Work done by the system in isothermal reversible process is `w_(rev.)= -2.303 nRT "log"(V_(2))/(V_(1))`. Also in case of adiabatic reversible process work done by the system is given by: `w_(rev.) = (nR)/(gamma -1) [T_2 - T_1]`. During expansion disorder increases and the increase in disorder is expressed in terms of change in entropy `DeltaS = q_(rev.)/T`. The entropy changes also occurs during transformation of one state to other end expressed as `DeltaS = DeltaH/T`. Both entropy and enthalpy changes obtained for a process were taken as a measure of spontaniety of process but finally it was recommended that decrease in free energy is responsible for spontaniety and `DeltaG=DeltaH - T DeltaS`.
Which statements are correct? (1) The expansion work for a gas into a vacuum is equal to zero. (2) `1` mole of a gas occupying `3` litre volume on expanding to `15` litre at constant pressure of `1atm` does expansion work `1.215 kJ`. (3) The maximum work done during expansion of `16gO_2` at `300K` from `5dm^3` to `25 dm^3` is `2.01 kJ`. (4) The `DeltaS` for `S to L` is almost negligible in comparision to `DeltaS` for `L to G`. (5) `DeltaS = 2.303 nR "log"(V_(2))/(V_(1)).` (at constant `T`)

Options

A. 2,3,4,5
B. 1,2,3,4,5
C. 1,2
D. 4,5

Answer

B

Explanation

(1) In free expansion `P_(ext) = 0 rarr w = 0`
(2) `W = -1 [15-3] = -12lit-atm`
`= -12 xx 101.3 J`
(3) `W_(rev) = -2.303 xx (16)/(32) xx 8.314 xx 300 xx log.(25)/(5)`
(4) `Delta S` for `L rarr G` is more
(5) `(Delta S)_(T) = 2.303nR log.(v_(2))/(v_(1))`