Height of atmosphere!!

This problem is taken from INPhO Problem.

From Ideal Gas equation:- $PV=nRT$ $PV=\dfrac{m}{M}RT$ $PM=\rho RT \ldots (1)$ Now From the assumption that atmosphere is in hydro-static equilibrium, we can write that:- $dP = -\rho gdz$ From equation $(1)$ $dP=-\dfrac{PMgdz}{RT} \ldots (2)$ Since all the processes are adiabatic, therefore:- $PV^{\gamma}=constant$ Using its analogue:- $T^{\gamma}P^{1-\gamma} = constant$ Differentiating it, we get:- $P^{1-\gamma}\gamma T^{\gamma-1}dT +T^{\gamma}(1-\gamma)P^{-\gamma}dP=0$ Simplifying it we get:- $\gamma PdT+(1-\gamma )TdP=0$ $\dfrac{dP}{P} = \dfrac{\gamma dT}{T(\gamma -1)} \ldots (3)$ From equation $(2)$ $\dfrac{dp}{P} = \dfrac{-Mgdz}{RT}$ From equation $(3)$ $\dfrac{\gamma dT}{T(\gamma -1)} = \dfrac{-Mgdz}{RT}$ Cancelling $T$ , integrating and applying limits, we get:- $\int_{T_{0}}^{T} \dfrac{\gamma}{\gamma-1}dT = \int_{0}^{h} \dfrac{-Mg}{R}dz$ Solving it we get:- $\dfrac{\gamma}{\gamma -1} (T-T_{0}) = \dfrac{-Mgh}{R}$ Now is some logic. For atmosphere to exist to some extent, The lowest temperature can be $0K$ . Hence if we put $T=0$ in above equation we will get required height. $\dfrac{-\gamma T_{0}}{\gamma-1} = \dfrac{-Mgh}{R}$ Now we can calculate $\boxed{ h = 30102.4m}$