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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a constant pressure to twice its initial volume.

Answer :

In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
W = 780000 J

Hope this answers the question. Have a nice day.

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