# Work done in an adiabatic process

## Homework Statement

An adiabatic process is performed on 15 moles of an ideal gas. The initial temperature is 320 K and the initial volume is
0.80 m3. The final volume is 0.40 m3. The adiabatic constant for the gas is 1.44. How much work is done by the gas?

## Homework Equations

W = Integrate[p dv, vInitial, vFinal] for adiabatic processes
pV = nRT

## The Attempt at a Solution

If p(v^y) initial == p(v^y) final where y = 1.44, and v varies, then p also varies, and I don't know how to use a varying pressure with the formula for work for adiabatic processes.

I try to find the initial pressure by
pV = nRT
p*V == 15*8.31*320
With Vinitial, p = 49860Pa
With Vfinal, p = 99720Pa

And I'm stuck.

ApexOfDE
Here is my solution: for adiabatic process, you can write: PV^(gamma) = C => P = C / V^(gamma). Plug it into W = int(PdV). After that, replace C by P and V.

p*((0.8)^1.44) = c
Which gives
p = 1.37896c

Integrating 1.37896 c dV from 0.8 to 0.4 gives
-0.551582 c

If I use the original p (that I found) and v (given) for c,
-0.551582(p*v) = -0.551582(49860*0.8) = -22kJ, which isn't the answer. Where did I go wrong?

ApexOfDE
No.

We have $$PV^ \gamma = C ==> P = C/V^ \gamma$$

$$W = \int PdV = \int CV^- ^\gamma dV = \frac{C V^ 1 ^- ^\gamma}{1 - \gamma}$$