Q and W for Van-der-Waals Gases

[krootox217]

Homework Statement

I have the following task:

Homework Equations

The Attempt at a Solution

I already managed to calculate Delta Um, but how do I calculate Q und W. Can I use the equations for the isothermic expansion for ideal gases, even if this are Van-der-Waals Gases?

 

[TSny]

Hello. Welcome to PF!

What is the most basic formula you know for calculating the work done by a system for any quasi-static process? Hint: This formula was probably presented when you first introduced the concept of work in thermodynamics.

 

[krootox217]

This one?

 

[Chestermiller]

This one?

Yes. They obviously want you to assume that the expansion is reversible. Your equation is correct if W represents the work done by the surroundings on the system.

Chet

 

[krootox217]

Ok, thanks a lot!

therefore, Q = -W?

 

[Chestermiller]

Ok, thanks a lot!

therefore, Q = -W?

No. ΔU_m is not equal to zero.

Chet

 

[krootox217]

This means, that delta U = Q + W and I have to subtract the Work from Delta U?I have another Question, the integral is according to Wolfram alpha -7397.55 J

http://www.wolframalpha.com/input/?...14*298.15)/(V-(3.2*10^-5))-(0.1105)/(V^2))+dV

And I calculated a Delta U of 104.98J

Therefore my Q is 7502.55J

Are these values posible?

 

[Chestermiller]

This means, that delta U = Q + W and I have to subtract the Work from Delta U?I have another Question, the integral is according to Wolfram alpha -7397.55 J

http://www.wolframalpha.com/input/?i=-+integral+from+0.001+to+0.020+((8.314*298.15)/(V-(3.2*10^-5))-(0.1105)/(V^2))+dV

And I calculated a Delta U of 104.98J

Therefore my Q is 7502.55J

Are these values posible?

You don't need to ask me this. Why don't you solve the same the same problem using the ideal gas law and see how the numbers compare? Incidentally, why did you need wolframalpha to do the integration for you? Why didn't you do the integration yourself?

Chet

 

[krootox217]

I calculated it by myself, but I often use wolfram alpha to check my results, and in this case, the result was the same, so it was easier just to paste the wolfram alpha link :)

Well, since delta U should be =0 for ideal gases, this small change seems possible?

 

[Chestermiller]

I calculated it by myself, but I often use wolfram alpha to check my results, and in this case, the result was the same, so it was easier just to paste the wolfram alpha link :)

Well, since delta U should be =0 for ideal gases, this small change seems possible?

Sure. Wouldn't you have expected that?

Chet

 

[krootox217]

I wasn't sure,

but now it makes sense.

Thanks a lot!