Kushagra Madhukar
Last Activity: 4 Years ago
Dear student,
Please find the solution to your problem below.
From 1st law of thermodynamics
qsys = ΔU – W = 0 – [–Pext.ΔV]
{SInce in isothermal process, ΔU = 0}
= 3 atm x (2 – 1) = 3 L–atm = 3 x 101.3 = 303.9 J
Now, ΔSsurr = qsurr/T = – qsys/T
= – 303.9/300
= – 1.013 J/K
Regarding your second doubt, since this is an expansion pressure at constant pressure, so it will be irreversible in nature.
The formula ∆S = nR*ln(v2/v1) is used when expansion is reversible.
Thanks and regards,
Kushagra