Question

The variation of pressure with elevation is $P=P_0{e}^{-\frac{Mgh}{RT}}$ where $h$ is the elevation, $M$ is the molecular weight of air, $R$ is the universal gas constant and $T$ is the temperature. What are the assumptions involved while calculating pressure with the above given expression?

• A. Density is constant from zero elevation to the point at higher elevation.
• B. Temperature is constant from zero elevation to the point at higher elevation.
• C. Air is assumed to follow ideal gas law.
• D. None of the above

Answer 1

Answer: (B), (C)

The correct options are
B Temperature is constant from zero elevation to the point at higher elevation.
C Air is assumed to follow ideal gas law.

The most basic expression of pressure variation in a static fluid in gravitation field is given by
$\frac{dP}{dz}=-\rho g....\left(1\right)$
We assume air to follow ideal gas law
i.e $PV=nRT$
or $P=\frac{m}{V}\frac{R}{M}T$
$\Rightarrow P=\rho \frac{R}{M}T......\left(2\right)$
For small variations in height, the density of a gas is assumed to be constant.
When we put expression $\left(2\right)$ in $\left(1\right)$, we get
$\frac{dP}{dz}=\frac{-PM}{RT}g$
Integrating on both sides,
$\Rightarrow \int \frac{dP}{P}=\int \frac{Mg}{RT}dz$
If we further assume temperature to be constant,
$\Rightarrow \ln P=-\frac{Mg}{RT}z+C.......\left(3\right)$
Let at $z=0,P=P_0$
$\therefore C=\text{ln}{P}_0......\left(4\right)$
Using equation $\left(4\right)$ in equation $\left(3\right)$, we get,
$\ln \frac{P}{P_0}=\frac{-Mgz}{RT}$
or we can say that, $P=P_0{e}^{-\frac{Mgz}{RT}}$
For any elevation $z=h$,
$P=P_0{e}^{-\frac{Mgh}{RT}}$
Thus, options (b) and (c) are the correct answers.