AP Physics B – Thermodynamics –Equations to be Remembered

Essential points to be remembered in respect of kinetic theory of gases were discussed in the post dated 13^th March 2008. It will be useful to have a glance at that post at the moment since you will have to deal with different types of gases when you consider different thermodynamic processes. You can access that post by clicking here

Here are the important points you need to remember in the section ‘thermodynamics’.

(1) Zeroth law of thermodynamics:

If two systems are separately in thermal equilibrium with a third system, then the two systems should be in thermal equilibrium.

(2) First law of thermodynamics:

In the case of any system, the energy (ΔQ) supplied to the system is used partly to increase the internal energy (ΔU) of the system and the rest to do external work (ΔW). Therefore,

ΔQ = ΔU + ΔW

(3) Isothermal process is one in which the temperature of the system is kept constant throughout.

In the case of an isothermal process of an ideal gas, PV = constant. (This is Boyle’s law)

If an ideal gas undergoes an isothermal change (at fixed temperature T) from its initial state (P₁, V₁) to its final state(P₂, V₂), the work done is given by

W = _V1 ∫^V2 PdV = _V1 ∫^V2 (RT/V) dV since PV = RT so that P = RT/V. [We have considered one mole of the gas and hence used the universal gas constant R].

Thus W = RT ln(V₂/V₁)

[Note that work is done only if the volume of the gas changes].

(4) Adiabatic process is one in which no heat flows between the system and the surroundings (ΔQ = 0).

In the case of an adiabatic process of an ideal gas, pressure and volume are related as

PV^γ = constant where γ is the ratio of specific heats of the gas.

Adiabatic expansion of a gas will result in lowering of the temperature of the gas where as adiabatic compression will result in rise in temperature of the gas. The volume and the temperature in an adiabatic change are related as

TV ^(γ–1) = constant

Pressure and temperature in an adiabatic change are related as

P^(1–γ) T ^γ = constant

The work done when one mole of an ideal gas undergoes an adiabatic process is given by

W = [1/(γ–1)][P₁V₁ – P₂V₂]

This can be written as W = [R/(γ–1)][T₁ – T₂] = C_v[T₁ – T₂]

(5) Isobaric process is one in which the pressure is constant. The work done by a gas in an isobaric process is given by

W = P(V₂ – V₁)

(6) Isochoric process is one in which the volume of the gas remains constant. Therefore, no work is done in an isochoric process.

(7) Cyclic process is one in which the system returns to its initial state. The change in internal energy (ΔU) is zero in a cyclic process.

(8) P–V diagram is the graph obtained by plotting the pressure P on the Y-axis and the volume V on the X-axis. Since the work done is ∫PdV, the area under a PV curve gives the work done.

In the case of a cyclic process, the P-V diagram will be a closed curve and the area enclosed by the curve will give the net amount of work done. Note that if the direction of the process as indicated in the P-V diagram of the cyclic process is clockwise, net amount of work is done by the gas. If the direction is anticlockwise, net amount of work is done on the gas.

(9) P–V diagram of a Carnot engine is shown in the adjoining figure.

The curve AB indicates the isothermal expansion of the working substance (ideal gas) at the (higher) source temperature T₁ and the pressure-volume values change from P₁,V₁ to P₂, V₂. The curve BC indicates the adiabatic expansion of the working substance and its state changes from P₂, V₂, T₁ (at B) to P₃, V₃, T₂ (at C).

The curve CD indicates the isothermal compression of the working substance (ideal gas) at the (lower) sink temperature T₂ and the pressure-volume values change from P₃, V₃ to P₄, V₄. The curve DA indicates the adiabatic compression of the working substance and its state changes from P₄, V₄, T₂ (at D) to P₁, V₁, T₁ (at A).

The net amount of work done by the working substance is equal to the area ABCDA

(10) Efficiency (η) of Carnot engine is given by

η = (Q₁ – Q₂)/Q₁ = (T₁ – T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at the source temperature T₁ and Q₂ is the quantity of heat liberated to the sink at the sink temperature T₂.

The maximum possible efficiency of one (100%) can be obtained only if the temperature of the sink is 0 K (which cannot be attained).

[Note that the temperatures in the above expression should be in degrees Kelvin].

(11) Coefficient of performance (β) of refrigerator is given by

β = Heat removed from the cold reservoir/ Work done by the pump

Note that a refrigerator is a heat engine operated in the reverse manner. The coefficient of performance is therefore given by

β = Q₂/ W = Q₂/ (Q₁ – Q₂) = T₂/(T₁ – T₂) where Q₂ is the heat taken from the cold reservoir, Q₁ is the higher amount of heat rejected to the surroundings by doing work W (by satisfying the equation, Q₁ = Q₂ + W), T₂ is the temperature of the cold reservoir and T₁ is the temperature of the surroundings.

An ideal refrigerator will have coefficient of performance equal to infinity (which is impossible since heat will have to be transferred from cold body to hot surroundings without the help of external energy source, in violation of the second law of thermodynamics).

We will discuss questions in this section in the next post.

Meanwhile, find some useful multiple choice questions at the following locations:
physicsplus: Heat engine and Refrigerator

physicsplus: Questions on Heat & Thermodynamics