Question

The atmospheric pressure at zero elevation is given as $1.01325\times {10}^5{\text{N/m}}^2$. What is the pressure at an elevation of $3\text{km}$? Given, molecular weight of air $=28\text{g/mol}$ and temperature of air is constant at $300\text{K}$. Also, acceleration due to gravity, $g=10{\text{m/s}}^2$.

• A. $0.64\times {10}^5{\text{N/m}}^2$
• B. ${10}^5{\text{N/m}}^2$
• C. $0.7235\times {10}^5{\text{N/m}}^2$
• D. $2\times {10}^5{\text{N/m}}^2$

Answer 1

The correct option is C $0.7235\times {10}^5{\text{N/m}}^2$

Pressure variation of a gas at higher elevation is given by
$P=P_0{e}^{\left(-\frac{mgh}{RT}\right)}.....\left(1\right)$
Given,
Pressure at zero elevation $\left(P_0\right)=1.01325\times {10}^5{\text{N/m}}^2$
Molecular weight of gas $\left(m\right)=28\times {10}^{-3}\text{kg/mol}$
Temperature of gas $\left(T\right)=300\text{K}$
Height at which pressure is to be determined $\left(h\right)=3000\text{m}$
So, by substituting the given data in $\left(1\right)$,
$P=1.01325\times {10}^5\times e^{(-\frac{28\times {10}^{-3}\times 10\times 3000}{8.315\times 300})}$
$=0.7235\times {10}^5{\text{N/m}}^2$
Thus, option (c) is the correct answer.