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Calculate the temperature of gas if it obeys Van der Waals equation from the following data. A flask of 25 litre contains 10 mole of a gas under 50 atm . Given $a = 5.46\; atm \;litre^2mol^{-2}$ and $b = 0.031\;litre\;mol^{-1}$

$(a)\;12.56^{\large\circ}C\qquad(b)\;1.256^{\large\circ}C\qquad(c)\;1256.93^{\large\circ}C\qquad(d)\;0.1256^{\large\circ}C$

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Given
P = 50 atm
V = 25 litre
n = 10
$a = 5.46\;atm\;litre^2\;mol^{-2}$
$b = 0.031\;litre\;mol^{-1}$
Now Van der Waal's equation for n mole of gas
$[P + \large\frac{n^2a}{V^2}] [ V - nb] = nRT$
$[50 + \large\frac{100\times5.46}{625}][25-10\times0.031]$
$= 10\times0.0821\times T$
$T = 1529.93K$
$=1256.93^{\large\circ}C$
Hence answer is (C)
answered Apr 2, 2014 by sharmaaparna1
 

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