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(A) The ideal gas equation in terms of density is given by $PV = nRT = \frac{m}{M}RT$,which implies $P = \frac{\rho RT}{M}$,where $\rho$ is the densit

Vedclass · Accepted Answer

$75/76$

Vedclass · Answer

(A) The ideal gas equation in terms of density is given by $PV = nRT = \frac{m}{M}RT$,which implies $P = \frac{ho RT}{M}$,where $ho$ is the density. Thus,$\frac{P}{ho T} = \frac{R}{M} = 	ext{constant}$. Therefore,$\frac{P_1}{ho_1 T_1} = \frac{P_2}{ho_2 T_2}$,which gives $\frac{ho_1}{ho_2} = \frac{P_1}{P_2} 	imes \frac{T_2}{T_1}$. Given: At the top: $P_{top} = 70 \, cm$ of $Hg$,$T_{top} = 7 + 273 = 280 \, K$. At the bottom: $P_{bottom} = 76 \, cm$ of $Hg$,$T_{bottom} = 27 + 273 = 300 \, K$. Substituting these values: $\frac{ho_{top}}{ho_{bottom}} = \frac{70}{76} 	imes \frac{300}{280} = \frac{70}{76} 	imes \frac{30}{28} = \frac{70}{76} 	imes \frac{15}{14} = \frac{5 	imes 15}{76} = \frac{75}{76}$.

At the top of a mountain,a thermometer reads $7^{\circ}C$ and a barometer reads $70 \, cm$ of $Hg$. At the bottom of the mountain,these read $27^{\circ}C$ and $76 \, cm$ of $Hg$ respectively. The ratio of the density of air at the top to that at the bottom is

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