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`, as shown in Fig. How much work is done by the expanding gas?

Explanation

The final pressure is `5 xx 4 = 20 atm = 2 MP a`.
The work done on the gas is the negative of the area under the curve `P = alpha V^(2)` From `V_(i)` to `V_(f)`. The work on the gas is negative, which means that the expanding gas does positive work. We find its value as
`W = - int_(i)^(f) PdV` with `V_(f) = 2V_(i) = 2(1.00 m^(3)) = 2.00 m^(3)`
`W = int_(v_(i))^(v^(f)) alpha V^(2) dV = - (1)/(3) alpha (V_(f)^(3) - V_(i)^(3))`
`W = - (1)/(5) (5.00 atm//m^(6)) (1.013 xx 10^(5) Pa//atm)`
`xx [(2.00 m^(3))^(3) - (1.00 m^(3))^(3)]`
`= 1.18 xx 10^(6) J`
One estimate was good. To do the integal of PdV, we start by indenifying how P depends on V, just as to do the integal of ydx in mathematics you start by identifying how y depends on x.