The Chemistry Maths Book, Second Edition

(Grace) #1

158 Chapter 5Integration


according as the external pressurep


ext

is greater than or less than p. Ifp


ext

1 > 1 pthen


the fluid is compressed, and the work done by the external forceF


ext

in moving the


piston from ato bis


(5.64)


Now, the external force has magnitude|F


ext

| 1 = 1 p


ext

A, and a length of cylinder |dx|


contains a volume|dV| 1 = 1 A|dx|. The work can therefore be written in ‘pressure–


volume’ form as


(5.65)


in which the limits of integration now refer to the volume, and the minus sign is


included to make the work positive for compression. It can be shown that this


expression for the mechanical work done on a thermodynamic system is independent


of the shape of the container.


To compress the fluid, it is necessary that the external pressure be greater than the


internal pressure of the fluid. Letp


ext

1 = 1 (p 1 + 1 ∆p)where∆pis a positive excess pressure


that, for simplicity, can be assumed to be constant throughout the compression.


Then, with V


a

1 > 1 V


b

,


(5.66)


Conversely, to allow the fluid to expand fromV


b

toV


a

it is necessary that the external


pressure be smaller than the internal pressure. Ifp


ext

1 = 1 (p 1 − 1 ∆p)then


(5.67)


>−Z


b

a

pdV


WpdVpVV


ba

b

a

ab

=−Z +∆()−


>−Z


a

b

pdV


W p dV p dV p dV p V V


ab

a

b

a

b

a

b

ab

=−ZZZ−∆∆=− + ()−


WpdV


ab

a

b

=−Z


ext

WFdx


ab

a

b

=Z


ext

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Figure 5.23

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