Question

Question: (I)A gas is at 20°C. To what temperature must it be raised to triple the rms speed of its molecules?

Short answer

The temperature required to triple the rms speed is \(2364^\circ {\rm{C}}\).

Step-by-step solution

Given data

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

The final rms speed is triple the initial rms speed.

Understanding the root mean square speed

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.

Evaluation of the increase in temperature of the gas

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.

\(\begin{aligned}{c}T' &= 293\;{\rm{K}} \times {\left( {\frac{{3{V_{{\rm{rms}}}}}}{{{V_{{\rm{rms}}}}}}} \right)^2}\\ &= 293\;{\rm{K}} \times 9\\ &= 2637\;{\rm{K}}\\ &= 2364^\circ {\rm{C}}\end{aligned}\)

Thus, the temperature required to triple the rms speed is \(2364^\circ {\rm{C}}\).