Calculate the work done during compression of 2 mol of an ideal gas from a volume of `1m^(3)` to `10 dm^(3)` 300 K against a pressure of 100 KPa .

Options

A. `-99 kJ`
B. `+99` kJ
C. `+22.98` kJ
D. `-22.98` kJ

Answer

B

Explanation

`n=2 moles, V_(1) = 1m^(3) = 10 + 3 dm^(3), V_(2) = 10 dm^(3), pi =100 k Pa`
`W=-P_(ex) (V_(2)-V_(1))`
`=-100 kPa (10 dm^(3) -1000 dm^(3))`
`=-100 kPa (-990 dm^(3))`
`+100 xx 990 J` `(therefore kPa xx dm^(3) =J)`
`=+99 kJ`