The volume of a gas at 21^{circ}C temperature | vedclass

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(C) The ideal gas equation is $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass. Since $V = \frac{m}{\rho}$,we have $P = \fra

Vedclass · Accepted Answer

$1.13$

Vedclass · Answer

(C) The ideal gas equation is $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass. Since $V = \frac{m}{ho}$,we have $P = \frac{ho RT}{M}$,which implies $\frac{P}{ho T} = \frac{R}{M} = 	ext{constant}$. Therefore,$\frac{P_1}{ho_1 T_1} = \frac{P_2}{ho_2 T_2}$. Given: At state $1$: $T_1 = 21^{\circ}C = 294.15 \,K$,$P_1 = 768 \,mm \,Hg$,$V_1 = 1 \,L$. At state $2$ $(NTP)$: $T_2 = 273.15 \,K$,$P_2 = 760 \,mm \,Hg$,$ho_2 = 1.2 \,g/L$. Using the relation $ho_1 = ho_2 	imes \frac{P_1}{P_2} 	imes \frac{T_2}{T_1}$: $ho_1 = 1.2 	imes \frac{768}{760} 	imes \frac{273.15}{294.15} \approx 1.13 \,g/L$. Since $V_1 = 1 \,L$,the mass $m = ho_1 	imes V_1 = 1.13 \,g/L 	imes 1 \,L = 1.13 \,g$.

The volume of a gas at $21^{\circ}C$ temperature and $768 \,mm$ pressure is $1 \,L$. If the density of the gas is $1.2 \,g/L$ at $NTP$,then its mass will be ..... $g$.

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