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 '17 at 18:39
  • $\begingroup$ Your method seems correct to me. $\endgroup$ – Mitchell Jun 4 '17 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.