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We have a cylinder closed by a moveable piston. A force will be applied on the piston which has an area of A.
The force can be expressed through product of pressure and section area. So:
F = pex∙A
If the piston is allowed to move a short distance ds, then the differential work done during this process is
We notice that by positive change of distance s is the change of volume closed in cylinder negative, so we get
Now we get the work for volume change:
δW= – pex∙dV
This is a general statement to calculate the work for volume change. We notice that in general expression, the external pressure is used to describe the work for volume change.
Now the reversible process is taken into account:
Since the system undergoes a reversible process, it is in equilibrium state at any moment. Considering the mechanical equilibrium, we know that the external pressure must be equal to the internal pressure which the gas enclosed in the cylinder occupies. So we can rewrite the expression:
δWrev= – pin∙dV
This equation is only available for reversible process!
Work for volume change for reversible isochoric process :
constant volume means dV=0
δWrev= – pin∙dV=0
No work for volume change is done for reversible isochoric process.
Work for volume change for reversible isobaric process:
Therefore the work for volume change for reversible isobaric process is:
Work for volume change for reversible isothermal process:
The ideal gas follows the equation: p∙V=m∙R∙T →p= m∙R∙T/V
Therefore the work for volume change for reversible isothermal process is:
The work for reversible isothermal process is represented by the area under the curve of the isothermal change of state
Work for volume change for isentropic process:
Since the isentropic process follows the equation:
then the internal pressure can be written as following:So the work for volume change is:
Now we get the work for volume change for isentropic process: