
At the top of the mountain, the thermometer reads ${0^ \circ }C$ and the barometer reads 710 mm Hg. At the bottom of the mountain the temperature is ${30^ \circ }C$ and the pressure is 760 mm Hg. The ratio of density of the air at the top with that of the bottom is:
A. $1:1$
B. $1.04:1$
C. $1:1.04$
D. $1:1.5$
Answer
545.4k+ views
Hint: Density is the measure of mass per unit of volume of a substance. It is denoted by the symbol $\rho $ (the greek letter rho). The average density of a substance can be calculated by its total mass divided by its total volume. A substance made from a dense material such as an iron will have less volume than a substance made from some less dense substance such as water. The density of a substance varies with the temperature and pressure of the substance. The variation is small for solids and liquids but greater for gases. When the pressure of the substance increases due to which the volume of the substance decreases and thus its density increases.
Mathematically we can say,
Density or $\rho = \dfrac{{Mass}}{{Volume}}$
Complete step by step answer:
In the question, the values of the pressure and temperature at the top and bottom are given and we have to calculate the ratio of the density of air at the top with that of the bottom. We will use the gas equation and calculate the ratio of densities.
Given,
At the top of the mountain,
Pressure $ = 710$mm Hg
Temperature $ = {0^ \circ }C = 273.15K$
(We have converted the degree Celsius into Kelvin)
At the bottom of the mountain,
Pressure $ = 760$mm Hg
Temperature $ = {30^ \circ }C = 273.15 + 30 = 303.15K$
(We have converted the degree Celsius into Kelvin)
We know,$PV = nRT$ (gas equation)
Where R is the gas constant
$ \Rightarrow PV = \dfrac{w}{m}RT$
$ \Rightarrow P = \dfrac{w}{V} \times \dfrac{{RT}}{m}$
$ \Rightarrow P = \dfrac{{\rho RT}}{m}$
$ \Rightarrow \rho = \dfrac{{P \times m}}{{RT}}$
(Now we have modified the gas equation and as we know the number of moles or molecules is equal to total mass divided by molecular mass and we have put it instead of n and got an equation for density)
(Here w is the total mass and m is the molecular mass)
As we know that density is dependent on pressure and temperature and independent of molecular mass in this case. So the modified equation for density will be:
$\rho = \dfrac{P}{T}$
Now putting the value in this equation we will find the ratio of the density of the air at the top with that of the bottom
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{{P_{Top}}}}{{{T_{Top}}}} \times \dfrac{{{T_{Bottom}}}}{{{P_{Bottom}}}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{710}}{{273.15}} \times \dfrac{{303.15}}{{760}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{215236.5}}{{207601.6}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{1.04}}{1}$
$ \Rightarrow {\rho _{Top}}:{\rho _{Bottom}} = 1.04:1$
(After calculation we got the ratio of the density of air of the top with that of the bottom)
Therefore, the ratio of the density of the air is $1.04:1$
So, the correct answer is Option B.
Note: The gas equation can be used to determine the volume of the gas at a certain temperature and pressure if its value at a particular temperature and pressure is known. The volume of the given mass of a gas varies with changing pressure and temperature as is evident from Boyle’s law and Charle’s law. If the volume at room temperature and pressure is given, it is generally converted to the volume at NTP (Normal temperature and pressure) or STP (Standard temperature and pressure).
Mathematically we can say,
Density or $\rho = \dfrac{{Mass}}{{Volume}}$
Complete step by step answer:
In the question, the values of the pressure and temperature at the top and bottom are given and we have to calculate the ratio of the density of air at the top with that of the bottom. We will use the gas equation and calculate the ratio of densities.
Given,
At the top of the mountain,
Pressure $ = 710$mm Hg
Temperature $ = {0^ \circ }C = 273.15K$
(We have converted the degree Celsius into Kelvin)
At the bottom of the mountain,
Pressure $ = 760$mm Hg
Temperature $ = {30^ \circ }C = 273.15 + 30 = 303.15K$
(We have converted the degree Celsius into Kelvin)
We know,$PV = nRT$ (gas equation)
Where R is the gas constant
$ \Rightarrow PV = \dfrac{w}{m}RT$
$ \Rightarrow P = \dfrac{w}{V} \times \dfrac{{RT}}{m}$
$ \Rightarrow P = \dfrac{{\rho RT}}{m}$
$ \Rightarrow \rho = \dfrac{{P \times m}}{{RT}}$
(Now we have modified the gas equation and as we know the number of moles or molecules is equal to total mass divided by molecular mass and we have put it instead of n and got an equation for density)
(Here w is the total mass and m is the molecular mass)
As we know that density is dependent on pressure and temperature and independent of molecular mass in this case. So the modified equation for density will be:
$\rho = \dfrac{P}{T}$
Now putting the value in this equation we will find the ratio of the density of the air at the top with that of the bottom
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{{P_{Top}}}}{{{T_{Top}}}} \times \dfrac{{{T_{Bottom}}}}{{{P_{Bottom}}}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{710}}{{273.15}} \times \dfrac{{303.15}}{{760}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{215236.5}}{{207601.6}}$
$ \Rightarrow \dfrac{{{\rho _{Top}}}}{{{\rho _{Bottom}}}} = \dfrac{{1.04}}{1}$
$ \Rightarrow {\rho _{Top}}:{\rho _{Bottom}} = 1.04:1$
(After calculation we got the ratio of the density of air of the top with that of the bottom)
Therefore, the ratio of the density of the air is $1.04:1$
So, the correct answer is Option B.
Note: The gas equation can be used to determine the volume of the gas at a certain temperature and pressure if its value at a particular temperature and pressure is known. The volume of the given mass of a gas varies with changing pressure and temperature as is evident from Boyle’s law and Charle’s law. If the volume at room temperature and pressure is given, it is generally converted to the volume at NTP (Normal temperature and pressure) or STP (Standard temperature and pressure).
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