A gas in a piston-cylinder assembly undergoes a compression process for which the relation between pressure and volume is given by pVn 5 constant. The initial volume is 0.1 m3, the final volume is 0.04 m3, and the final pressure is 2 bar. Determine the initial pressure, in bar, and the work for the process, in kJ, if (a) n 5 0, (b) n 5 1, (c) n 5 1.3.

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 * (V2/V1) ^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar.

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1) ^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant * dV

Work = 2 * 10^5 * [ 0.04 - 0.1 ]

Work = 200000 * - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 * (0.04/0.1) ^1

P1 = 2 * 0.4 = 0.8 bar

Work = ∫ PdV = constant * ∫dV/V

Work = P1V1 * ln (V2/V1)

Work = 0.8 * 10^5 * 0.1 * ln 0.4

Work = - 7330.3J

Work = - 7.33 kJ

C.) When n = 1.3

P1 = 2 * (0.04/0.1) ^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1) / (1 - 1.3)

Work = (2*10^5*0.04) - (0.608 10^5*0.1) / (1 - 1.3)

Work = (8000 - 6080) / - 0.3

Work = - 1920/0.3

Work = - 6400 J

Work = - 6.4 kJ