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

In Isothermal process,
`DeltaU=0`, then `q_("irr")= - W_("irr")`
`q_("irr")= - (-P_(ect) DeltaV)`
`= 3` L atm `=3xx101.3 J=303.9 J`
`DeltaS_(surr)=q_(surr)/T=(-q_(sys))/T=(-303.9 J)/(300 K)=-1.013 J K^(-1)`
`C(s)+O_(2)(g) rarrCO_(2)(g), DeltaH=-393.5 kJ mol^(-1)` ...(i)