Question

At the top of the mountain, a thermometer reads $ {{7^\circ C}} $ and a barometer reads 70 cm of Hg. At the foot of the mountain, they read $ {{27^\circ C}} $ and 76 cm of Hg respectively. The ratio of density of the air at the top to that at the bottom of the mountain is:
A) $ {\text{0}}{\text{.885}} $
B) $ {\text{0}}{\text{.987}} $
C) $ {\text{0}}{\text{.75}} $
D) $ {\text{10}} $

Answer 1

Hint: In the above equation, pressure and temperature at the bottom and at the top of the mountain is given and we have to find out the density of the air ratio. So, first we have to find the relation between ideal gas law and density. We shall rearrange the ideal gas equation and the value of the number of moles to find the relation between density and pressure.
Formula Used: $ {\text{PV = nRT}} $
 $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $
Where n is the number of moles, P is the pressure, V is the volume, m is the mass, M is the molecular mass, R is a constant and T is the temperature.

Complete Step by step solution
Since, we can relate the pressure and temperature using the ideal gas law, we have $ {\text{PV = nRT}} $
We know that the number of moles can be written as, $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $ , where m is given mass and M is molar mass. So, now we can write the ideal gas equation as:
 $ {\text{PV = }}\dfrac{{\text{m}}}{{\text{M}}}{\text{RT}} $
  $ \dfrac{{\text{m}}}{{\text{V}}}{\text{ = }}\dfrac{{{\text{PM}}}}{{{\text{RT}}}} $
We know that $ \dfrac{{\text{m}}}{{\text{V}}} $ gives density of the air. So,
Density ( $ {\text{d}} $ )= $ \dfrac{{{\text{PM}}}}{{{\text{RT}}}} $
So, now we can find the ratio of density of air at the top ( $ {{\text{d}}_{\text{1}}} $ ) to the density of air at the bottom ( $ {{\text{d}}_{\text{2}}} $ ) as:
 $ \dfrac{{{{\text{d}}_{\text{1}}}}}{{{{\text{d}}_{\text{2}}}}}{\text{ = }}\dfrac{{\dfrac{{{{\text{P}}_{\text{1}}}{{\text{M}}_{\text{1}}}}}{{{\text{R}}{{\text{T}}_{\text{1}}}}}}}{{\dfrac{{{{\text{P}}_{\text{2}}}{{\text{M}}_{\text{2}}}}}{{{\text{R}}{{\text{T}}_{\text{2}}}}}}} $
Rearranging:
 $ \dfrac{{{{\text{d}}_{\text{1}}}}}{{{{\text{d}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{1}}}{{\text{M}}_{\text{1}}}{\text{R}}{{\text{T}}_{\text{2}}}}}{{{{\text{P}}_{\text{2}}}{{\text{M}}_{\text{2}}}{\text{R}}{{\text{T}}_{\text{1}}}}} $
The molar mass of the gas and the value of R is constant and hence get cancelled out, which reduces the equation to:
 $ \dfrac{{{{\text{d}}_{\text{1}}}}}{{{{\text{d}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{1}}}}} $
 $ {{\text{T}}_1}{{ = 7^\circ C = 7 + 273K = 280K}} $
 $ {{\text{T}}_{\text{2}}}{{ = 27^\circ C = 27 + 273K = 300K}} $
Substituting the values, we get
 $ \dfrac{{{{\text{d}}_{\text{1}}}}}{{{{\text{d}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{70 \times 300}}}}{{{{280 \times 76}}}}{\text{ = 0}}{\text{.987}} $
So, the ratio of density of air at the top ( $ {{\text{d}}_{\text{1}}} $ ) to the density of air at the bottom ( $ {{\text{d}}_{\text{2}}} $ ) is $ {\text{0}}{\text{.987}} $ .
Hence, option B is the correct option.

Note
-The ideal gas models tend to fail at lower temperature or higher pressure, when intermolecular forces and molecular size become important.
-It also fails for most heavy gases, such as many refrigerants, and for gases with strong intermolecular forces.