At a constant pressure of 10^(4)Nm^(-2)...

At a constant pressure of `10^(4)Nm^(-2)` , a gas expands by `0.25m^(3)` work done by the gas is

2500 J

250 J/kg K

25 J

2.5 J

Text Solution

Verified by Experts

The correct Answer is:

`W=PdV`
`=10^(4)xx0.25`
=2500 J.

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