one mole of an ideal gas at 300k in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant presses of 3.0 atm. In this process. The change in entropy of surrroundings `(DeltaS)` in `J^(-1)` is
(1 L atm = 101.3 J)

Options

A. `5.763`
B. `1.013`
C. `-1.013`
D. `-5.763`

Answer

C

Explanation

For isothermal expansion, `DeltaU = 0`
Since pressure is constant , process is irreversible
`q_(irrev) = -w_(irrev) = -(p DeltaV)`
` = -[-3(2-1)] = 3 Latm`
` = 3 xx 101.3 J`
`DeltaS_(surr) = (-q_(irrev))/(T) = -(3 xx 101.3 J)/(300K)`
` = -1.013 JK^(-1)`.