Question: (I)The rms speed of molecules in a gas at 20.0°C is to be increased by 4.0%. To what temperature must it be raised?

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

The percentage increase is \(p = 4\% \).

The root-mean-square speed of gas molecules depends on the molecular mass and absolute temperature of the gas.

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

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

Here, R is the universal gas constant, T is the temperature, and M is the molecular mass.

Since the rms speed increased by 4%,

\(\begin{aligned}{c}{{v'}_{{\rm{rms}}}} &= {v_{{\rm{rms}}}} + \frac{4}{{100}}{v_{{\rm{rms}}}}\\ &= 1.04{v_{{\rm{rms}}}}\end{aligned}\)

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

\(\begin{aligned}{c}\frac{{{{v'}_{{\rm{rms}}}}}}{{{v_{{\rm{rms}}}}}} &= \sqrt {\frac{{T'}}{T}} \\T' &= T{\left( {\frac{{{{v'}_{{\rm{rms}}}}}}{{{v_{{\rm{rms}}}}}}} \right)^2}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}T' &= \left( {273 + 20} \right)\;{\rm{K}} \times {\left( {\frac{{1.04{V_{{\rm{rms}}}}}}{{{V_{{\rm{rms}}}}}}} \right)^2}\\ &= 293\;{\rm{K}} \times 1.0816\\ \approx 317\;{\rm{K}}\\ &= 44^\circ {\rm{C}}\end{aligned}\)

Thus, the temperature required to increase the rms speed by 4.0% is \(44^\circ {\rm{C}}\).