What is the density of N_2 gas at 227,^circ C | vedclass

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(C) The ideal gas equation is $PV = nRT$.Since $n = \frac{W}{M}$,where $W$ is mass and $M$ is molar mass,we have $PV = \frac{W}{M} RT$.Rearranging for

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$3.41$

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(C) The ideal gas equation is $PV = nRT$.Since $n = \frac{W}{M}$,where $W$ is mass and $M$ is molar mass,we have $PV = \frac{W}{M} RT$.Rearranging for density $d = \frac{W}{V}$,we get $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.Given: $P = 5.00\ atm$,$M(N_2) = 28\ g/mol$,$T = 227 + 273 = 500\ K$,and $R = 0.082\ L\ atm\ K^{-1}\ mol^{-1}$.Substituting the values: $d = \frac{5.00 	imes 28}{0.082 	imes 500} = \frac{140}{41} \approx 3.41\ g/L$.Note: The density is $3.41\ g/L$. Since $1\ L = 1000\ mL$,the density in $g/mL$ is $0.00341\ g/mL$. Given the options,the numerical value $3.41$ is expected.

What is the density of $N_2$ gas at $227\,^\circ C$ and $5.00\ atm$ pressure? (Given: $R = 0.082\ L\ atm\ K^{-1}\ mol^{-1}$)

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