Van der Waals, temperature decrease

[Gregg]

Homework Statement

I need to find a p-V region where the temperature is decreasing

## (p + \frac{a}{V^2} ) (V-b) = RT ##

The Attempt at a Solution

So I thought that a place to start would be

## dT = \frac{\partial T}{\partial P} dP + \frac{\partial T}{\partial V} dV < 0 ##

So ##\frac{\partial T}{\partial P} dP < - \frac{\partial T}{\partial V} dV ##

Do I now, ## \int \frac{\partial T}{\partial P} dP < - \int \frac{\partial T}{\partial V} dV ##?

## \frac{\partial T}{\partial P} = V - b ##

##\frac{\partial T}{\partial V} = p + a \log (V) -2 \frac{a b}{V^3} ##

Is this the way to do it? The integrals are complicated and I can't really do them

 

[mark726]

without knowing what values of p and V I am working with.

Hi there,

First of all, great start on tackling this problem! Your approach seems to be on the right track. However, I would suggest making a few adjustments to your strategy.

Instead of trying to find a specific p-V region, let's think about this problem in terms of a temperature-pressure-volume (T-P-V) diagram. This will give us a visual representation of the system and make it easier to identify the region where the temperature is decreasing.

Now, let's recall the ideal gas law: PV = nRT. Using this equation, we can rewrite your first equation as:

##(p + \frac{a}{V^2}) (V-b) = \frac{nR}{V}T##

Since we are interested in finding a region where the temperature is decreasing, we can rewrite this equation as:

##\frac{nR}{V}T = (p + \frac{a}{V^2}) (V-b)##

Now, let's differentiate this equation with respect to V:

##\frac{nR}{V}\frac{dT}{dV} = (p + \frac{a}{V^2}) - \frac{2a}{V}(V-b)##

Since we are looking for a region where the temperature is decreasing, we want the derivative ##\frac{dT}{dV}## to be negative. Therefore, we can set this derivative to be less than zero:

##\frac{nR}{V}\frac{dT}{dV} < 0##

Now, we can substitute in the partial derivatives you have already calculated in your attempt:

##\frac{nR}{V}(V-b) < -\frac{nR}{V}(p + a\log(V) - \frac{2ab}{V^2})##

Simplifying this equation, we get:

##V-b < -(p + a\log(V) - \frac{2ab}{V^2})##

Now, we can rearrange this equation to get:

##p + a\log(V) - \frac{2ab}{V^2} < V - b##

This inequality gives us the desired region on the T-P-V diagram where the temperature is decreasing. We can plot this inequality on the T-P-V diagram and find the corresponding p-V region.

I hope this helps! Good luck with your calculations.