1.

The work done during the expansion of a gas from a volume of 4dm^(3) to 6dm^(3) against a constant external pressure of 3 atm is (1L atm = 101.32 J)

Answer»

`+ 304 J`
`- 304 J`
`- 6 J`
`-608 J`

SOLUTION :`W=-pDeltaV, W=-3xx(6-4)`
`W=-6xx101.32(THEREFORE "1 L atm=101.32 J")`
W=-608 J


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