Question

At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Cl2 molecules have at 22 ∘C?

Answer 1

Answer:

Explanation:

Given

kinetic Energy of $Cl_2$ molecule

$\frac{1}{2}mv^2=\frac{3}{2}RT$

$v=\sqrt{\frac{3RT}{M}}$

For same temperature both xenon and chlorine have same average speed

$v_{Cl_2}=\sqrt{\frac{3RT}{M_{Cl_2}}}$

$v_{Xe}=\sqrt{\frac{3RT}{M_{Xe}}}$

$v_{Cl_2}=v_{Xe}$

$\sqrt{\frac{3RT}{M_{Cl_2}}}=\sqrt{\frac{3RT}{M_{Xe}}}$

$\frac{T{Cl_2}}{M{Cl_2}}=\frac{T{Xe}}{M_{Xe}}$

$\frac{295}{71}=\frac{T}{131.3}$

$T=545.54\ K\approx 272.54^{\circ}$

Answer 2

At "270°C" xenon atoms will have the same average speed. A further explanation is below.

As we know the formula,

→ $\frac{1}{2}Mv^2 = \frac{3}{2}RT$

or,

→ $v^2=\frac{3}{2} RT\times \frac{2}{M}$

       $= \frac{3RT}{M}$

Since,

The speeds are going to be same, then

→ $\frac{3RT_1}{yM_1} = \frac{3RT_2}{M_2}$

By cancelling out the factors, we get

→ $\frac{T_1}{M_1} = \frac{T_2}{M_2}$

or,

→ $T_2 = \frac{T_1M_2}{M_1}$

By substituting the values, we get

→      $= \frac{293\times 131.3}{70.9}$

→      $= 542.6 \ K$

or,

→      $= 542.6-273$

→      $= 270^{\circ} C$

Thus the response above is right.

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