Q72P - University Physics with Modern Physics

Question

A long, closed cylindrical tank contains a diatomic gas that is maintained at a uniform temperature that can be varied. When you measure the speed of sound v in the gas as a function of the temperature T of the gas, you obtain these results:   

   

                                                      

(a) Explain how you can plot these results so that the graph will be well fit by a straight line. Construct this graph and verify that the plotted points do lie close to a straight line. (b) Because the gas is diatomic, g = 1.40. Use the slope of the line in part (a) to calculate M, the molar mass of the gas. Express M in grams/mole. What type of gas is in the tank?

Step-by-Step Solution

Answer

A) The equation that states the straight line of the graph is ${\mathrm{v}}^2=\left(\sqrt{\frac{\mathrm{\gamma R}}{\mathrm{M}}}\right)\mathrm{T}$                             

 B) $\mathrm{M}=46.5\mathrm{g}/\mathrm{mol}$

STEP 1 Relation between the speed of sound v and the temperature T

The speed of sound v and the temperature T is $\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{\gamma }\mathbf{R}\mathbf{T}}{\mathbf{M}}}$ where $\mathit{\gamma }$ is the ratio of heat capacities, M is the molar mass and R is gas constant

STEP 2 Graph between v and T

Take the square of both sides of the above equation, so we can plot a graph between v and T

${\mathrm{v}}^2=\left(\sqrt{\frac{\mathrm{\gamma R}}{\mathrm{M}}}\right)\mathrm{T}$ 

Plot a graph between $v^2$ and T. As shown in the figure below, the graph is a straight line.

     

   

STEP 3 Calculate the slope of the graph

The slope of the graph is $\left(\sqrt{\frac{\mathrm{\gamma R}}{\mathrm{M}}}\right)$ For the diatomic gas $\gamma =1$ and $\mathrm{R}=8.314\mathrm{J}/\mathrm{mol}.\mathrm{k}$ Now, we find the slope from the curve and solve the equation for M

$$\begin{gathered} \mathrm{Slope}=\frac{14*{10}^4{\mathrm{m}}^2/{\mathrm{s}}^2-13.5{\mathrm{m}}^2/\mathrm{s}}{340\mathrm{K}-320\mathrm{K}} \\ =250{\mathrm{m}}^2/{\mathrm{Ks}}^2 \\ \mathrm{M}=\left(\frac{1.40\times \left(8.3114\mathrm{J}/\mathrm{molK}\right)}{250{\mathrm{m}}^2/{\mathrm{Ks}}^2}\right) \\ =0.0465\mathrm{Kg}/\mathrm{mol} \\ =46.5\mathrm{g}/\mathrm{mol} \end{gathered}$$