Question

Two kg of air at 500kPa, 80°C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100kPa and 5°C. For this process, determine 1) the maximum work 2) the change in a availability and 3) the irreversibility.​

Answer 1

Heya!!

Answer:

M = 2kg; P1 = 500000 Pa;

T1 = 353K; V2 = 2V1; P0=P2 = 100000 Pa;

T2 = 278K

Solution:

S2-s1 = mRln(p1/p2) + m Cpln(T2/T1)

= 2 x 280 x ln(5/1) + 2 x 1000 ln(278/353)

= 422 J/K

V1 = mRT1/P1 = 2 x 280 x 353/ 500000 = 0.4 m^3

V2 = 0.8 m^3

Wmax = ∆U+ ∆Q+ ∆W

= mCv(T1-T2) + T0(S2-S1) + P0(V1-V2)

= 2 x 720 x (353-278) + 278 x 422 + (0.4-0.8)

= 185 kJ

A = Wmax = 185 kJ

I = Q- T∆s = 0 – 278 ( -422) = 116 kJ.

Glad help you,

thanks