Question

Prove that cp-cv=R for 1 mole of ideal gas

Answer 1

We know dQ =dU +dW
for 1 mole we take n=1
now, nCpdT=nCvdT+nRdT
or, n (Cp_Cv)=R
as n=1, then Cp_Cv=R (proved )

Answer 2

Proof for $C_p-C_v=R$ for 1 mole of ideal gas

• No work is done at constant volume, whereas at constant pressure, some work is done as a result of increase in volume.
• Thus $C_p-C_v=$ work done by one mole of ideal gas due to expansion of gas when heated through $1^0 C$

∴ For one mole of an ideal gas, PV=RT  ......... (1)

• On raising the temperature by $1^0 C$, temperature changes from T(T+1) and the volume changes from V to (V+ΔV).

Thus, at (T+1)°C, P(V+ΔV) = R(T+1) ........... (2)

• Subtracting (1) from (2),

P(V+ΔV)-PV = R(T+1)-RT

PΔV = R

• But PΔV is the work of expansion done by one mole of the gas when heated through 1°C.
• Thus $C_p-C_v=R$, hence proved.

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