PhysEduc › Class 11 Physics › Chapter 11: Thermodynamics

Heat, work and the first law

NCERT Class 11 Physics, Chapter 11: Thermodynamics, §11.4–11.5. A short visual lesson: 2 narrated 3D videos, 1 interactive simulation and 2 quiz questions.

Internal energy, heat and work

Narrated video. Transcript:

Zoom into a gas in a cylinder. Its molecules fly about at random, collide, and pull on each other a little. Each one has kinetic energy and some potential energy.

Add up all these molecular energies, measured in the frame where the gas as a whole is at rest. The total is the internal energy of the gas, U.

Moving the whole cylinder on a train does not change U. Internal energy counts only the disordered motion inside the system.

U depends only on the state of the system: its pressure, volume and temperature. It does not matter how the system got there. So internal energy is a state variable. For an ideal gas, U depends on temperature alone.

There are two ways to change U. First, put the cylinder on a hot plate. Energy flows in because the plate is hotter than the gas. Energy that moves because of a temperature difference is called heat.

Second, push the piston down quickly. You do work on the gas, and its molecules speed up, so U rises with no temperature difference needed. This is energy transfer as work.

Heat and work are two modes of energy transfer, like two doors into the same room. Once the energy is inside, you cannot say which door it came through.

So a common mistake is to say that a hot body “contains a lot of heat”. It does not. A body has internal energy. Heat and work only describe energy on the move.

The first law of thermodynamics

Narrated video. Transcript:

Supply heat ΔQ to the gas in the cylinder. Part of it raises the internal energy by ΔU. The rest pushes the piston out, so the gas does work ΔW on the surroundings.

Energy is conserved, so ΔQ equals ΔU plus ΔW. This is the first law of thermodynamics: the law of conservation of energy, applied to heat and work.

Watch the signs. ΔQ is positive when heat goes into the system. ΔW is positive when the system does work on its surroundings, and negative when work is done on it.

Now find ΔW. The gas presses on a piston of area A with pressure p. The force is p A. If the piston moves out a small distance Δx, the work is p A Δx.

A times Δx is the change in volume, ΔV. So the work done by the gas is ΔW equals p ΔV, and the first law becomes ΔQ equals ΔU plus p ΔV.

Let’s boil water at atmospheric pressure. One gram of water at 100 °C needs 2256 joules of latent heat to become steam at 100 °C.

The volume grows from about 1 cubic centimetre to 1671. The work done pushing back the atmosphere is 1.013 × 10⁵ pascals times 1.671 × 10⁻³ cubic metres, about 169 joules.

So ΔU is 2256 minus 169, about 2087 joules. Most of the heat goes into pulling the molecules apart, not into pushing the air.

Finally, take the gas from state 1 to state 2 by two different paths. ΔU is the same, because U is a state variable. But ΔQ and ΔW are different. Heat and work depend on the path.

Try it: heat a gas at constant pressure

Interactive simulation. Things to try:

  • Press Run process (or Space). The load keeps the pressure fixed while the heater warms the gas.
  • Watch the three bars: Q (heat in) is always the sum of ΔU and W. That is the first law.
  • The shaded rectangle on the P-V diagram is the work, W = pΔV.
  • Try the other processes: in each one, check that ΔQ = ΔU + ΔW still holds.

Same start, same finish

A gas is taken from state 1 to state 2 along two different paths on a P-V diagram. Which quantity must be the same for both paths?

  1. The heat ΔQ absorbed by the gas
  2. The work ΔW done by the gas
  3. The change in internal energy ΔU
  4. All three are the same
Answer

C. Internal energy is a state variable: it depends only on the state, so ΔU = U₂ − U₁ for every path. Work is the area under the path, which changes from path to path, and so ΔQ = ΔU + ΔW changes too.

Balancing the energy

200 J of heat is supplied to a gas, and the gas does 80 J of work on its surroundings as it expands. What is the change in its internal energy?

  1. +280 J
  2. +120 J
  3. −120 J
  4. +80 J
Answer

B. First law: ΔQ = ΔU + ΔW, so ΔU = ΔQ − ΔW = 200 − 80 = +120 J. Heat in is positive and work done by the gas is positive. Adding them (280 J) treats the work as if it were done on the gas.

Summary

  • First law of thermodynamics: ΔQ = ΔU + ΔW
  • Work done by a gas: ΔW = p ΔV
  • Combined: ΔQ = ΔU + p ΔV
  • Internal energy U is the sum of the molecular kinetic and potential energies, in the frame where the system is at rest. It is a state variable.
  • Heat is energy transferred because of a temperature difference; work is energy transferred by other means (e.g. moving a piston). Neither is “contained” in a body.
  • Sign convention: ΔQ > 0 when heat is supplied to the system; ΔW > 0 when work is done by the system.
  • ΔU is the same for every path between two states; ΔQ and ΔW depend on the path.
  • Boiling 1 g of water at 1 atm: ΔQ = 2256 J, ΔW ≈ 169 J, so ΔU ≈ 2087 J.