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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?
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Solution
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
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