The gas is in the standard temperature and pressure condition i.e. at S.T.P
Hence V1 = 22.4 dm3 and V2 have to be calculated.
As given expansion is isothermal and reversible therefore, ∆U = 0
We know that ∆U = q + w
But ∆U = 0
Hence q = – w = 2000 cal
= 2000 X 4.184 J = 8368 J
As work done in reversible isothermal expansion is given by:
w= -nRT ln (V2/ V1)
Therefore nRT ln (V2/ V1) = – w = 8368 J
(I mol) X (8.314 J K– mol–) X (273 K) ln (V2/ 22.4 dm3) = 8368 J
V2 = 242.50 dm3
Hence the final volume of one mole of an ideal gas initially at 0oC and 1 atm pressure is equal to 242.50 dm3
This is one of the best lessons I ever had. I liked it very much, and I want you to keep this way. Thanks a lot.