Thermodynamics

CBSE Class 11 Physics â€¢ Chapter 11

Chapter Overview

This chapter covers the fundamental concepts of Thermodynamics. Ensure you understand the definitions and derivations thoroughly.

11.1 Laws of Thermodynamics

Zeroth Law: If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other. This leads to the concept of Temperature.

11.2 First Law of Thermodynamics

Statement: Energy can neither be created nor destroyed. The heat supplied to a system is used partly to increase its internal energy and partly to do work against surroundings.

$$ \Delta Q = \Delta U + \Delta W $$

$\Delta Q$: Heat supplied (+ve if gained, -ve if lost in chemistry convention, but here +ve if absorbed).
$\Delta U$: Change in Internal Energy ($U \propto T$).
$\Delta W$: Work done by the system ($+P\Delta V$).

11.3 Thermodynamic Processes

Figure 11.1: Thermodynamic Processes on P-V Diagram

Isothermal: Temperature constant ($T=const$). $\Delta U = 0$. $\Delta Q = \Delta W$. ($PV = \text{const}$). Work $W = nRT \ln(V_2/V_1)$.
Adiabatic: No heat exchange ($\Delta Q = 0$). $\Delta U = - \Delta W$. ($PV^\gamma = \text{const}$). Work $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
Isobaric: Pressure constant. $W = P(V_2 - V_1)$.
Isochoric: Volume constant. $W = 0$. $\Delta Q = \Delta U$.

Practice Problem 1

Q: A gas expands from volume $V_1$ to $V_2$ isothermally. If the temperature is $300 K$, find work done for 2 moles doubling the volume.

Ans: Isothermal Work $W = 2.303 nRT \log(V_2/V_1)$.
$W = 2.303 \times 2 \times 8.31 \times 300 \times \log(2)$.
$W = 11486 \times 0.3010 \approx \mathbf{3457 \text{ J}}$.

11.4 Mayer's Relation

For an ideal gas, $C_p - C_v = R$.

Monoatomic: $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$, $\gamma = 1.67$.
Diatomic: $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, $\gamma = 1.4$.

11.5 Heat Engines & Second Law

Heat Engine: A device that converts heat energy into mechanical work along a cyclic process.

Efficiency ($\eta$): $\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}$
$Q_1$: Heat absorbed from Source at $T_1$.
$Q_2$: Heat rejected to Sink at $T_2$.

Second Law of Thermodynamics:

Kelvin-Planck Statement: No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. ($\eta < 100\%$).
Clausius Statement: No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. (External work is needed).

11.6 Carnot Engine

Figure 11.2: Heat Engine Schematic

Carnot Efficiency: For a reversible engine operating between temperatures $T_1$ and $T_2$:

$$ \eta = 1 - \frac{T_2}{T_1} $$

Note: Temperatures must be in Kelvin. Efficiency is always $< 1$.

Refrigerator (Coefficient of Performance $\beta$): $\beta = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$.

Practice Problem 2 (Carnot)

Q: A Carnot engine absorbs 1000 J of heat from a reservoir at $127^\circ C$ and rejects 600 J of heat. Find the efficiency and sink temperature.

Ans: $Q_1 = 1000 J$, $Q_2 = 600 J$. $T_1 = 127 + 273 = 400 K$.
Efficiency $\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{600}{1000} = 1 - 0.6 = \mathbf{0.4 \text{ or } 40\%}$.
$\eta = 1 - \frac{T_2}{T_1} \Rightarrow 0.4 = 1 - \frac{T_2}{400} \Rightarrow \frac{T_2}{400} = 0.6$.
$T_2 = 0.6 \times 400 = 240 K = -33^\circ C$.

Important Question Bank

Q3. Conceptual

Question: Can a room be cooled by leaving the door of an electric refrigerator open?

Answer: No. A refrigerator pumps heat from inside to outside (room). If the door is open, it takes heat from the room and releases it back into the room plus the work done by the compressor (heat). Thus, the room temperature will actually increase.

Study Tip: Master the derivations in this chapter as they are frequently asked in exams. Practice numericals based on the formulas.