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.

  • $\begingroup$ You should show your derivation, otherwise this question will be considered a homework question and put on hold. $\endgroup$
    – Mitchell
    Jun 4, 2017 at 18:39
  • $\begingroup$ Your method seems correct to me. $\endgroup$
    – Mitchell
    Jun 4, 2017 at 19:12
  • $\begingroup$ 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. $\endgroup$ Jun 4, 2017 at 21:15

1 Answer 1


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.


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