Calculating work from graph

1. Jul 22, 2007

trah22

1. The problem statement, all variables and given/known data

a sample of ideal gas is expanded to twice its original volume of 1.00m^3 in a quasi-static process for which P=alphaV^2, with alpha=5.00 atm/m^6. How much work is done on the expanding gas.

2. Relevant equations

w=-intergral sign Pdv from intial to final volume( i dont know how to type this out)

3. The attempt at a solution

w=-aplha(v^3/3) from 2m^3 to 1m^3
=(-5.065 x10^5/3)(2^3-1^3)=-1.2x10^6

this actually the correct answer, however i dont understand why -5.065x10^5 is being divided by3 i thought it would be : -5.065x10^5((2^3/3)-(1^3/3)=w/e could someone explain why your suppose to set it up like that....

2. Jul 22, 2007

nicktacik

Hint:

What is

$$\int\alpha x^2 dx$$

3. Jul 22, 2007

trah22

alpha)(x^3/3).... did i forget something

Last edited: Jul 22, 2007
4. Jul 22, 2007

maverick280857

What is the work done in a reversible quasistatic process? It is given by

$$W = \int P dV$$

Thats how the factor of 1/3 comes about. Maybe the answer is wrong?

(PS--Which book is this?)

5. Jul 22, 2007

trah22

the text is principles of physics serway and jewett 4th edition, is the correct setup to the anser then -5.065x10^5(2^3/3-1^3/3) or the way they put it? Should the 1/3 still be factored out?

Last edited: Jul 22, 2007