Answer in Brief

**Answer in brief.**

Derive the expression for PV work.

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#### Solution

**Pressure-volume work**

- Consider a certain amount of gas at constant pressure P is enclosed in a cylinder fitted with a frictionless, rigid movable piston of area A. Let the volume of the gas be V
_{1}at temperature T. This is shown in the adjacent diagram. - On expansion, the force exerted by a gas is equal to area of the piston multiplied by pressure with which the gas pushes against piston. This pressure is equal in magnitude and opposite in sign to the external atmospheric pressure that opposes the movement and has its value - P
_{ext}.

Thus,

f = - P_{ext }× A .....(1)

where, P_{ext}is the external atmospheric pressure. - If the piston moves out a distance d, then the amount of work done is equal to the force multiplied by distance.

W = f × d .....(2)

Substituting equation (1) in (2) gives

W = - P_{ext }× A × d .....(3) - The product of area of the piston and distance it moves is the volume change (ΔV) in the system.

ΔV = A × d .....(4)

Combining equation (3) and (4), we get

W = - P_{ext }ΔV

W = - P_{ext }(V_{2}- V_{1})

Where V_{2}is the final volume of the gas.

Is there an error in this question or solution?

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