Answer to Question #301801 in Molecular Physics | Thermodynamics for Dhanush

Question

Calculate the temperature at which root mean square speed of gas molecules is double of its speed at 27°c pressure remaining constant

Answer 1

$v=\sqrt{\frac{3RT}M},$

$v\sim \sqrt T,$

$T_2=2^2T_1=4T_1=1200~K=927°C.$