Answer to Question #297966 in Molecular Physics | Thermodynamics for Dhanush

Question

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

Answer 1

The rms speed (u) is directly proportional to the square root of temperature.

$\frac{u'}{u}=\sqrt{\frac{T'}{T}}$

$\frac{2u}{u}=\sqrt{\frac{T'}{27+273}}$

$2=\sqrt{\frac{T'}{300}}$

$4=\frac{T'}{300}$

$T'=1200K$

Thus, at 1200 K would the rms speed of a gas molecule have twice its value at 27 deg C