1.

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 surrounding (DeltaS_("surr")) in JK^(-1) is (1 L atm = 101.3 J)

Answer»

5.763
1.013
`-1.013`
`-5.763`

SOLUTION :Isothermal PROCESS, `DeltaU=0`
`dq=-dW=P_(ext)(V_(2)-V_(1))=3L-atm=3xx101.3" JOULE"`
`DeltaS_("SURROUNDING")=-(3xx101.3)/(300)" Joule "K^(-1)=-1.013 " Joule "K^(-1)`
`therefore DeltaS_("SURR")=-1.013" Joule "K^(-1)`


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