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For one mole of an ideal gas $(\large\frac{\delta P}{\delta T})_V.(\large\frac{\delta V}{\delta T})_P . (\large\frac{\delta V}{\delta T})_T $ is equal to ....

$(a)\;\large\frac{-R^2}{P^2}\qquad(b)\;\large\frac{R^2}{P^2}\qquad(c)\;+1\qquad(d)\;-1$

1 Answer

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  • The behaviour of gases has been expressed in terms of various gas laws obtained experimentally i.e Boyle's law , Charles law , Pressure-temperature law , Dalton's law of partial pressure , Graham's law of diffusion and Avogadro's hypothesis . Later on Maxwell derived kinetic equation $PV = \large\frac{1}{3}mnu^2$ , theoretically by assuming the concept of molecules and their motion . The term u represents root mean square speed of molecule.
$PV = RT$ for 1 mole of gas
At constant T : $(\large\frac{\delta V}{\delta P})_T = -\large\frac{V}{P}$
At constant V : $(\large\frac{\delta P}{\delta T})_V = \large\frac{R}{V}$
At constant P : $(\large\frac{\delta V}{\delta T})_P = \large\frac{R}{P}$
$\therefore (\large\frac{\delta P}{\delta T})_V.(\large\frac{\delta V}{\delta T})_P . (\large\frac{\delta V}{\delta T})_T = \large\frac{-R^2V}{P^2V} = -\large\frac{R^2}{P^2}$
Hence answer is (A)
answered Apr 7, 2014 by sharmaaparna1
edited Sep 22, 2014 by pady_1
 

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