# Why do I get negative work done for adiabatic compression of air?

I am solving the following problem:

Air having an initial pressure of 6516 kPa and an initial volume of 0.113 m3 is compressed adiabatically to a final volume of 0.057 m3. Calculate the work done by the gas as it compresses to a final pressure of 17237 kPa.

The given answer is -175.9 kJ.

I started from: $$PV^k = C$$

Work is therefore: $$W = \int\limits_1^2 P\,dV$$

$$\frac{(P_2V_2 - P_1V_1)} {(1-k)} =\frac{(17237*0.057)-(6516*0.113)} {(1-1.4)} = 615.025 \:\mathrm{kJ}$$

Is my relation correct? What am I missing and why is the answer negative?

• It's $p_1 V_1 - p_2 V_2$ then you get a negative value. I would get the same value as you if I disregard the error on $p_2$. Adiabatically you get to 16985 kPa. If you compress to 17237 kPa the process is isochor. Therefore no work is done here. You would need to know the mass and dT to calculate $\Delta Q$. Which by the way would have a positiv value. Can you clear that up? Jul 5 '15 at 11:49
• im pretty sure its p2v2 - p1v1 for non flow work; it is also expressed "adiabatic" Jul 5 '15 at 11:58
• The question should say "work done on the gas as it is compressed". The usual convention is that work done by the gas is positive, while work done on the gas is negative. Jul 5 '15 at 12:13
• You are right, I messed up and didn't realize that you had $(1-\kappa)$ instead of $(\kappa-1)$, my fault. As Carlton already explained it's convention that expansion of a gas releases energy thus having a negative work and compression (work on the gas) results in a positive work. If you wonder why I brought up isochor, according to my calculations 17237 kPa is not the exact pressure for an adiabatic compression. I wondered if the task means to (1-2) compress adiabatically to $0.057m^3$ and then (2-3) compress to 17237 kPa. Either way I can't solve for -175.9 kJ. Jul 6 '15 at 11:05
• Maybe I'm missing something but you seem to be doing too much. W=(U1-U2), you have the volume, you can find all the properties you need from the tables. K relates cp and cv, which are also found in the thermal tables .. use ideal gas law. Dec 31 '16 at 19:38

Is my relation correct?

Yes, it's correct. And here is the derivation:

mechanical work is defined as: $$W = \int_{V_i}^{V_f}P\,dV$$

multiplying by $\frac{V^{k}}{V^{k}}$: $$W = \int_{V_i}^{V_f}\frac{PV^{k}}{V^{k}}\,dV$$

Since $PV^{k}$ is constant, we can safely put it out the integration, yielding: $$W = PV^k\int V^{-k}\,dV = PV^k\left[\frac{V^{-k+1}}{1-k}\right]^{V_f}_{V_i}$$

Finally, putting $V_f = V_2$ and $V_i = V_1$: $$W = \frac{(P_2V_2 - P_1V_1)} {(1-k)}$$