The maximum work done in expanding `1.6 xx 10^(-2)` oxygen at 300 k and occupying a volume of `10 cm^ 3` isothermally until the volume of `10 dm^3` isothermally unti the volume becomes `50 dm^2 is ( R= 8.314 JK6^(-1) mol^(-1))`

Options

A. `-2.0 xx 10^3J`
B. `-2.81 xx 10^3 J `
C. `2.0 xx 10^(-3) J`
D. `-2.0 xx 10^(-6) J`

Answer

A

Explanation

Wmax =-2.303 nRT `"log" (V_2)/(V_1)`
`= - 2.303 xx (1.6 xx 10^(-2))/(32 xx 10^(-3)) xx 8.314 xx 300 xx log ""(50)/(10)`
`=- 2.303 xx 0.5 xx 8.314 xx 300 xx 0.6990 `
`= -2007 . 5 j`