Finding critical temperature from equation of state

[lkh1986]

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.

 

[Hunt_]

Yes you plot P versus V and then at the point of inflection you set the 1st and 2nd derivatives to zero. This is where the critical pt exists. When you only have an equation and no graph whatsoever, you work out the derivatives and see if the critical pt exists.

If you're not sure about your answer , post the equation of state you're working on. If you're very sure that your calculation is correct and the results are contradictory then yes no critical pt can be predicted by this equation.

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.

Suppose an equation of state cannot predict Tc for a gas. Does that mean the gas cannot be liquified at Tc? Absolutely not. The equation itself is limited and cannot explain liquifaction for this gas at Tc but that doesn't mean that the gas in reality has no Tc.

 

[Blueshift5]

I would like to clarify a few points about finding critical temperature from an equation of state. First, it is important to note that the equation of state PV=nRT is only applicable to ideal gases, which do not exist in the real world. Therefore, using this equation to find the critical temperature may not accurately reflect the behavior of real gases.

Secondly, the method of equating the first and second derivatives to 0 to find the critical temperature assumes that the gas follows a certain type of equation of state, such as the van der Waals equation. This may not be applicable to all types of equations of state, as you have mentioned.

Furthermore, the critical temperature refers to the temperature at which a gas undergoes a phase transition from gas to liquid. This is typically represented on a phase diagram as the point where the liquid and gas phases coexist. It is not necessarily the same as a critical point in mathematics, although the concept of a stationary point can be applied in both cases.

In conclusion, while the method of equating derivatives to 0 can be useful in some cases, it may not always give accurate results for real gases. It is important to consider the limitations and assumptions of the equation of state being used. Additionally, the concept of critical temperature should not be confused with a critical point in mathematics.