1
$\begingroup$

I hope you are excellent. I'd like you to help me make sense of the integral $ \oint PdV \neq 0 $ for some thermodynamic process. What can it mean for the integral to be nonzero? I can only interpret it as if there is work, however my deep understanding is very limited. I appreciate your comments.

$\endgroup$

2 Answers 2

1
$\begingroup$

In thermodynamics, the differential work done on a system is defined as the following: $$\delta W= -P dV.$$

Work done is an inexact differential, denoted by the symbol $\delta$. This means that the total work depends not only on the initial and endpoints but also on the path taken along the process. To find this total work done for a particular thermodynamic process, one must integrate this differential along the path of the process:

$$W=-\int_{\textrm{process}} P dV.$$

So what does $\oint PdV\neq 0$ mean? This is simply a statement of the path-dependent property of work done. You can see this by breaking up the loop integral into two integrals. $$\oint PdV=\int_1^2PdV+\int_2^1PdV ,$$ where each integral is taken along a different path. If work done did not depend on the path taken, then we could simply write the following, as a property of integrals: $$\int_1^2PdV=-\int_2^1PdV \tag{1}$$ yielding, $$\oint PdV=0.$$

However, since work done depends on the path taken, equation $(1)$ is not true in general, thus

$$\oint PdV \neq 0.$$

$\endgroup$
0
$\begingroup$

Given integral repersent the work done to a closed path . Which is non zero means think as a non conservative force you will get intuition. Since force is directly related to pressure and then work done work done. hope this answer might help you to why work done through a closed path is non zero.

$\endgroup$

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.