Question

A piston having 0.033 mol of gas at 35.0 C expands from 0.77 L to 2.00 L. Calculate the work performed if the expansion occurs (a) against an external pressure of 0.455 atm, and (b) reversibly

Answer 1

a) We know

work done for a system against an external pressure

     dW = - Pext dV

Here,

            Pext = 0.455 atm and dV = (Final Volume)V2 - (Initial Volume)V1 = 2 - 0.77 = 1.23 L

So, dW = - 0.455 x 1.23 = 0.56 atm L = 0.56 x 0.101 kJ = 0.057 kJ

b) For a reversible process Pext = Pint = P = nRT /V

So, work done dW = -Integration of PdV = - nRT ln V2/V1

Here, n = no of mole = 0.033 mol

R = 0.0821 Latm K-1 mol-1

T = 35 ^oC = 273 + 35 = 308 K

V2 = Final volume = 2 L

V1 = Initial volume = 0.77 L

dW = - 0.033 x 0.0821 x 308 ln 2 / 0.77

       = 0.834 x 0.954 = 0.796 atm L = 0.796 x 0.101 kJ = 0.080 kJ