Answer to Question #179158 in Molecular Physics | Thermodynamics for Anand

Question

Calculate the temperature at which the root mean square velocity of a chlorine molecule will be equal to its escape velocity from the earth’s gravitational field. 

Given that the radius of earth is 6400 km.

Answer 1

Solution.

$M(Cl)=35.5\sdot10^{-3}kgmol^{-1};$

$R_e=6.4\sdot 10^{6}m;$

$k=1.38\sdot10^{-23}JK^{-1};$

$N_A=6.02\sdot 10^{23}mol^{-1};$

$v-?;$

$W_k=\dfrac{mv^2}{2}; W_k=\dfrac{3kT}{2}; \implies$ $v=\sqrt{\dfrac{3kT}{m}};$

$m=\dfrac{M}{N_A}; v=\sqrt{\dfrac{3kN_AT}{M}};$

$v_{escape}=\sqrt{2gR_e};$

$\sqrt{\dfrac{3kN_AT}{M}}=\sqrt{2gR_e} \implies T=\dfrac{2gR_eM}{3kN_{A}};$

$T=\dfrac{2\sdot 10mc^{-2}\sdot 6.4\sdot 10^6m\sdot 35.5\sdot 10^{-3}kgmol^{-1}}{3\sdot1.38\sdot10^{-23}JK^{-1}\sdot6.02\sdot 10^{23}mol^{-1}}=1.8\sdot10^5K;$

Answer: $T=1.8\sdot10^5K.$