1.

Calculate the values of ∆U and ∆H for an ideal gas in terms of CP and CV.

Answer»

Calculation of ∆U and ∆H:

For one mole of an ideal gas, 

we have, CV = dU / dT

dU = C.dT 

For a finite change, 

we have, ∆U = (U2 – U1) = CV (T2 – T1)

and for n moles of an ideal gas we get

∆U = nCV (T2 – T1 ) …….(1)

We know,

∆H = ∆(U + PV) 

∆H = ∆U + ∆(PV) 

∆H = ∆U + ∆RT [∴ PV= RT] 

∆H = ∆U + R∆T 

∆H = CV (T2 – T1) + R (T2 – T1) 

∆H = (CV + R) (T2 – T1 ) 

∆H = CP(T2 – T1) [∴C – C = R] 

For n moles of an ideal gas we get 

∆H = nCP (T2 – T1) ……..(2)


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