Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.2

Adiabatic Change.

Pump air into a bicycle tyre hard and fast and the barrel of the pump warms up. You never held it to a flame and you gave it no heat. So where did the warmth come from? This note answers that, with three relations and the sums that go with them.

1. Warm without any heat given

A process in which the system neither takes in heat nor gives it out is an adiabatic process. No heat comes or goes, yet the pressure and volume of the gas change, and so does its temperature.

Compress a gas suddenly and the heat produced has no time to escape, so the temperature rises. Expand it suddenly and the gas loses heat, so the temperature falls. No heat is exchanged with the outside, yet the internal energy still rises or falls. In short: ΔQ = 0 but ΔT ≠ 0, and with no heat exchanged the entropy stays unchanged. If 350 J of work is done on the gas, ΔW = −350 J, so Δu = +350 J. Take the start as 298 K (25°C), 1 atm, γ = 1.4, and drag the handle to shrink the volume.

The word adiabatic means heat not passing through: heat does not enter the system and the system does not give heat out.

2. The relations, the curve and the conditions

Boyle's law does not apply to an adiabatic change. The relation between pressure and volume is PV^γ = constant, and that between temperature and volume is TV^γ−1 = constant.

The result shows up on a graph. Take one gas, one start (4 atm, 10 L, 300 K), and double the volume to 20 L. Once slowly, taking in heat (isothermal). Once fast, with no heat (adiabatic, γ = 1.4). Drag the handle to compare the paths: the adiabatic curve is steeper, so the pressure falls further for the same expansion.

An adiabatic change needs three conditions. One, the gas sits in a non-conducting container. Two, the medium around the container has a low heat capacity. Three, the pressure changes so quickly that heat has no chance to be exchanged with the outside.

So an adiabatic change is a very fast process. The temperature changes in it, and the container must be a heat non-conductor.

3. Where PV^γ = K comes from

Give one mole of ideal gas dQ of heat and the first law says dQ = C_V dT + P dV (1.7). Adiabatically dQ = 0, so C_V dT + P dV = 0 (1.8). The rest goes one step at a time.

The integral is ln P + γ ln V = constant, that is PV^γ = K (1.9). Between the initial and final states P₁V₁^γ = P₂V₂^γ = constant (1.10), where γ = C_P/C_V.

Put P = RT/V into PV^γ = constant and you get the relation between volume and temperature: TV^γ−1 = constant. Between two states T₁V₁^γ−1 = T₂V₂^γ−1.

Put V = RT/P and you get the relation between pressure and temperature: T^γP^1−γ = constant, that is TP^(1−γ)/γ = constant.

The slopes explain the steepness. From PV^γ = K, dP/dV = −γP/V, so the adiabatic curve has slope −γP/V. The isothermal curve has slope −P/V. The ratio of the two, isothermal over adiabatic, is 1/γ. Since γ is greater than 1, the adiabatic curve is the steeper one.

4. The sums: one law, different numbers

First a small sum. In an adiabatic change (γ = 1.4) of a diatomic gas, raising the pressure by 0.5% gives V₁ = V × (1/1.005)^1/1.4 ≈ 0.9964 V. So the volume falls by about 0.36%.

The first three of the book's Example 1.4 are the same sum. Put the temperature in kelvin first. (1) Air at 25°C = 298 K is compressed adiabatically to half its volume: T₂ = 2^0.4 × 298 = 393.18 K = 120.18°C, and P₂ = 2^1.4 × 1 = 2.64 atm. (2) For air at 100°C, T₂ = 373 × 2^0.4 = 492.18 K, a rise of 119.18°C. (3) Air expands adiabatically to three times its volume: P₂ = P₁/3^1.4 = 1.013 × 10⁵/4.656 = 2.176 × 10⁴ N m⁻². Change the numbers in the calculator below.

Example 1.4(4): on the Carnot graph P₁ = 3 atm, T₁ = 600 K, V₁ = 2 × 10⁻³ m³, V₂ = 6 × 10⁻³ m³, T₂ = 300 K. Isothermally P₂ = P₁V₁/V₂ = 3 × 2/6 = 1 atm. Then adiabatically T₁V₂^γ−1 = T₂V₃^γ−1 (γ = 1.41), so V₃ = V₂ × 2^1/0.41 ≈ 0.0325 m³. Note that V₂ goes in here, not V₁.

Example 1.4(5): gas at 300 K, 10⁶ Pa and 0.001 m³ expands first isothermally and then adiabatically, doubling its volume each time (γ = 1.4). With nRT = PV = 1000 J, the isothermal work is W₁ = 1000 ln 2 = 693 J. Adiabatically T₂ = 300 × (1/2)^0.4 = 227.4 K. Since nR = PV/T = 3.33 J/K, W₂ = nR(T₁ − T₂)/(γ − 1) = 3.33 × 72.6/0.4 = 605 J. The total work is 693 + 605 = 1298 J.

5. Check yourself

The first question is in the exam's style.

In the second, watch the exponents.

The third checks a condition or a calculation.

The whole note on eight cards.

Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.3.2.