Question about work done as pressure and volume increase?

Question

Say we have He in a cylinder with an initial volume of 1 liter and an intial pressure of 1 atm. Somehow, the He expands to a final volume of 3 liters with the pressure rising in direct proportionto its volume. How would you calculate the work done during this process? How could you cause the pressure to rise as the gas expands?

Answer 1

Well, the only way this will happen (assuming the number of moles of He is constant) is if the temperature increases. Remember our ideal gas law: PV=nRT, where R (and in this case n) are constant.

Since P is proportional to V, we'll say that P = kV where k is our proportionality constant.

I'm not sure which physics you're in, but Work = Integral of PdV. This is sometimes expressed positive, sometimes negative, depending on convention. If you're not in calculus-based physics, then you're probably provided with the Work after its integrated:
W = kV^2/2 from V_initial to V_final. Thus, Work = k/2 * (V_final^2 - V_initial^2), where k is still your proportionality constant.

Also, as you're dealing with liters and atmospheres, you'll end up with units of L*atm, NOT Joules.