If the density of a mixture of nitrogen and o | vedclass

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(A) Given: $P = 1 \ atm$,$T = 400 \ K$,$\delta = 0.920 \ g \ L^{-1}$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$.Using the ideal gas equation $PV = nRT$

Vedclass · Accepted Answer

$0.456$

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(A) Given: $P = 1 \ atm$,$T = 400 \ K$,$\delta = 0.920 \ g \ L^{-1}$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$.Using the ideal gas equation $PV = nRT$ and $n = \frac{m}{M}$,we get $PM = \delta RT$.$M = \frac{\delta RT}{P} = \frac{0.920 	imes 0.082 	imes 400}{1} = 30.176 \ g \ mol^{-1}$.Let $x$ be the mole fraction of $N_2$. Then the mole fraction of $O_2$ is $(1-x)$.The average molar mass $M = x M_{N_2} + (1-x) M_{O_2}$.$30.176 = x(28) + (1-x)(32)$.$30.176 = 28x + 32 - 32x$.$4x = 32 - 30.176 = 1.824$.$x = \frac{1.824}{4} = 0.456$.Thus,the mole fraction of nitrogen is $0.456$.

If the density of a mixture of nitrogen and oxygen gases at $400 \ K$ and $1 \ atm$ pressure is $0.920 \ g \ L^{-1}$,what is the mole fraction of nitrogen in the mixture? ($R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$; assume ideal gas behavior for oxygen and nitrogen)

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