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2.8 g of N(2) gas at 300 K and 20 atm wa...

`2.8 g` of `N_(2)` gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. Calculate W for the gas.

A

`+236.95 J`

B

`+136.95 J`

C

`-236.95 J`

D

`136.95 J`

Text Solution

Verified by Experts

The correct Answer is:
C
`W_(irr) = -P_(ex)(V_(2)-V_(1))`
`= -P_(ex)((nRT)/(P_(2))-(nRT)/(P_(1)))`
`= -P_(ex)nRT((1)/(P_(2))-(1)/(P_(1)))`
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