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Thermodynamics

Question
CBSEENCH11006140

Derive an expression for pressure-volume work (PV – work).

Solution
Let us consider a gas enclosed in a cylinder fitted with a frictionless piston having an area of cross-section equal to ‘a’. Let the gas expand against the external pressure P facing the piston to move through a small distance d. Then work of expansion is given by,

straight W space equals space Force space cross times space displacement
space space space space space equals straight F space cross times space straight d
But space Force space left parenthesis straight F right parenthesis space equals space Pressure space left parenthesis straight P right parenthesis space cross times space area
therefore space space space space space space space straight W space equals space straight P space cross times space straight a space cross times space straight d
Since space straight a space cross times straight d space equals space Change space in space volume space left parenthesis increment straight V right parenthesis
       straight i. straight e. space space space space space space left parenthesis straight V subscript 2 minus straight V subscript 1 right parenthesis
therefore space space space space space straight W space equals space straight P cross times increment straight V
During expansion, work is done by the system and ∆V(V2 – V1) is positive. Therefore, in order to satisfy the sign conventions, a negative sign is put in the expression for work.
straight W space equals space minus straight P space cross times space increment straight V
During compression, Work is done on the system. ∆V(V2 – V1) for compression is negative. Therefore work, as obtained from the above expression, comes out to be positive and satisfies the conventions.
The general expression for all types of PV-work is given by
straight W space equals space minus straight P increment straight V