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

Short Answer

Expert verified

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

Step by step solution

01

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.

02

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:

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

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

03

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 \(2364^\circ {\rm{C}}\).

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