Use \, (or similar) to add space to differential elements in integrals. Good habit in general, and helps readability, particularly with e.g. dx dy dz, etc.
Air
• 3.2k
• 4
• 22
• 46

Is my relation correct?

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

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

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

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

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

It's just a sign convention, if work is done by a system work is positive, if work is done on a system work is negative (as in your case).

what am I missing?

When I tried to solve the problem I got the same result as yours. Are you sure there is nothing missing in the problem description?

deleted 14 characters in body
Algo
• 2.1k
• 1
• 14
• 31

Is my relation correct?

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

mechanical work by definition is defined as: $$W = \int_{V_i}^{V_f}PdV$$

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 $$V_i = V_1$$: $$W = \frac{(P_2V_2 - P_1V_1)} {(1-k)}$$

It's just a sign convention, if work is done by a system work is positive, if work is done on a system work is negative (as in your case).

what am I missing?

When I tried to solve the problem I got the same result as yours. Are you sure there is nothing missing in the problem description?

Algo
• 2.1k
• 1
• 14
• 31

Is my relation correct?

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

mechanical work by definition is defined as: $$W = \int_{V_i}^{V_f}PdV$$

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 $$V_i = V_1$$: $$W = \frac{(P_2V_2 - P_1V_1)} {(1-k)}$$