JEE Physics Thermodynamics Notes Set 04

Systems, Equilibrium and the Zeroth Law

A thermodynamic system is a large collection of molecules inside a boundary, described by its pressure, volume and temperature. An open system exchanges both matter and energy with its surroundings, a closed system exchanges only energy, and an isolated system exchanges neither. The zeroth law says that if two bodies are each in thermal equilibrium with a third, they are in thermal equilibrium with each other, which is what allows temperature to be defined. For an ideal gas the state variables are linked by \( PV=nRT \).

Work Done by a Gas

When a gas pushes a piston through a small distance, the work is \( dW=P\,dV \), so the work between two states is \( W=\int P\,dV \), the area under the P–V curve. It is positive for expansion and negative for compression. For a cycle the net work equals the area enclosed by the loop: positive if the loop is traversed clockwise and negative if anticlockwise. Work depends on the path, not just on the end states. For a gas expanding against a spring and the atmosphere, the work done by the gas is the sum of the energy given to the spring, to the atmosphere and to lifting the piston.

Standard Processes

Isothermal: \( T \) constant, \( PV=\text{const} \), \( \Delta U=0 \) and \( W=nRT\ln\dfrac{V_f}{V_i} \). Isochoric: \( V \) constant, \( W=0 \) and \( Q=\Delta U=nC_v\Delta T \). Isobaric: \( P \) constant, \( W=P\Delta V=nR\Delta T \) and \( Q=nC_p\Delta T \). Adiabatic: \( Q=0 \), \( PV^{\gamma}=\text{const} \), \( TV^{\gamma-1}=\text{const} \), \( W=\dfrac{P_iV_i-P_fV_f}{\gamma-1} \). On a P–V diagram an adiabatic curve is steeper than an isotherm through the same point, which is why an adiabatic expansion ends cooler and at lower pressure than an isothermal one.

First Law of Thermodynamics

The first law is conservation of energy for heat and work: \( Q=\Delta U+W \). Heat absorbed by the system, work done by the system and increase of internal energy are taken as positive. Internal energy is a state function, so \( \Delta U \) is the same along every path joining two states, while Q and W change with the path. For an ideal gas \( \Delta U=nC_v\Delta T \) in every process. In a cycle \( \Delta U=0 \), so the net heat absorbed equals the net work done. In a free expansion into a vacuum, Q = 0 and W = 0, so the temperature of an ideal gas does not change.

Specific Heats of Gases

The molar heat capacities of an ideal gas are \( C_v=\dfrac f2R \) and \( C_p=C_v+R \), so \( \gamma=\dfrac{C_p}{C_v}=1+\dfrac2f \). A monatomic gas has \( f=3 \) and \( \gamma=5/3 \), a diatomic gas \( f=5 \) and \( \gamma=7/5 \). Because the heat needed depends on the process, a gas has infinitely many heat capacities. At constant pressure the extra heat goes into the work of expansion, which is why \( C_p>C_v \). For a solid the expansion is so small that \( C_p \) is only slightly larger than \( C_v \). When gases at different temperatures are mixed without loss of energy, the final temperature is the weighted average of the initial temperatures, with weights \( nC_v \).

Polytropic Processes

A process obeying \( PV^n=\text{const} \) has molar heat capacity \( C=C_v+\dfrac{R}{1-n} \). Setting n = 0, 1, \( \gamma \) and \( \infty \) gives the isobaric, isothermal, adiabatic and isochoric processes. When \( 1\lt n\lt \gamma \) the heat capacity is negative: the gas absorbs heat while its temperature falls. The bulk modulus in such a process is \( K=nP \) for \( PV^n=\text{const} \); it equals P for an isothermal change and \( \gamma P \) for an adiabatic one. For a general process such as \( T=T_0+aV^3 \), write the pressure from the gas law and use calculus to find extremes.

Heat Engines, Refrigerators and the Second Law

A heat engine takes heat \( Q_1 \) from a hot source, converts part to work W and rejects \( Q_2 \) to a cold sink; its efficiency is \( \eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1} \). A reversible Carnot engine between absolute temperatures \( T_1>T_2 \) has \( \eta=1-\dfrac{T_2}{T_1} \), the maximum possible. A refrigerator runs the cycle backwards, and its coefficient of performance is \( \dfrac{Q_L}{W}=\dfrac{T_L}{T_H-T_L} \). The second law forbids heat flowing by itself from a cold body to a hot one and forbids an engine with 100 % efficiency. A refrigerator with its door open heats the room, because the work supplied is also released as heat.

Worked Example

Two moles of a monatomic gas at 300 K are compressed adiabatically to 400 K. Since \( Q=0 \), the work done by the gas is \( W=-\Delta U=-nC_v\Delta T=-2\times\tfrac32\times\tfrac{25}{3}\times100=-2500 \) J, so 2500 J of work is done on the gas. Another example: a Carnot engine takes heat at 627 °C and rejects it at 27 °C, so \( T_1=900 \) K, \( T_2=300 \) K and \( \eta=1-\dfrac{300}{900}=\dfrac23 \). If it absorbs 3000 J per cycle it delivers 2000 J of work and rejects 1000 J.