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One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (Ssurr) in JK1is (1 L atm = 101.3 J). Why cant we use ∆S = nR*ln(v2/v1)

qwerty , 7 Years ago
Grade 12th pass
anser 2 Answers
kolahalakhila

Last Activity: 7 Years ago

From 1st law of thermodynamics
qsys = U – w = 0 – [–Pext.V]
= 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm
 Ssurr =(qrev)surr/t = -qsys/t
      =3.0 101.3J/300k
         = – 1.013 J/K

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

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