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Boyle's law may be represented as

$(a)\;(\large\frac{dp}{dv})_T = \large\frac{K}{V}\qquad(b)\;(\large\frac{dp}{dv})_T = \large\frac{-K}{V^2}\qquad(c)\;(\large\frac{dp}{dv})_T = \large\frac{-K}{V}\qquad(d)\;None\;of\;these$

1 Answer

PV = Constant
On differentiating at constant T we get
Pdv + Vdp = 0
(Or) $(\large\frac{dp}{dv})_T = \large\frac{-P}{V}$
$(\large\frac{dp}{dv})_T = \large\frac{-K}{V^2}$
Since PV = K
Hence answer is (b)
answered Feb 28, 2014 by sharmaaparna1
 

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