Question #301801

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

Expert's answer

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

vT,v\sim \sqrt T,

T2=22T1=4T1=1200 K=927°C.T_2=2^2T_1=4T_1=1200~K=927°C.


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