Question

Suppose the density of air at Madras is

ρ_{0}

and atmospheric pressure is

P_{0} .

If we go up, the density and the pressure both decrease. Suppose we wish to calculate the pressure at a height

10 km

above Madras. If we use the equation

P_{0} - P = ρ_{0} g z,

will we get a pressure more than the actual or less than the actual? Neglect the variation in

g .

Does your answer change if you also consider the variation in

g

?

Accepted Answer

When we calculate pressure difference between two points, i.e, $P_{0} - P = ρ_{0} g h$ , density and acceleration due to gravity remains constant.
Case 1:
If we go to a large height like $10 km$ above the Earth surface, density of air decreases (let $ρ$ , which is less than $ρ_{0}$ ), then using the above formula, the pressure at this height is $P_{actual} = P_{0} - ρ g h$ , which is more than $P$ (keeping acceleration due to gravity constant) calculated from the above equation.

Case 2:
Yes. If the value of $g$ changes, then also pressure at this height changes. As $g$ decreases with height, from the pressure difference equation. The actual pressure is more than that calculated by using the equation. However, the pressure is not calculated for such a height by using this formula. The atmospheric pressure change with height is calculated by using Gas equation at isothermal condition and keeping variation of $g$ constant, which comes to be $P = P_{0} e^{(\frac{- ρ_{0} g h}{P_{0}})}$ , where $P_{0}$ and $ρ_{0}$ are the pressure and density at the ground level (reference level) and $P$ is the pressure at the height $h$ .
Hence, in both the case we will get more pressure than the calculated value using the given equation.

H C Verma Solutions for Chapter: Fluid Mechanics, Exercise 1: QUESTIONS FOR SHORT ANSWER

H C Verma Physics Solutions for Exercise - H C Verma Solutions for Chapter: Fluid Mechanics, Exercise 1: QUESTIONS FOR SHORT ANSWER

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