Chapter 11: Thermodynamics
Unit VIII – Thermodynamics
11.1 Thermal Equilibrium and Zeroth Law
Thermal equilibrium: Two bodies are in thermal equilibrium when they are at the same temperature and no net heat flows between them.
Zeroth Law of Thermodynamics:
If body A is in thermal equilibrium with body C, and body B is also in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other.
This defines temperature as a fundamental measurable quantity.
11.2 Internal Energy, Heat, and Work
- Internal energy (U): Total kinetic + potential energy of all molecules of a system.
- Heat (Q): Energy transferred due to temperature difference between system and surroundings.
- Work done by gas (W): \(W = \int P,dV\) = area under P-V curve.
Sign Convention:
- Q positive: heat given to the system
- W positive: work done by the system
11.3 First Law of Thermodynamics
The change in internal energy of a system equals the heat supplied to it minus the work done by it.
\[ \boxed{\Delta U = Q - W} \]
Consequences:
- For a cyclic process: \(\Delta U = 0 \Rightarrow Q = W\)
- For isochoric (\(\Delta V = 0\)): \(W = 0 \Rightarrow \Delta U = Q\)
- For isobaric (\(\Delta P = 0\)): \(W = P\Delta V\)
11.4 Thermodynamic Processes
11.4.1 Isothermal Process (constant temperature)
For ideal gas: \(PV = \text{constant}\) (Boyle’s Law)
\[ W = nRT\ln\frac{V_2}{V_1} \]
Since \(T\) is constant: \(\Delta U = 0 \Rightarrow Q = W\)
11.4.2 Adiabatic Process (no heat exchange, Q = 0)
\[ PV^\gamma = \text{constant}, \quad TV^{\gamma-1} = \text{constant} \]
\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1} \]
Since \(Q = 0\): \(\Delta U = -W\)
11.4.3 Isochoric Process (constant volume)
\[ W = 0 \Rightarrow \Delta U = Q = nC_V\Delta T \]
11.4.4 Isobaric Process (constant pressure)
\[ W = P\Delta V = nR\Delta T \]
\[ Q = nC_P\Delta T, \quad \Delta U = nC_V\Delta T \]
Table: \(C_P - C_V = R\) and \(\gamma = C_P/C_V\)
11.5 P-V Diagrams
11.6 Second Law of Thermodynamics
Kelvin-Planck Statement:
No heat engine can convert heat entirely into work without rejecting some heat to a colder reservoir.
Clausius Statement:
Heat cannot spontaneously flow from a colder to a hotter body.
Both statements are equivalent.
11.7 Heat Engine and Efficiency
A heat engine takes heat \(Q_1\) from a hot source, does work \(W\), and rejects heat \(Q_2\) to a cold sink:
\[ W = Q_1 - Q_2 \]
\[ \eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} \]
Carnot engine (maximum efficiency):
\[ \eta_{Carnot} = 1 - \frac{T_2}{T_1} \]
where \(T_1\) = temperature of hot source, \(T_2\) = temperature of cold sink (in Kelvin).
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(\Delta U = Q - W\) | First Law |
| \(W = \int P,dV\) | Work done by gas |
| \(PV^\gamma = \text{const}\) | Adiabatic process |
| \(W_{iso} = nRT\ln(V_2/V_1)\) | Isothermal work |
| \(\eta = 1 - T_2/T_1\) | Carnot efficiency |
| \(C_P - C_V = R\) | Mayer’s relation |
Practice Questions
Section A – MCQ (1 mark each)
Q1. In an isothermal expansion of an ideal gas:
(a) Temperature increases (b) Internal energy increases (c) Internal energy remains constant (d) Heat exchanged is zero
Answer
(c) Internal energy remains constant — For ideal gas, U depends only on T; isothermal means T = const.
Q2. A Carnot engine works between 500 K and 300 K. Its efficiency is:
(a) 20% (b) 30% (c) 40% (d) 60%
Answer
(c) 40% — \(\eta = 1 - 300/500 = 1 - 0.6 = 0.4 = 40%\)
Section B – Short Answer (2–3 marks)
Q3. In an adiabatic process, a gas is compressed from V to V/2. If \(\gamma = 1.4\) and initial pressure is \(10^5\) Pa, find the final pressure.
Answer
\(PV^\gamma = P’V’^\gamma\)
\[ P’ = P\left(\frac{V}{V’}\right)^\gamma = 10^5 \times (2)^{1.4} = 10^5 \times 2.639 \approx \mathbf{2.64 \times 10^5 \text{ Pa}} \]
Section D – Competency-Based Questions
Q4. (Case Study) A gas undergoes the cycle: Isothermal expansion → Isochoric cooling → Isobaric compression → Isochoric heating.
(i) Sketch the P-V diagram.
(ii) In which process is work done by the gas positive?
(iii) State the First Law for the adiabatic process.
(iv) Why is the efficiency of a heat engine always less than 100%?
Answer
(i) P-V diagram: A→B is hyperbola (isothermal expand), B→C is vertical line going down (isochoric cool), C→D is horizontal line going left (isobaric compress), D→A is vertical line going up (isochoric heat).
(ii) Work done by gas is positive during isothermal expansion (A→B) — gas expands, does positive work.
Work done during isobaric compression is negative (work done on gas).
(iii) For adiabatic: \(Q = 0 \Rightarrow \Delta U = -W\) — if gas is compressed adiabatically, internal energy increases (temperature rises).
(iv) By the second law of thermodynamics, some heat (\(Q_2\)) must always be rejected to the cold sink. Hence, \(\eta = 1 - Q_2/Q_1 \lt 1\) always.