You have an ideal gas in an insulated container. The volume of the container increases from 10 m^3 to 20 m^3 under a constant pressure of 50. × 10^5 Pa. What was the work done on the gas?

Answer 1

The ∆V = 20 m³ - 10 m³ = 10 m³

10 m³ = 10,000 liters = 1x10⁴ L

work = P∆V at constant pressure and here P = 50x10⁵ Pa.

Now we need to get all of our units to match. If we want our answer to be in Pa m³, (the SI unit of work), then we can leave the units as they are: work = (50x10⁵ Pa)(10m³) = 500x10⁸ Pa m³ = 5x10¹⁰ Pa m³

If you want to express the answer in the more common units of L-atm, then we need to convert:

work = (1x10⁴ L)(50x10⁵ Pa)(1 atm/101325 Pa) = 4.93x10⁸ L-atm

Further, another common unit is joules: 4.93x10⁸ L-atm x 101.325 J/L-atm = 4.9x10⁶ J