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What is the pressure of HCl gas at $-40^{\large\circ}C$ if its density is $8.0 kgm^{-3}? (R = 8.314 JK^{-1}mol^{-1})$

$(a)\;424.58\times10^2 Pa\qquad(b)\;424.58\times10^4 Pa\qquad(c)\;424.58\times10^5 Pa\qquad(d)\;424.58\times10^3 Pa$

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Equation for ideal gas,
$PV = \large\frac{w}{M}RT$
or $P = \large\frac{w}{V} \times \large\frac{RT}{M}$
$\;\;\; = d \times \large\frac{RT}{M}\;\;\; (\large\frac{w}{V} = density = d)$
Given
$d = 8.0 kg m^{-3} ; R = 8.314 JK^{-1}mol^{-1};$
$T = -40 + 273 = 233K$
And $M = 36.5 g mol^{-1} = 36.5\times 10^{-3}kg\;mol^{-1}$
Substituting the values in the above equation ,
$P = \large\frac{8.0\times8.314\times233}{36.5\times10^{-3}}$
$\;\;\;\; = 424.58\times10^3 Pa$
Hence answer is (D)
answered Apr 8, 2014 by sharmaaparna1
 

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