Chapter 13, Problem 51P

To determine

To Calculate:The raised temperature.

Answer to Problem 51P

If the rms speed is increased by one percent, then the temperature is raised by $\text{25}\text{.9}\text{°C}$ .

Explanation of Solution

Given info:

Temperature, $T_1=20\text{°C}=20+273=293\text{K}$

Formula used:

The root mean square speed of the gas molecules is,

  ${\text{v}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3kT}}{\text{m}}}$

Here, $k$ is the Boltzmann constant,

  $T$ is the temperature of the gas,

  $m$ is mass of the gas.

Calculation:

The rms value for the first case,

  ${\left({\text{v}}_{\text{rms}}\right)}_{\text{1}}\text{=}\sqrt{\frac{{\text{3kT}}_{\text{1}}}{\text{m}}}$......... Equation (1)

And, also for the second case,

  ${\left({\text{v}}_{\text{rms}}\right)}_{\text{2}}\text{=}\sqrt{\frac{{\text{3kT}}_{\text{2}}}{\text{m}}}$......... Equation (2)

Now divide these equations by each other, then

  $\begin{array}{l}\frac{{\left({\text{v}}_{\text{rms}}\right)}_{\text{2}}}{{\left({\text{v}}_{\text{rms}}\right)}_{\text{1}}}\text{=1}\text{.010=}\frac{\sqrt{\frac{{\text{3kT}}_{\text{2}}}{\text{m}}}}{\sqrt{\frac{{\text{3kT}}_{\text{1}}}{\text{m}}}}\text{=}\sqrt{\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}\\ \Rightarrow \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=1}\text{.010}\\ \Rightarrow {\text{T}}_{\text{2}}\text{=1}{\text{.010T}}_{\text{1}}\\ \Rightarrow {\text{T}}_{\text{2}}\text{=1}\text{.010×}\left(\text{293K}\right)\\ \Rightarrow {\text{T}}_{\text{2}}\text{=299}\text{.1K}\text{=}\text{25}\text{.9°C}\end{array}$

Conclusion:

The raised temperature is $\text{25}\text{.9°C}$ .