91Ó°ÊÓ

• Fromthegraph, the most probable speed${\text{V}}_P=400\text{″¾}/\text{s}$
• Speed${\text{V}}_s=1200\text{″¾}/\text{s}$

The expression for the most probable speed is given by,

$V_P=\sqrt{\frac{2RT}{M}}$

Here $V_P$is the most probable speed,$R$is the universal gas constant,$T$is the temperature and $M$is the mass.

The expression for the RMS speed is given by,

$V_{rms}=\sqrt{\frac{3RT}{M}}$

Here$V_{rms}$ is the RMS speed.

Gas temperature of nitrogen:

From the graph, we see that,${\text{V}}_P=400\text{m}/\text{s}$.

We know that the mass of nitrogen molecule is
$\text{M}=28\frac{\text{g}}{\text{mol}}=0.028\text{Kg}/\text{mol}$

The formula for most probable speed is

$V_P=\sqrt{\frac{2RT}{M}}$

Squaring both sides, we get

$V_p^2=\frac{2RT}{M}$

$T=M\frac{V_p^2}{2R}$……. (i)

Here$R$is the gas constant, and its value is

$\text{R}=8.31\text{ J}/\text{mol}.\text{K}$

Substituting all the values in equation (i), we get

$\begin{array}{c}\text{T}=\frac{0.028\frac{\text{kg}}{\text{mol}}×\left(\frac{400\text{″¾}}{\text{s}}\right){}^2}{2×8.31\frac{\text{J}}{\text{mol}.\text{K}}}\\ =2.7×{10}^2\text{K}\end{array}$

Therefore the temperature of the gas is $2.7×{10}^2\text{K}$.

Root mean square speed of the molecules:

Using the equations of most probable speed and rms speed,

$V_P=\sqrt{\frac{2RT}{M}}$and$V_{rms}=\sqrt{\frac{3RT}{M}}$

Comparing these two equations, we get

$V_{rms}=\sqrt{\frac{3}{2}}{V}_p$……. (ii)

From the graph, we have$V_P=400\text{″¾}/\text{s}$,

Substitute $400\text{″¾}/\text{s}$for $V_P$into the equation (ii)

$\begin{array}{c}{\text{V}}_{rms}=\sqrt{\frac{3}{2}}(400\text{″¾}/\text{s})\\ =4.9×{10}^2\text{m}/\text{s}\end{array}$

Therefore the RMS speed of the molecules is $4.9×{10}^2\text{m}/\text{s}$.