Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.5-1.6

Internal Energy, Heat and Work.

Blow up a balloon and squeeze it between your hands. It pushes back from the inside. Behind that push is energy stored in the gas, ready to do work. Keeping the books on this energy, on heat and on work is what the first law of thermodynamics is about.

1. The energy inside

Every system holds a definite amount of energy that can do work or change into another form. That is why a gas-filled balloon pushes back on your hand. This energy is the internal energy. By definition, internal energy is the kinetic energy of the molecules and atoms of a body together with the energy due to the intermolecular forces between them.

Internal energy is the sum of two parts: thermal energy, the kinetic energy of the randomly moving molecules, and molecular potential energy. In symbols, E = K.E. + P.E.

2. Heat is not internal energy

When heat flows from a hot body to a cold one, the internal energy of the hot body falls and that of the cold body rises. The word heat names this transfer of energy.

So a body holds internal energy, not heat; saying that a body contains heat is not correct. One more point: the total internal energy of a body cannot be measured, but when heat is applied its change can be measured exactly.

3. How internal energy changes

The first law gives dQ = du + dW = du + PdV, so du = dQ − PdV. At constant volume dV = 0 and du = dQ: the rise of internal energy equals the heat supplied.

For m moles at constant volume, raising the temperature by dT needs dQ = mC_V dT; for 1 mole, du = C_V dT. The volume does not have to stay constant, though, because the internal energy depends only on temperature.

This is Mayer's hypothesis: the internal energy of a fixed quantity of gas depends only on its temperature, not on its pressure or volume. So if the pressure or volume changes but the temperature stays fixed, the internal energy does not change.

The change of internal energy depends on the initial and final states of the system, not on the path taken to reach the final state. In a reversible or cyclic process the working body returns to its initial state, so the change of internal energy is zero.

Here is a familiar case of heat appearing. When a bullet hits its target, its speed is checked and part of its kinetic energy changes into heat, so the bullet and the target both warm up.

4. When a gas does work

With gas pressure P and piston area A, the force on the piston is F = PA. When the piston moves dx, the work is dW = F dx = PA dx = P dV (1.19), since A dx = dV. Work = pressure × change of volume, and it is called the external work.

The work done by a gas in expansion is positive, and in compression it is negative.

At constant pressure, with the volume going from V₁ to V₂, the work done by the gas is dW = P(V₂ − V₁) (1.20).

Take pressure in N m⁻² and volume in m³ and the work comes out in joules (J).

For a constant-pressure process the relation among heat, internal energy and work is dQ = du + PdV (1.22), or dQ = du + P(V₂ − V₁) (1.23).

5. Work is an area

A gas goes from A to B. Move B and change the path: the work is the area under the path on the P-V graph.

The P-V graph is called the indicator diagram. The work done in any thermal process equals the area under the curve on it (CABD in the book's figure 1.11). Below are two pictures on the same axes: constant pressure and falling pressure. Tap the shaded part, and the wrong places too.

6. Where the first law stops

The first limitation of the first law: heat flows from a hot body to a cold one, but never by itself from a cold body to a hot one. Yet if it did, neither the first law nor the conservation of energy would be broken.

The second limitation: part of the heat supplied becomes work, but the first law does not tell us whether all of it can become work.

Because of these two limits a further law of thermodynamics, the second law, is needed.

7. Try the sums

First sum. At a constant 2 × 10⁵ Pa, with the volume going from 0.01 m³ to 0.04 m³, W = P(V₂ − V₁) = 2 × 10⁵ × 0.03 = 6000 J.

Second sum. When 2 mol of a monatomic gas warm by 50 K, du = n C_V ΔT = 2 × (3/2)(8.31) × 50 ≈ 1247 J, whatever happens to pressure or volume.

Third sum. A lead bullet has specific heat 130 J kg⁻¹ K⁻¹. Stopping from 100 m s⁻¹ with all its kinetic energy becoming heat gives mv²/2 = m c Δθ, so Δθ = v²/(2c) = 10000/260 ≈ 38.5 K.

Pick a process and change the numbers. Nothing you type goes anywhere. The notebook is yours.

8. Check yourself

Three questions in the exam's style. Read the first one slowly.

The second is much the same.

In the last one, keep an eye on signs and units.

The whole note on eight cards.

Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.5-1.6.