Why Is V = 0 Ignored in Partial Derivatives?

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Suppose we are given : PV = nRT, where n and R are constants.

We are told to find the partial derivative dP/dV.

Am I allowed to do this :

P = nRT/V

Then differentiate this w.r.t. to V.

I disregarded the fact that V = 0 makes the RHS undefined.


# This question came from Princeton Review's "Cracking the GRE Math Subject Test" page 160, qns 13.

The solution given uses the above method but I do not understand why V = 0 is not taken into account.

 

[Hootenanny]

Suppose we are given : PV = nRT, where n and R are constants.

We are told to find the partial derivative dP/dV.

Am I allowed to do this :

P = nRT/V

Then differentiate this w.r.t. to V.

I disregarded the fact that V = 0 makes the RHS undefined.


# This question came from Princeton Review's "Cracking the GRE Math Subject Test" page 160, qns 13.

The solution given uses the above method but I do not understand why V = 0 is not taken into account.

When would the volume of an ideal gas ever be zero?

 

[HallsofIvy]

Suppose we are given : PV = nRT, where n and R are constants.

We are told to find the partial derivative dP/dV.

Am I allowed to do this :

P = nRT/V

Then differentiate this w.r.t. to V.

I disregarded the fact that V = 0 makes the RHS undefined.


# This question came from Princeton Review's "Cracking the GRE Math Subject Test" page 160, qns 13.

Doing it that way gives
$$\frac{\partial P}{\partial V}= -nRT/V^2$$
You don't need to worry about the fact that V= 0 would make nRT/V undefined because it also makes the -nRT/V² undefined. That is, the formula you got does not give an answer in exactly the situation where there is no answer.

The solution given uses the above method but I do not understand why V = 0 is not taken into account.

I would have been inclined to use "implicit" differentiation: from PV= nRT, (dP/dV)V+ P= 0 so dP/dV= -P/V= -(nRT/V)/V= -nRT/V^2.

 

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Thanks a lot guys~ That really helped!