Could someone give me a proof to justify that pressure is a state function?
Thanks in advance.
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2 Answers
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Although we can define a state function but ther also exists some mathematical criterias for any thermodynamic variable to be a state function.This site enlists them. They are :
- State functions are exact differentials.
- The cyclic integral of state function is zero.
- If the state function is a dependent variable , then the differential of the function can be represented as the partial differentials of the independent state functions on which the function depends.
- The mixed derivative(second partial derivatives) of the independent variables are equal.
Based on the above four criterias , you can check whether pressure is a state function or not.
The second one is much helpful if mathematics is not your subject. So I will expand it a bit.
We define a cyclic process as a process in which the thermodynamic parameters of the system reaches its initial values and pressure is one of the known thermodynamic parameters. So in cyclic process, the change in pressure is zero and hence it is a state function.
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1$\begingroup$ 1,2,4 are equivalent by greens theorem. $\endgroup$ Nov 20, 2020 at 7:48
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$\begingroup$ @JoseAf glad to know. Btw if any of the two answers are really satisfactory then you can accept any one of them . $\endgroup$– AnkitNov 23, 2020 at 6:52
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$\begingroup$ I didn't do it because it seemed to me that it was unfair to both answers xD Thanks a lot!! $\endgroup$– JoseAfNov 23, 2020 at 10:14
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A state function in thermodynamics is a property that changes irrespective of the path taken. Thus anything that is used to represent states in thermodynamics has to be a state function, and Pressure is one of those state variables. Hence Pressure is a state function.
