Question

Calculate the change in energy for the
expression of one mole of an ideal gas from 2
atm to 0.2 atm at 25 ºC ( R = 8.314 J/K.mol)

Answer 1

Answer:

Amount of work done in reversible isothermal expansion,

w=−2.303nRTlogV1V2

given, n=2,R=8.314JK−1mol−1,T=298K,V2=20L and V1=10L

Substituting the values in above equation

w=−2.303×2×8.314×298×log1020

w=−2.303×2×8.314×298×0.3010=−3434.9J

Explanation:

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