At the top of the mountain- the thermometer reads 0- C and the barometer reads 710 mm Hg- At the bottom of the mountain the temperature is 30- C and the pressure is 760 mm Hg- The ratio of the density of air at the top with that at the bottom is-

The correct option is B 1.04: 1 PV= nRT=w/mRT P=dRT/m d∝P/T d_(top )/d_bottom=710/273×303/760=1.04: 1 ...

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Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

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Question

At the top of the mountain, the thermometer reads $0^{\circ} C$ and the barometer reads 710 $m m H g$ . At the bottom of the mountain the temperature is $30^{\circ} C$ and the pressure is 760 $m m H g$ . The ratio of the density of air at the top with that at the bottom is:

1:1

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1.04: 1

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1:1.04

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1:1.5

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Solution

The correct option is B 1.04: 1
$P V = n R T = \frac{w}{m} R T$
$P = \frac{d R T}{m}$
$d \propto \frac{P}{T}$
$\frac{d_{(t o p)}}{d_{b o t t o m}} = \frac{710}{273} \times \frac{303}{760} = 1.04 : 1$

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At the top of the mountain, the thermometer reads 0∘C and the barometer reads 710 mmHg. At the bottom of the mountain the temperature is 30∘C and the pressure is 760 mmHg. The ratio of the density of air at the top with that at the bottom is:

The correct option is B 1.04: 1PV=nRT=wmRTP=dRTmd∝PTd(top)dbottom=710273×303760=1.04:1

At the top of the mountain, the thermometer reads $0^{\circ} C$ and the barometer reads 710 $m m H g$ . At the bottom of the mountain the temperature is $30^{\circ} C$ and the pressure is 760 $m m H g$ . The ratio of the density of air at the top with that at the bottom is:

The correct option is B 1.04: 1
$P V = n R T = \frac{w}{m} R T$
$P = \frac{d R T}{m}$
$d \propto \frac{P}{T}$
$\frac{d_{(t o p)}}{d_{b o t t o m}} = \frac{710}{273} \times \frac{303}{760} = 1.04 : 1$