i) index of process ii) work transfer iii) Heat transfer iv) Change in internal Energy

Given :

m = 1kg

M = 28 ( molecular weight of nitrogen )

P1 = 1 bar; T1 = 300K

P2 = 5bar; T2 = 400K

i) Finding index

P1/P2 = (T1/T2)^(n/n-1)

1/5 = (300/400)^ (n/n-1)

On solving in calci we get, n = 1.218

ii) Work

W = m R (T1-T2) / (n-1) = mRo/M (T1-T2) /(n-1)

= 1 x 8314 / 28 x (300 – 400) / (1.218 – 1) = - 136380 J

iii) Change in internal energy

∆U = m Cv (T2-T1)

But Cp- Cv = R and Cp = Cv x 1.4 (Note that though the index of process is 1.218, the adiabatic index will remain 1.4 only)

Cv = R/0.4 = Ro /0.4M = 8314/ (0.4 x 28) = 742.3 J

∆U = 1 x 742.3 (400-300) = 74230 J

iv) Heat transfer

Q = ∆U + W

= 74230 – 136380 = -62150 J