A 1.0 mol sample of an ideal gas is kept at 0.0^0 C during an expansion from 3.0L TO 10.0L.

how much work is done by the gas during the expansion?

Answer ::

As, the temperature of gas is fixed.
This process is isothermal.

Now,
work done (W) = nRT ln ( V2/V1)
= (1)(8.31)(273)ln ( 3/10)
= - 2.7 × 10^3 J.

Thanks