ay which point cp=cv for real gases?

Well here’s a mathematical proof. But it can probably be explained in simpler terms. Basically, if you follow the proof, you can see that the ‘nb’ term makes no difference to either the specific heat at constant pressure or at constant volume.

At constant pressure the amount of heat, Q, needed to produce a temperature change ΔT at constant pressure for n moles of ANY gas is:
Q = nCpΔT
Where Cp is the molar specific heat at constant pressure.

The energy supplied, Q, can be considered as increasing in internal energy (ΔU) and doing work at constant pressure, PΔV. So
ΔU + PΔV = nCpΔT
ΔU = nCpΔT - PΔV (equation 1)
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Consider an ideal gas obeying PV=nRT, where P is constant.

If V increases by ΔV and T increases by ΔT
P(V+ΔV) = =nR(T+ΔT
PV+ PΔV = =nRT+ nRΔT
Subtracting ‘PV = nRT’ gives:
PΔV = nRΔT

From equation 1
ΔU = n.Cp. ΔT - nR ΔT
= nΔT(Cp - R)

ΔU/ΔT = n(Cp - R) (equation 2)
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Now repeat the above calculation for a real gas obeying P(V-nb)=nRT. where P is constant.
PV – Pnb = nRT

If V increases by ΔV and T increase by ΔT
P(V+ΔV + nb) =nR(T+ΔT)
PV+ PΔV + Pnb =nRT+ nRΔT)
Subtracting ‘PV – Pnb = nRT’ gives:
PΔV = nRΔT
Note that the ‘nb’ term has disappeared and we are left with the same formula as for an ideal gas.

The same calculation used above for an deal gas then applies so
ΔU/ΔT = n(Cp - R) (same equation 2 as above)
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At constant volume, a gas does no work. So only internal energy (U) changes.
If Cv is the molar specific heat at constant volume
Q = ΔU = nCvΔT
ΔU/ ΔT = nCv.
This is true for the ideal gas and for the real gas.

From equation 2
nCv. = n(Cp - R)
Cp – Cv – R

So Cp – Cv = R is true for both the ideal and the real gas. Therefore
(Cp – Cv)real / (Cp – Cv)ideal = R/R = 1