Question: (I)By what factor will the rms speed of gas molecules increase if the temperature is increased from 20°C to 160°C?

The initial temperature is \(T = 20^\circ {\rm{C}} = 293\;{\rm{K}}\).

The final temperature is \(T' = 160^\circ {\rm{C}} = 433\;{\rm{K}}\).

The square root of the average of the squares of the speeds of molecules is known as root-mean-square speed.

The root-mean-square speed is given as follows:

\({v_{rms}} = \sqrt {\frac{{3RT}}{M}} \) … (i)

Here,Ris the universal gas constant; T is the temperature; M is the molecular mass.

For a particular gas, M is constant. Then from equation (i),

\(\frac{{{{v'}_{{\rm{rms}}}}}}{{{v_{{\rm{rms}}}}}} = \sqrt {\frac{{T'}}{T}} \)

Substitute the values in the above equation.

\(\begin{aligned}{c}\frac{{{{v'}_{{\rm{rms}}}}}}{{{v_{{\rm{rms}}}}}} &= \sqrt {\frac{{433\;{\rm{K}}}}{{293\;{\rm{K}}}}} \\{{v'}_{{\rm{rms}}}} &= 1.2{v_{{\rm{rms}}}}\end{aligned}\)

Thus, the rms speed of gas molecules will increase by a factor of 1.2.