The amount of work done, when `20 xx 10^(-3)` kg of Argon (Mol. wt. = 40) expands reversibly from a pressure of 10 atm. to 1 atm. at a temperature `t^(@)C` is :

Options

A. `-(2.303 R (272 +t)xx10^(-3))/(2)`
B. `-(2.303 R(273+t) log 0.1)/(2)`
C. `-(2.303 Rt xx 10^(-3))/(2)`
D. `-(2.303 R(273+t))/(2)`

Answer

D

Explanation

`n=("Weight of Ar")/("Molecular weight")=(20xx10^(-3))/(40xx10^(-3))=(1)/(2)`
`W_("max") =-2.303 nRT log_(10) (P_(1))/(P_(2))`
`=-2.303 xx(1)/(2) xx RT log_(10) (10)/(1)`
`W_("max") =(-2.303 xx R xx (273 +t))/(2) ( because T=t+273)`