# Constant Heat capacities during a quasi-static adiabatic expansion

How to prove the work done by an ideal gas with constant heat capacities during a quasi-static adiabatic expansion is equal to W=-C(Ti-Tf).

I know we can use 1st law thermodynamic, Q=U-W where, Q = Heat, U = Internal Energy, W = Work

However, my derivation/prove leads to wrong and mess-up equation.

W = ΔU
W = -PdV
W = -(K/V^Y)*dV
W = -K∫(1/V^Y)*dV
W = -K[V^(1-Y)/(1-Y)]*∫dV
W = -(K/(1-Y))[Vf^(1-Y) - Vi^(1-Y)]
W = -(K/(1-Y))[Vf^(-Y)*Vf - Vi^(-Y)*Vi]
W = -(1/(1-Y))[((Vf*K)/(Vf^Y)) - ((Vi*K)/(Vi^Y))]


Then i confuse.

• You should show your derivation, otherwise this question will be considered a homework question and put on hold. – Mitchell Jun 4 '17 at 18:39
• Your method seems correct to me. – Mitchell Jun 4 '17 at 19:12
• Use the ideal gas law in conjunction with $PV^{\gamma}$ to see how the temperature variation is related to the volume variation, and then substitute this into your equation for the total work. – Chet Miller Jun 4 '17 at 21:15

The derivation to get the term $PV^{\gamma}=constant$ for an adiabatic process, uses the heat capacity at constant volume in its initial steps.

$\Delta U=nC_v\Delta T$ $\tag1$

For an adiabatic process $q=0$,

$\Delta U=W$

$nC_v\Delta T=-PdV$ $\tag 2$

Equation $(2)$ is all you need.

One might argue why heat capacity at constant volume is used when there actually a observable change in volume of the system.

The term $\Delta U=nC_v \Delta T$, is independent of the process that the ideal gas goes through, even when $\Delta V \neq 0$.

In you wish to know more about $\Delta{U}$, check out these links : When is $\Delta U=nC_V \Delta T$ true? and Work done in adiabatic process.