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Calculating the work done during an isothermal expansion using integration

  • Thread starter bsmm11
  • Start date
  • #1
14
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Homework Statement


In calculus, the work done when a gas expands from volume V1 to volume V2 is given by
W = ∫V2V1 P dV
Use this expression to show that the work done by n moles of gas at temperature T during an isothermal expansion from volume V1 to V2 is
W = nRT ln(V2/V1)


Homework Equations


Q = ΔU + W
PV = nRT


The Attempt at a Solution


W = [VP]V2V1 = PV2 - PV1 = PΔV
But I think it should be ΔPΔV since this is an isothermal expansion. W = PΔV is for isobaric since P is constant.
Then I can't even guess where the ln comes from. :frown:
 

Answers and Replies

  • #2
37
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You got P as a constant because you treated it like one when you took the integral.

But if you look at the ideal gas law you can see that pressure is a function of volume. So then you can put that expression into the integral and n, R, and T are constants, then integrate.
 
  • #3
296
0
Hey,

Unfortunately You have got it wrong.


See work is defined as
dW =PdV , where P is external pressure and V is small volume change.

This comes from the fact that dW=Force * displacement

dW=(External)Pressure*Area*displacement

However area * displacement is change in volume so
dW=PdV

You have to integrate this expresion to get the value of work.

Now in isothermal reversible conditions , you have to find work done by system which is a GAS
In such cases pressure external =pressure of the gas.

Remember, Ideal gas Equation.?

How will you integrate PdV now with T being constant.
 

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