Work done when volume is increase or decrease

Consider a gas in a container with a movable piston on top. If the gas expands, the piston moves out and work is done by the system on the surroundings.

 

Alternatively, if the gas inside contracts, the piston moves in and work is done by the surroundings on the system. Why would the gas inside contract or expand?

 

It would if the external pressure, P_ex, and the internal pressure, P_in, were  different.  To  calculate the work done in moving the piston, we know that the  force =  pressure times area and then work equals pressure times area times distance or work equals pressure times the change in volume. So, W = the integral of (P_ex) dV

 

The differential work done (dW) associated with a differential displacement (dl) is given by

 

For a piston cylinder assembly,

dW = F dl

dW = F dl = PA (dl) = P dV

If the gas is allowed to expand reversibly from the initial pressure P to final pressure P,  then  the work done is given by

                                                                         W =  ∫ p dV

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