Adiabatics everywhere!

Relevant wiki: First Law of Thermodynamics

The point here is that the temperature of air won't be constant.The expansion being an adiabatic process we can compute the lapse rate(fall of temperature $dT$ with heigh $dz$ ).Here is a way to do it.as usual for an adiabatic process $dQ=pdV+n(Cv)dT=0$ and from ideal gas law $PV=nRT$ .So one can write $dV=d(nRT/P)=(nR/P^2)(PdT-TdP)$ .putting in the equation and using $dP=-ugdz$ where u is the density air or $dP=-(PM/RT)gdz$ .yields dT/dz=(Mg/R)(\(\gamma -1)/ $\gamma$ ).Now we move to Laplace equation v=√\(yRT/M) or dz/dt=√\(Ry/M(To-(Mg/R)((\gamma -1)/ $\gamma$ *z).now that we have the differential we may integrate it to obtain the required ans.Its worth noting that it differs from the usual methods technique by abt. $17.7$ percent.As the height given is quite large as compared to everydays examples such variations do occur.Note that y is the adiabatic exponent. $\gamma$

i did it using S.H.M type of a thing lol :P temperature of air won't be constant. let i move up an x-distance and move additional dx how would temperature change ? first use pv^(5/3)=k considering a sphere of radius x with additional dx radius extension , then t^5/3=Cp^2/3..........if u do it like this temperature change deltaT=T(2/3)*dx/x .......now it can be done easily, nice Q. hope u r happy now @Spandan Senapati coz the Q is lvl 5 now :P btw how much time it took u to solve ?