Find the ratio of the density of air at the t | vedclass

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(A) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,we have $P = \frac{\rho RT}{M}$,where $\rho$ is the density of the gas.Thus,$\frac{P}{\rho

Vedclass · Accepted Answer

$75/76$

Vedclass · Answer

(A) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,we have $P = \frac{ho RT}{M}$,where $ho$ is the density of the gas.Thus,$\frac{P}{ho T} = \frac{R}{M} = 	ext{constant}$.This implies $\frac{ho_1}{ho_2} = \frac{P_1}{P_2} 	imes \frac{T_2}{T_1}$.Given at the top: $P_{	ext{top}} = 70 	ext{ cm of Hg}$,$T_{	ext{top}} = 7^{\circ}	ext{C} = 280 	ext{ K}$.Given at the bottom: $P_{	ext{bottom}} = 76 	ext{ cm of Hg}$,$T_{	ext{bottom}} = 27^{\circ}	ext{C} = 300 	ext{ K}$.Therefore,the ratio of density at the top to the density at the bottom is:$\frac{ho_{	ext{top}}}{ho_{	ext{bottom}}} = \frac{P_{	ext{top}}}{P_{	ext{bottom}}} 	imes \frac{T_{	ext{bottom}}}{T_{	ext{top}}} = \frac{70}{76} 	imes \frac{300}{280} = \frac{70}{76} 	imes \frac{30}{28} = \frac{70}{76} 	imes \frac{15}{14} = \frac{5 	imes 15}{76} = \frac{75}{76}$.

Find the ratio of the density of air at the top of a mountain to the density of air at the bottom of the mountain,given the conditions shown in the image.

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