Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific volume is one-tenth of the specific volume at state 1. Applying the ideal gas model and assuming variations in specific heat, determine (a) T2, in °C, and (b) the work, in kJ/kg.

(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg

Explanation:

Solution

T₁ = 35°C = 308

the first step to take is to Use the Table A-17: Ideal gas  properties for air:

Now,

At T₁ = 308 K

V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)

So,

V₁ =219.818 kJ/kg

Thus,

Vr₁ = 596 + [308-305/310-305] (572.3 - 596)

= 581.78

so,

Vr₂/Vr₁ = 1/10

Vr₂ =58.178

Applying Table A-17, at Vr₂ = 58.178

Then,

(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)

T₂ = 747.5 and K = 474.5 °C

V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)

so,

V₂ =549.996 kJ/kg

Hence,

(b) q - w = v₂ -v₁

= 0 -w  = 549.996- 219.818

w = 330.178 kJ/kg

answer:

what work

explanation: