1.

1 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 (DeltaS_("sur")) in J K^(-1) is (1 L atm=101.3 J)

Answer»

5.763
1.013
-1.013
-5.763

Solution :`DeltaU=Q+w`. For ISOTHERMAL PROCESS, `DeltaU=0`
`:. Q=-w=-PDeltaV=-3(2-1)=-3L "ATM"=-303.9` J
Now apply `DeltaS=(q)/(T)`


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