To implement isothermal compression of 0.8 kg of air at p1 = 0.1 MPa and t = 25 ° C, work of 100 kJ was expended.

To implement isothermal compression of 0.8 kg of air at p1 = 0.1 MPa and t = 25 ° C, work of 100 kJ was expended. Find the pressure p2 of compressed air and the amount of heat that must be removed from the gas?

m = 0.8 kg.

M = 0.029 kg / mol.

p1 = 0.1 MPa = 0.1 * 10 ^ 6 Pa.

t = 25 ° C.

A = 100 kJ = 100000 J.

R = 8.31 m2 * kg / s2 * ° K * mol.

p2 -?

Isothermal is a process in which the temperature of the thermodynamic system does not change: T = const.

The work of gas A during an isothermal process is determined by the formula A = m * R * T * ln (V2 / V1) / M.

p1 * V1 = m * R * T / M.

V1 = m * R * T / M * p1.

T = 273 + t = 273 + 25 = 298 K.

V1 = 0.8 kg * 8.31 m2 * kg / s2 * ° K * mol * 298 K / 0.029 kg / mol * 0.1 * 10 ^ 6 Pa = 0.068 m3.

ln (V2 / V1) = A * M / m * R * T.

ln (V2 / V1) = 100000 J * 0.029 kg / mol / 0.8 kg * 8.31 m2 * kg / s2 * ° K * mol * 298 K = 1.46.

V2 / V1 = 4.31.

The isothermal process is described by the Boyle-Mariotte law: p1 * V1 = p2 * V2.

p2 = p1 * V1 / V2.

p2 = 0.1 * 10 ^ 6 Pa / 4.31 = 23201.8 Pa = 0.23 * 10 ^ 6 Pa.

In an isothermal process, the internal energy does not change, therefore A = Q.

Answer: the pressure will become p2 = 0.23 * 10 ^ 6 Pa, it is necessary to remove Q = 100000 J of thermal energy from the air.