The work done when two mole of an ideal gas is compressed form a volume of `5 m^(3)` to `1 dm^(3)` at 300 K , under a pressure of 100 kPa is

Options

A. `499.9` kJ
B. `-499.9 `kJ
C. `-99.5` kJ
D. `42495`kJ

Answer

A

Explanation

Work done , W = `- P_(ext") (V_(2) - V_(1))`
Here , `P_("ext") = 100 k Pa`
`V_(1) = 5 m^(3) = 5 xx 10^(3) L implies V_(2) = 1 dm^(3) = 1 L`
`therefore W = -100 xx (1 - 500)`
` = - 499900 k PaL`
`= - 499900 J = - 499.9 kJ ( because 1 k Pa L = 1 J)`