One kilogram of air is compressed in a cylinder for each of

One kilogram of air is compressed in a cylinder for each of the quasi-equilibrium processes listed in the following table. Fill in the missing quantities for the selected process.

Solution

a)

Given Volume is constant

so W = 0

Given

dU = 200 kJ

From the first law

Q = dU + W

so

Q = 200 kJ

Given

T2 = 200 C = 473 K

Now from

dU = m*Cv*dT =1*0.718*(T2-T1)

1*0.718*(T2-T1) = 200

1*0.718*(473-T1) = 200

T1 = 194.5 K

T1 = -78.5 C

Now

dH = m*Cp*dT

dH = 1*1.005*(473-194.5)

dH = 279.89 kJ

From gas law

P1/T1 = P2/T2

P1 = P2*(T1/T2) = 200*(194.5/473)

P1 = 82.24 kPa

From

Pv = RT

P1*v1 = R*T1

82.24*v1 = 0.287*194.5

v1 = 0.679 m^3/kg

P2*v2 = R*T2

200*v2 = 0.287*473

v2 = 0.679 m^3/kg