The ideal gas equation is $$pV=nRT$$
Show that the pressure exerted by a fixed mass of gas is given by the
equation $$p=\frac{\rho RT}{M}$$Where $\rho$ is the density of the gas and $M$ is the mass of one mole
of gas.
Proof: $\rho = \frac{m}{V}$ where $m,V$ is mass and volume of the gas respectively.
$$p=\frac{nRT}{V}=\frac{n\rho RT}{m}$$
But we know that, $m=nM \implies \frac{n}{m}=\frac{1}{M}$
$$\therefore p=\frac{\rho RT}{M}$$
(b) The Earth’s atmosphere may be treated as an ideal gas whose
density, pressure and temperature all decrease with height.In 1924, Howard Somervell and Edward Norton set a new altitude record
when attempting to climb Mount Everest. They managed to climb to a
vertical height of $8570 m$ above sea level by breathing in natural
air. At this height, the air pressure was $0.35$ times the pressure at
sea level and the temperature was $−33 °C$. At sea level, air has a
temperature $20 °C$ and density $1.3~kg~m^{-3}$Calculate the density of the air at a height of $8570 m$ at the time
the record was set.
As I was just asked to prove the above equation, I am going to assume that it will be useful for this part of the problem.
So, we have (at sea level) $$p=\frac{1.3 \times R \times 293}{M_1}$$
and at our height of $8570m$, $$p=\frac{\rho \times R \times 240}{M_2}$$
Now onto my question. Would $M's$ in my two equations be equal? I.e. would the mass of one mole of air at sea level and at our height of $8750m$ be equal? If so, could I have an explanation why please?
Best Answer
Would the mass of ten apples change if you moved the apples to another location? The principle is essentially the same here - a mole of a gas is just a particular number of gas atoms.