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  • If temperature of the atmosphere varies with height as <img alt="\mathrm{T=\left(T_{0}-a h\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BT%3D%5Cleft%28T_%7B0%7D-a%20h%5C

If temperature of the atmosphere varies with height as \mathrm{T=\left(T_{0}-a h\right)}, where \mathrm{a} and \mathrm{T_{0}} are positive constants, then the pressure as a function of height \mathrm{h}  is (assume atmospheric pressure at sea level \mathrm{(h=0)} is \mathrm{p_{0}} and molecule mass \mathrm{M}of the air and acceleration due to gravity \mathrm{g} be constant)

Option: 1

\mathrm{p=p_{0}\left(\frac{T_{0}-a h}{T_{0}}\right)^{M g / R}}


Option: 2

\mathrm{p=p_{0}\left(\frac{T_{0}-a h}{T_{0}}\right)^{2 M g / R a}}


Option: 3

\mathrm{p=p_{0}\left(\frac{T_{0}-a h}{T_{0}}\right)^{3 M g / R a}}


Option: 4

\mathrm{p=p_{0}\left(\frac{T_{0}-a h}{T_{0}}\right)^{4 M g / R a}}


Answers (1)

best_answer

\mathrm{\frac{d p}{d h}=-\rho g=-\left(\frac{p M}{R T}\right) g=-\left[\frac{p M}{R\left(T_{0}-a h\right)}\right] g}

\mathrm{\frac{d p}{p} =-\left(\frac{M g}{R}\right) \frac{d h}{T_{0}-a h}}

\mathrm{\int_{\rho_{0}}^{p} \frac{d p}{p} =-\left(\frac{M g}{R}\right) \int_{0}^{h} \frac{d h}{\left(T_{0}-a h\right)} }

\mathrm{p =p_{0}\left(\frac{T_{0}-a h}{T_{0}}\right)^{\frac{M g}{R a}}}.

Posted by

Divya Prakash Singh

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