...against a constant external pressure of 4.0 atm.
-
W = -Pext X dV
Pext = 4 atm
dV = 3.6-3.2 = 0.4 L
W = - 4 X 0.4 = -1.6 L atm
1 L atm = 101.3 J
so W = -1.6 X 101.3 = -162.08 J
-ve sign shows that work is done by the system
Pext = 4 atm
dV = 3.6-3.2 = 0.4 L
W = - 4 X 0.4 = -1.6 L atm
1 L atm = 101.3 J
so W = -1.6 X 101.3 = -162.08 J
-ve sign shows that work is done by the system
-
Work done is P(V2-V1)..As is pressure is constant.So its 4x(3.6-3.2)=1.6 lt-atm,,,In joules,its 1.6x101=161.6J.