Question

Volume of a gas is expanded from 3dm^3 to 6dm^3 against a constant pressure of 1 atm. Calculate the work done by the system.​

Answer 1

As gas is expending under constant external pressure.
W = - Pext (Vf - Vi)  
  { P = 3atm  and  Vf = 5dm3 , Vi = 3 dm3, c = 4.184 J/K/g , n = 10 mol , Ti = 290 K }
= -3 atm ( 5-3)dm3
= -6 atm dm3  as { 1 L atm = 101.3 J}
W = -607.8 J
As all work is transferred as heat therefore q = 607.8 J
Now q = m c (Tf-Ti)     { using n = m/ molar mass }
607.8 = (10 × 18) × 4.184 × (Tf - 290)
Tf =  290.8 K