Work done when volume is increase or decrease

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, Pex,
and the internal pressure, Pin, 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 (Pex) 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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