☰ Contents · Physics

The first and second laws of thermodynamics

Lessons 27 · 1 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 10, 1st edition. Niso Poligraf Publishing House, Tashkent, 2017
27

Irreversibility of thermal processes. The laws of thermodynamics

Textbook: pp. 98–101
GoalKnow reversible and irreversible processes, internal energy and the first and second laws of thermodynamics, and apply the relation of ΔU, Q and A in problems.
New words
internal energy · ichki energiyairreversible process · qaytmas jarayonfirst law of thermodynamics · termodinamikaning birinchi qonuniperpetual motion machine · abadiy dvigatel
Explanation

If a system can return from the final state to the initial state through the same intermediate states, the process is reversible; real processes (friction, heat exchange) are irreversible, and a reversible process is an idealisation. The internal energy U of a thermodynamic system is the sum of the kinetic energies of the random motion of its molecules and their interaction potential energies; it depends only on the state, so what matters in a process is its change ΔU. The first law (conservation of energy): the heat given to a system goes to increasing the internal energy and to the work the system does against external forces: Q = ΔU + A. Heat received Q > 0, heat given away Q < 0; if the system does work A > 0, if work is done on it A < 0. This law shows that a perpetual motion machine of the first kind (one giving work without using energy) is impossible. The second law fixes the direction of processes: Clausius – heat does not pass by itself from a colder to a hotter body; Planck – no process converts heat completely into work. So a perpetual motion machine of the second kind is also impossible.

Worked examples
A gas receives 800 J of heat and does 300 J of work: ΔU = Q – A = 800 – 300 = 500 J (the internal energy rose).
External forces do 200 J of work on a gas (A = –200 J) and the gas gives out 120 J of heat (Q = –120 J): ΔU = Q – A = –120 + 200 = 80 J. In isobaric expansion (p = 100 kPa) with a volume increase of 0.002 m³, A = pΔV = 100 000·0.002 = 200 J.
Class activity

Only with the teacher: place containers of warm and cold water together and watch the temperatures equalise; explain the direction in which heat passes by itself. Do not use boiling water.

Practice
1
Write the first law of thermodynamics as a formula.
2
A gas is given 900 J of heat and does 400 J of work. By how many J does the internal energy change?
3
In an isobaric process p = 200 kPa (200000 Pa) and the volume grows by 0.003 m³. What is the gas’s work (J)?
4
Why does heat not pass by itself from a cold body to a hot one?