JEE Physics Thermodynamics Notes Set 03

Gas Laws and Kinetic Theory

1. Ideal gas: \( PV=nRT=NkT \). At constant mass: Boyle's law \( PV=\text{const} \) (T fixed), Charles's law \( V\propto T \) (P fixed). Density \( \rho=\dfrac{PM}{RT} \), so at constant pressure \( \rho T \) is constant.
2. Average translational kinetic energy per molecule \( =\dfrac{3}{2}kT \), the same for every gas at the same temperature. \( v_{rms}=\sqrt{\dfrac{3RT}{M}} \).
3. Each degree of freedom carries \( \dfrac{1}{2}kT \) per molecule. Monatomic gas f = 3, diatomic f = 5 (vibration frozen), and \( C_V=\dfrac{f}{2}R \), \( C_P=\left(\dfrac{f}{2}+1\right)R \), \( \gamma=1+\dfrac{2}{f} \).
4. Speed of sound in a gas \( v=\sqrt{\dfrac{\gamma RT}{M}} \), so \( \dfrac{v}{v_{rms}}=\sqrt{\dfrac{\gamma}{3}} \).

First Law of Thermodynamics

\( \Delta Q=\Delta U+W \), where W is the work done by the gas. Internal energy U is a state function: \( \Delta U=nC_V\Delta T \) for an ideal gas in any process. Heat and work depend on the path. Over a complete cycle \( \Delta U=0 \), so the net heat equals the net work, and the work is the area of the loop on the P-V diagram (positive if clockwise).

Processes of an Ideal Gas

Heat Engines and Refrigerators

1. Efficiency of an engine \( \eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1} \). A Carnot engine between kelvin temperatures \( T_1 \) and \( T_2 \) has \( \eta=1-\dfrac{T_2}{T_1} \), the maximum possible.
2. Coefficient of performance of a refrigerator \( =\dfrac{Q_2}{W}=\dfrac{T_2}{T_1-T_2} \) for the Carnot case.
3. Free expansion of a gas into vacuum: Q = 0, W = 0, so \( \Delta U=0 \) (the temperature of an ideal gas does not change).

Worked Examples

1. A gas follows \( P\propto V^{2} \) and is diatomic. Then x = −2 and \( C=\dfrac{5R}{2}+\dfrac{R}{1+2}=\dfrac{17R}{6} \).
2. A Carnot engine works between 500 K and 375 K and takes 600 kcal per cycle. \( \eta=0.25 \), work = 150 kcal, heat rejected = 450 kcal.
3. A mixture of 2 mol of O\(_2\) and 4 mol of Ar at temperature T has \( U=2\times\dfrac{5}{2}RT+4\times\dfrac{3}{2}RT=11RT \).
4. A monatomic gas receives Q and does \( Q/2 \) of work. Then \( \Delta U=Q/2=\dfrac{3}{2}R\Delta T \) per mole, so \( C=\dfrac{Q}{\Delta T}=3R \).