Question

If p₀ = 1.03 × 10⁵ Nm^–2, ρ₀ = 1.29 kg m^–3 and g = 9.8 ms^–2, at what height will the pressure drop to (1/10) the value at the surface of the earth?

Answer 1

As p = `p_0e^((ρ_0gh)/p_0)`,

⇒ In `p/p_0 = - (ρ_0gh)/p_0`

By question, `p = 1/10 p_0`

⇒ In `((1/10 p_0)/p_0) = - (ρ_0g)/p_0 h`

⇒ In `1/10 = - (ρ_0g)/p_0 hρ_0`

∴ h = `- p_0/(ρ_0g)` In `1/10`  `- p_0/(p_0g)` In `(10)^-1`

= `p_0/(ρ_0g)` In 10`

= `p_0/(ρ_0g) xx 2.303`   .....[∵ In (x) = 2.303 log₁₀(x)]

= `(1.013 xx 10^5)/(1.22 xx 9.8) xx 2.303`

= 0.16 × 10⁵ m

= 16 × 10³ m