# The air was heated at 30 K at a constant pressure of 10 ^ 5 Pa. At the same time, its volume changed from 1.5 to 1.67 m3.

**The air was heated at 30 K at a constant pressure of 10 ^ 5 Pa. At the same time, its volume changed from 1.5 to 1.67 m3. Determine the work done by the air during expansion and the increment of internal energy.**

To find the values of the work of air and the increment of its internal energy, we will use the formulas: A = P * ΔV = P * (Vк – Vн) and ΔU = i * P * ΔV / 2 = i * P * (Vк – Vн) / 2.

Constants and variables: P – constant air pressure (P = 10 ^ 5 Pa); Vк – air volume after heating (Vк = 1.67 m3); Vн – initial volume (Vн = 1.5 m3); i is the number of degrees of freedom of air molecules (i = 5, since air mostly consists of diatomic molecules).

Calculation: a) Work: A = 10 ^ 5 * (1.67 – 1.5) = 17 * 10 ^ 3 J = 17 kJ;

b) Increase in internal energy: ΔU = 5 * 10 ^ 5 * (1.67 – 1.5) / 2 = 42.5 * 10 ^ 3 J = 42.5 kJ.