Work done by a gas with only pressure and volume

1. Apr 15, 2013

PhizKid

1. The problem statement, all variables and given/known data
An ideal diatomic gas undergoes an adiabatic compression. Its initial pressure and volume are 1.20 atm and 0.200 m^3 and its final pressure is 2.40 atm. How much work is done by the gas?

2. Relevant equations
$W = -\int p \cdot dV$

3. The attempt at a solution
So after integrating I got: $W = -pV^{\gamma} \cdot \frac{{V_f}^{1 - \gamma} - {V_i}^{1 - \gamma}}{1 - \gamma}$

My first question: How do I know what $pV^{\gamma}$ is? Do I use the final pressure and volume for this constant term?

Second: How do I find $V_f$? I don't see any equation I can use to solve for it using only final pressure.

2. Apr 16, 2013

Staff: Mentor

You know the initial pressure and the initial volume. Why not use those?

How about $p V^\gamma = \mathrm{const.}$?

3. Apr 16, 2013

PhizKid

I wasn't sure if that was a valid substitution. How do you know you are allowed to use the initial conditions and not something else? Can I use either initial or final conditions (provided I know the final conditions) for that substitution?

Oh...is $p_{i} V_{i}^{\gamma} = p_{f} V_{f}^{\gamma}$? Or are you implying something else? If $p_{i} V_{i}^{\gamma} = p_{f} V_{f}^{\gamma}$ is true, how do you know it's true? I don't see why we can assume the initial adiabatic pressure and volume conditions must equal the final ones.

Last edited: Apr 16, 2013
4. Apr 16, 2013

Staff: Mentor

Both points I was making rest on the same idea: during an adiabatic process, $V^\gamma p = \mathrm{const.}$ $\gamma$ is not called the adiabatic exponent for nothing! You should find the proof for this in any thermodynamics textbook. The conclusion is that you can use the initial conditions as well as the final conditions in the equation for work, and the initial conditions you already have. Also, you can use $p_{i} V_{i}^{\gamma} = p_{f} V_{f}^{\gamma}$ to find $V_f$.

5. Apr 16, 2013

PhizKid

Thank you very much DrClaude

6. Apr 16, 2013

Andrew Mason

For work done BY the gas, W = ∫PdV. For work done ON the gas W = -∫PdV

You can also solve this problem by using W = -ΔU = nCVΔT (since Q = 0). You cannot determine ΔT or n but you can determine nΔT quite easily.

AM