One mole of an ideal gas at 300 K in thermal contact with its...
One mole of an ideal gas at 300 K in thermal contact with its 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 the surroundings (∆Ssurr) in J K-1 is:
(1 L atm = 101.3 J)
(1 L atm = 101.3 J)
Solution:
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)surrT=-qsysT
=-3.0×101.3 J300 K
=-1.013 J/K
Alternate solution:
∆Ssurr=qsurrT=-qsysT=WsysT
∵ For the isothermal process:
∆U=0 qsys=-Wsys
∆Ssurr.=-Pext(Vf-Vi)T
=-3(2-1)300×101.3
=-1.013 J/K
qsys=∆U-w=0-[-Pext.∆V]
=3.0 atm×(2.0 L-1.0 L)=3.0 L atm
∴∆Ssurr=(qrev)surrT=-qsysT
=-3.0×101.3 J300 K
=-1.013 J/K
Alternate solution:
∆Ssurr=qsurrT=-qsysT=WsysT
∵ For the isothermal process:
∆U=0 qsys=-Wsys
∆Ssurr.=-Pext(Vf-Vi)T
=-3(2-1)300×101.3
=-1.013 J/K
Join the conversation