# A closed vessel with a capacity of 0.6 m ^ 3 contains air at a pressure of 0.5 MPa and a temperature of 20 C.

**A closed vessel with a capacity of 0.6 m ^ 3 contains air at a pressure of 0.5 MPa and a temperature of 20 C. As a result of cooling the vessel, the air contained in it loses 105 kJ. Taking the heat capacity of the air constant, determine what pressure and what temperature is then established in the vessel.**

Data: V (capacity of a closed vessel taken) = 0.6 m3; Pн (initial air pressure) = 0.5 MPa = 5 * 10 ^ 5 Pa; Tн (initial temperature) = 20 ºС = 293 К; Q (heat loss by air) = 10 ^ 5 kJ = 105 * 10 ^ 3 J.

Constants: Sv (specific heat capacity of air) = 1005 J / (kg * K); M (molar mass of air) = 2.9 * 10 ^ -2 kg / mol; R (universal gas constant) = 8.31 J / (K * mol).

1) Air mass: m = P * V * M / (R * T) = 5 * 10 ^ 5 * 0.6 * 2.9 * 10 ^ -2 / (8.31 * 293) = 3.57 kg …

2) Change in air temperature: Q = Cw * m * Δt and Δt = Q / (Cw * m) = 10 ^ 5 * 10 ^ 3 / (1005 * 3.57) = 29.3 K.

3) Final temperature: Tc = Tn – Δt = 293 – 29.3 = 263.7 K.

4) Steady-state pressure (isochoric process): Pc = Pn * Tc / Tn = 5 * 10 ^ 5 * 263.7 / 293 = 4.5 * 105 Pa.

Answer: The temperature will be 263.7 K and the pressure 4.5 * 105 Pa.