V_2 – V_1) + 0$$
$$W_{ACB} = (1.0 \times 10^5\text{ Pa}) \times (6.0 \times 10^{-3} – 2.0 \times 10^{-3}\text{ m}^3) = 1.0 \times 10^5 \times 4.0 \times 10^{-3} = 400\text{ J}$$
From the First Law:
$$\Delta U = U_B – U_A = Q_{ACB} – W_{ACB} = 1100\text{ J} – 400\text{ J} = \mathbf{700\text{ J}}$$
Because internal energy is a state function, $\Delta U = 700\text{ J}$ for all paths connecting $A$ to $B$.
Step 2: Path 2 ($ADB$) analysis:
$$W_{ADB} = W_{AD} + W_{DB} = 0 + P_2(V_2 – V_1) = (3.0 \times 10^5\text{ Pa}) \times (4.0 \times 10^{-3}\text{ m}^3) = \mathbf{1200\text{ J}}$$
$$Q_{ADB} = \Delta U + W_{ADB} = 700\text{ J} + 1200\text{ J} = \mathbf{1900\text{ J}}$$
Step 3: Direct linear Path 3 ($AB$):
The area under a straight-line segment on a $P$-$V$ plane is a trapezoid:
$$W_{AB} = \frac{P_1 + P_2}{2} (V_2 – V_1) = \frac{1.0 \times 10^5 + 3.0 \times 10^5}{2} \times (4.0 \times 10^{-3}) = (2.0 \times 10^5) \times (4.0 \times 10^{-3}) = \mathbf{800\text{ J}}$$
$$Q_{AB} = \Delta U + W_{AB} = 700\text{ J} + 800\text{ J} = \mathbf{1500\text{ J}}$$
Step 4: Return Path ($BA$):
$$\Delta U_{BA} = U_A – U_B = -\Delta U = -700\text{ J}$$
$$Q_{BA} = \Delta U_{BA} + W_{BA} = -700\text{ J} + (-500\text{ J}) = \mathbf{-1200\text{ J}}$$
The negative sign confirms that $1200\text{ J}$ of heat is released to the surroundings.
Key Takeaway: While work varies between $400\text{ J}$, $800\text{ J}$, and $1200\text{ J}$, the change in internal energy remains strictly $700\text{ J}$, anchoring energy conservation across all thermodynamic routes.
Example 2: Work and Heat in a Closed Triangular Cycle
Problem Statement: One mole of a monoatomic ideal gas undergoes a closed cyclic process $A \to B \to C \to A$ on a $P$-$V$ diagram:
- $A \to B$: Isobaric expansion at $P_A = 4.0 \times 10^5\text{ Pa}$ from $V_A = 1.0\text{ L}$ to $V_B = 4.0\text{ L}$.
- $B \to C$: Isochoric depressurization at $V_B = 4.0\text{ L}$ from $P_B = 4.0 \times 10^5\text{ Pa}$ to $P_C = 1.0 \times 10^5\text{ Pa}$.
- $C \to A$: Linear compression returning directly from $(V_B, P_C)$ to $(V_A, P_A)$.
Calculate:
- (a) The work done in each of the three individual steps.
- (b) The net work $W_{\text{net}}$ performed during one complete cycle.
- (c) The net heat $Q_{\text{net}}$ absorbed by the gas per cycle.
Step-by-Step Solution:
Step 1: Convert units to SI:
$$V_A = 1.0\text{ L} = 1.0 \times 10^{-3}\text{ m}^3, \quad V_B = 4.0\text{ L} = 4.0 \times 10^{-3}\text{ m}^3$$
$$\Delta V = V_B – V_A = 3.0 \times 10^{-3}\text{ m}^3$$
Step 2: Calculate work for each step:
Step $A \to B$ (Isobaric expansion):
$$W_{AB} = P_A (V_B – V_A) = (4.0 \times 10^5\text{ Pa}) \times (3.0 \times 10^{-3}\text{ m}^3) = +1200\text{ J}$$
Step $B \to C$ (Isochoric cooling):
$$W_{BC} = 0 \quad (\because dV = 0)$$
Step $C \to A$ (Linear compression):
Volume decreases from $V_B$ to $V_A$, so work is negative (area of trapezoid under $CA$):
$$W_{CA} = -\frac{P_C + P_A}{2} (V_B – V_A) = -\frac{1.0 \times 10^5 + 4.0 \times 10^5}{2} \times (3.0 \times 10^{-3}) = -(2.5 \times 10^5) \times (3.0 \times 10^{-3}) = -750\text{ J}$$
Step 3: Calculate net work $W_{\text{net}}$:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 1200 + 0 – 750 = \mathbf{+450\text{ J}}$$
Geometric Verification: Net work is the area of triangle $ABC$:
$$W_{\text{net}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (V_B – V_A) \times (P_A – P_C) = \frac{1}{2} \times (3.0 \times 10^{-3}) \times (3.0 \times 10^5) = \mathbf{450\text{ J}}$$
Since the cycle is clockwise, $W_{\text{net}} > 0$.
Step 4: Calculate net heat absorbed $Q_{\text{net}}$:
For any complete cycle returning to starting state $A$:
$$\Delta U_{\text{net}} = 0 \implies Q_{\text{net}} = W_{\text{net}} = \mathbf{+450\text{ J}}$$
Key Takeaway: Over a complete cycle, net work equals enclosed geometric area, and total heat absorbed equals net work performed.
Example 3: Internal Energy Change in a Diatomic Gas Transition (JEE Advanced Level)
Problem Statement: Two moles ($n = 2.0\text{ mol}$) of an ideal diatomic gas ($\gamma = 1.40 = 7/5$, $C_v = \frac{5}{2}R$) are subjected to a thermodynamic expansion. The initial state of the gas is characterized by $P_1 = 2.0 \times 10^5\text{ Pa}$ and $V_1 = 0.010\text{ m}^3$. The final state is $P_2 = 1.0 \times 10^5\text{ Pa}$ and $V_2 = 0.030\text{ m}^3$. During this process, the gas performs $W = 1800\text{ J}$ of mechanical work.
- (a) Calculate the change in internal energy $\Delta U$ of the gas.
- (b) Determine the net quantity of heat $Q$ exchanged with the surroundings.
- (c) Find the initial and final temperatures $T_1$ and $T_2$ of the gas (take $R = 8.314\text{ J/(mol}\cdot\text{K)}$).
Step-by-Step Solution:
Step 1: Calculate $\Delta U$ using macroscopic state variables:
For an ideal gas with adiabatic index $\gamma$:
$$\Delta U = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}$$
Compute products $P V$ at both states:
$$P_1 V_1 = (2.0 \times 10^5\text{ Pa}) \times (0.010\text{ m}^3) = 2000\text{ J}$$
$$P_2 V_2 = (1.0 \times 10^5\text{ Pa}) \times (0.030\text{ m}^3) = 3000\text{ J}$$
$$\Delta U = \frac{3000 – 2000}{1.40 – 1} = \frac{1000}{0.40} = \mathbf{+2500\text{ J}}$$
Because $\Delta U > 0$, the temperature of the gas increased.
Step 2: Calculate heat exchanged $Q$:
$$Q = \Delta U + W = 2500\text{ J} + 1800\text{ J} = \mathbf{+4300\text{ J}}$$
The system absorbs $4300\text{ J}$ of heat energy.
Step 3: Calculate initial and final temperatures:
$$T_1 = \frac{P_1 V_1}{n R} = \frac{2000}{2.0 \times 8.314} = \frac{2000}{16.628} \approx \mathbf{120.3\text{ K}}$$
$$T_2 = \frac{P_2 V_2}{n R} = \frac{3000}{16.628} \approx \mathbf{180.4\text{ K}}$$
Key Takeaway: The formula $\Delta U = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}$ bypasses the need to know the specific path equation, delivering the exact internal energy change directly from endpoint state coordinates.
Example 4: Work and Heat Along a Non-Linear Path $P = a V^2$ (Olympiad / JEE Advanced)
Problem Statement: One mole of a monoatomic ideal gas ($\gamma = 5/3$, $C_v = \frac{3}{2}R$) expands from an initial volume $V_1 = 1.0\text{ m}^3$ to a final volume $V_2 = 2.0\text{ m}^3$ according to the non-linear process equation $P = a V^2$, where $a = 10.0\text{ Pa/m}^6$. Determine:
- (a) The work done $W$ by the gas during expansion.
- (b) The change in internal energy $\Delta U$ of the gas.
- (c) The total heat $Q$ absorbed.
- (d) The effective molar heat capacity $C$ of the gas along this path in terms of the universal gas constant $R$.
Step-by-Step Solution:
Step 1: Calculate work done $W$ by direct integration:
$$W = \int_{V_1}^{V_2} P \, dV = \int_{1.0}^{2.0} a V^2 \, dV = \frac{a}{3} \left[ V^3 \right]_{1.0}^{2.0} = \frac{10.0}{3} \left(2.0^3 – 1.0^3\right)$$
$$W = \frac{10.0}{3} (8.0 – 1.0) = \frac{70.0}{3} \approx \mathbf{23.33\text{ J}}$$
Step 2: Calculate endpoint pressures and $\Delta U$:
$$P_1 = a V_1^2 = 10.0 \times (1.0)^2 = 10.0\text{ Pa} \implies P_1 V_1 = 10.0 \times 1.0 = 10.0\text{ J}$$
$$P_2 = a V_2^2 = 10.0 \times (2.0)^2 = 40.0\text{ Pa} \implies P_2 V_2 = 40.0 \times 2.0 = 80.0\text{ J}$$
$$\Delta U = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1} = \frac{80.0 – 10.0}{\frac{5}{3} – 1} = \frac{70.0}{\frac{2}{3}} = 70.0 \times \frac{3}{2} = \mathbf{105.0\text{ J}}$$
Step 3: Calculate total heat absorbed $Q$:
$$Q = \Delta U + W = 105.0 + \frac{70.0}{3} = \frac{315.0 + 70.0}{3} = \frac{385.0}{3} \approx \mathbf{128.33\text{ J}}$$
Step 4: Calculate molar heat capacity $C$:
The temperature change is $\Delta T = \frac{P_2 V_2 – P_1 V_1}{n R} = \frac{70.0}{1.0 \times R} = \frac{70.0}{R}$.
$$C = \frac{Q}{n \Delta T} = \frac{\frac{385.0}{3}}{1.0 \times \left(\frac{70.0}{R}\right)} = \left(\frac{385.0}{210.0}\right) R = \frac{11}{6}R \approx \mathbf{1.833\, R}$$
Key Takeaway: For any arbitrary polytropic or non-linear process, the molar heat capacity $C = C_v + \frac{P \, dV}{n \, dT}$ can be uniquely derived by combining calculus work integration with the first law.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. According to Joule’s law, the internal energy of an ideal gas depends solely upon which of the following variables?
- (A) Pressure only
- (B) Volume only
- (C) Absolute Temperature only
- (D) Both Pressure and Volume
Q2. Which of the following groups consists exclusively of thermodynamic state functions?
- (A) Heat, Work, Internal Energy
- (B) Pressure, Volume, Work
- (C) Internal Energy, Pressure, Temperature, Volume
- (D) Heat, Temperature, Enthalpy
Q3. In an isochoric thermodynamic process performed on an ideal gas, the work done $W$ and heat exchange $Q$ satisfy:
- (A) $W = Q$
- (B) $W = 0, \quad Q = \Delta U$
- (C) $Q = 0, \quad W = -\Delta U$
- (D) $\Delta U = 0, \quad Q = W$
Q4. An ideal gas completes one full cycle of an expansion and compression process represented by a closed loop on a standard $P$-$V$ diagram. Over the entire cycle:
- (A) The change in internal energy is zero, and net heat absorbed equals net work performed.
- (B) Net work done must always be zero.
- (C) Net heat exchanged is always zero.
- (D) The change in internal energy equals the enclosed area.
Q5. During an adiabatic expansion of an ideal gas, the work done by the gas is positive ($W > 0$). Consequently, the internal energy and temperature of the gas must:
- (A) Increase; Increase
- (B) Decrease; Decrease
- (C) Remain unchanged; Remain unchanged
- (D) Decrease; Increase
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. Which of the following statements regarding heat, work, and internal energy are fundamentally correct?
- (A) $\delta Q$ and $\delta W$ are inexact differentials, whereas $dU$ is an exact differential.
- (B) For any arbitrary cyclic process, $\oint dU = 0$ and $\oint \delta Q = \oint \delta W$.
- (C) Even though $Q$ and $W$ depend on the path, the quantity $(Q – W)$ is strictly path-independent.
- (D) A system can contain a definite quantity of heat in a given equilibrium state.
Q7. For an ideal gas undergoing a quasi-static transition from state $(P_1, V_1)$ to state $(P_2, V_2)$, the change in internal energy $\Delta U$ can be written as:
- (A) $\Delta U = n C_v \Delta T$ (valid only for an isochoric process)
- (B) $\Delta U = n C_v \Delta T$ (valid for any arbitrary process)
- (C) $\Delta U = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}$ (valid for any arbitrary process)
- (D) $\Delta U = \frac{f}{2}(P_2 V_2 – P_1 V_1)$ (where $f$ is the degree of freedom)
Q8. An ideal gas expands from volume $V_1$ to volume $V_2$ along three different paths: (1) Isobaric, (2) Isothermal, and (3) Adiabatic. Starting from the exact same initial state $(P_1, V_1)$:
- (A) The work done satisfies $W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$.
- (B) The final temperature is lowest for the adiabatic path.
- (C) The internal energy remains unchanged along the isothermal path.
- (D) The heat absorbed is zero along the isochoric path.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. In a thermodynamic process, the internal energy of a system increases by $300\text{ J}$ ($\Delta U = +300\text{ J}$) while the system simultaneously releases $120\text{ J}$ of heat to its surroundings ($Q = -120\text{ J}$). Calculate the mechanical work done on the system (in Joules).
Q10. An ideal gas is taken from state $A$ to state $C$ along path $ABC$, during which it absorbs $Q_{ABC} = 80.0\text{ J}$ of heat while performing $W_{ABC} = 30.0\text{ J}$ of work. If the gas is instead taken from $A$ to $C$ along an alternative path $ADC$ where the work done by the gas is $W_{ADC} = 10.0\text{ J}$, calculate the heat $Q_{ADC}$ (in Joules) absorbed by the gas along path $ADC$.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (C) | Joule’s Law of Internal Energy |
| Q2 | (C) | State vs Path Functions |
| Q3 | (B) | Isochoric Work-Energy Relations |
| Q4 | (A) | Cyclic Work-Heat Equivalence |
| Q5 | (B) | Adiabatic Expansion Cooling |
| Q6 | (A), (B), (C) | First Law Differentials & Exactness |
| Q7 | (B), (C), (D) | Internal Energy Invariant Formulations |
| Q8 | (A), (B), (C) | Work Comparison Across Expansion Paths |
| Q9 | 420 | Work Done ON System Calculation |
| Q10 | 60 | Multi-Path Heat Exchange Calculation |
Step-by-Step Solutions
Solution 1:
In an ideal gas, intermolecular potential energy is identically zero ($E_p = 0$). Internal energy consists purely of kinetic energy, which depends strictly on absolute temperature $T$: $U = f(T)$. Correct Option: (C).
Solution 2:
A state function depends only on the current state coordinates ($P, V, T, U$). Heat ($Q$) and work ($W$) are energy transfers that exist only during a process and are path functions. Hence, internal energy, pressure, temperature, and volume are all state functions. Correct Option: (C).
Solution 3:
In an isochoric process, volume is held constant ($dV = 0$), so $W = \int P \, dV = 0$. By the first law, $Q = \Delta U + W \implies Q = \Delta U$. Correct Option: (B).
Solution 4:
For any cyclic process, the system returns to its initial thermodynamic state ($U_f = U_i$), so $\Delta U_{\text{net}} = 0$. By the First Law, $Q_{\text{net}} = \Delta U + W_{\text{net}} = W_{\text{net}}$. Correct Option: (A).
Solution 5:
In an adiabatic process, $Q = 0$. The first law states $Q = \Delta U + W \implies \Delta U = -W$. During expansion, $W > 0 \implies \Delta U < 0$. Since $U = n C_v T$, a decrease in internal energy causes a corresponding decrease in temperature (cooling effect). Correct Option: (B).
Solution 6:
- (A) is Correct: Heat and work depend on path (inexact), whereas internal energy is a point property (exact differential).
- (B) is Correct: The integral of any exact differential over a closed cycle is zero ($\oint dU = 0$), implying $\oint \delta Q = \oint \delta W$.
- (C) is Correct: By the First Law, $Q – W = \Delta U$, which is path-independent.
- (D) is Incorrect: Heat is energy in transit, not a stored property.
Correct Options: (A), (B), (C).
Solution 7:
- (A) is Incorrect: The formula is NOT restricted to isochoric processes; it applies to all processes.
- (B) is Correct: For an ideal gas, $U$ depends only on $T$, so $\Delta U = n C_v \Delta T$ is universally valid.
- (C) is Correct: $n C_v \Delta T = n \left(\frac{R}{\gamma – 1}\right) \Delta T = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1}$.
- (D) is Correct: Since $C_v = \frac{f}{2}R$, $\Delta U = \frac{f}{2} n R \Delta T = \frac{f}{2}(P_2 V_2 – P_1 V_1)$.
Correct Options: (B), (C), (D).
Solution 8:
- (A) is Correct: On a $P$-$V$ diagram, the isobaric curve is horizontal (highest pressure), isothermal drops as $1/V$, and adiabatic drops steepest as $1/V^\gamma$. Area under the curve: $W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$.
- (B) is Correct: In adiabatic expansion, temperature drops ($T_f < T_i$), while in isothermal expansion $T_f = T_i$, and in isobaric expansion $T_f = T_i(V_2/V_1) > T_i$.
- (C) is Correct: For isothermal process of an ideal gas, $\Delta T = 0 \implies \Delta U = 0$.
- (D) is Incorrect: In an isochoric process, heat absorbed is $Q = n C_v \Delta T \neq 0$ (work is zero, not heat).
Correct Options: (A), (B), (C).
Solution 9:
By the First Law of Thermodynamics:
$$Q = \Delta U + W_{\text{by}}$$
$$-120\text{ J} = 300\text{ J} + W_{\text{by}} \implies W_{\text{by}} = -120 – 300 = -420\text{ J}$$
The work done on the system is the negative of work done by the system:
$$W_{\text{on}} = -W_{\text{by}} = -(-420\text{ J}) = \mathbf{420\text{ J}}$$
Final Answer: 420
Solution 10:
Along path $ABC$:
$$\Delta U = U_C – U_A = Q_{ABC} – W_{ABC} = 80.0\text{ J} – 30.0\text{ J} = 50.0\text{ J}$$
Along alternative path $ADC$ connecting the same endpoints:
$$\Delta U = 50.0\text{ J}$$
$$Q_{ADC} = \Delta U + W_{ADC} = 50.0\text{ J} + 10.0\text{ J} = \mathbf{60.0\text{ J}}$$
Final Answer: 60