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lkh1986
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Homework Statement
Given the equation of state PV=nRT. So, I get the first derivative and second derivative (P with respect to V) and equate them to 0. Then, I found out that they cannot be equal to 0, so I make the conclusion that no critical temperatue exists for an ideal gas.
If the van der Waals equation is given, I can find the critical temperature by using the same way, namely equate the first and second derivatives to 0. Yeah, critical temperature exists for this case.
But how about other kind of equation of state? Let say P(V-nb)=nRT or (P+a)(V)=nRT. Can we use the same method? (I think we can). And is the CRITICAL temperature here refers to the CRITICAL point on the graph of the plot P versus V? As in the critical/stationary point in the field of mathematics?
Homework Equations
The Attempt at a Solution
I am given an equation in a book. Then first I assume there exists a critical point (although it may not exist). Then, I get the first and second derivatives. Then I equate them to 0. And after some calculations, I get 0.25 = 0.125, which is clearly wrong. So, this contradicts with my previous assumption that a critical temperature exists. Therefore, NO critical temperature exists this gas. Is this kind of contradiction method accepted? Because I am told that this popular contradiction method is widely used in mathematics.
Since critical temperature is the highest temperature in which a gas can be liquefy, so if a gas doesn't have a critical temperature, I can say that the gas cannot be liquefy. Is this correct? Thanks.
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