Nitrogen gas is at temperature $300 ° C$ . The temperature (in $K$ ) at which the $r m s$ speed of a $H_{2}$ molecule would be equal to the $r m s$ speed of a nitrogen molecule is

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Solution

Step 1. Given data:
The temperature of nitrogen gas $T_{N} = 300^{\circ} C$
Step 2. Formula used:
The $r . m . s$ speed for gas is $v_{r m s =} \sqrt{\frac{3 R T}{M}}$ where, $R = u n i v e r s a l g a s c o n s \tan t, M = m o l e c u l a r m a s s$
Step 3. Calculating the temperature at which the $r . m . s$ speed of hydrogen will be equal to the $r . m . s$ speed of nitrogen gas.
According to the given question,
Let at the temperature $T_{H}$ of hydrogen, the $r . m . s$ speed of hydrogen will be equal to the $r . m . s$ speed of nitrogen gas.
For hydrogen $v_{r m s} = \sqrt{\frac{3 R T_{H}}{M_{H}}} M_{H =}$ Molecular mass of hydrogen molecule
For nitrogen $v_{r m s} = \sqrt{\frac{3 R T_{N}}{M_{N}}} M_{N} =$ Molecular mass of nitrogen molecule
now,
$\begin{array}{rcl} \sqrt{\frac{3 R T_{H}}{M_{H}}} & = & \sqrt{\frac{3 R T_{N}}{M_{N}}} \\ \Rightarrow & \frac{T_{H}}{M_{H}} = \frac{T_{N}}{M_{N}} \\ \Rightarrow & T_{H} = (\frac{T_{N}}{M_{N}}) M_{H} \\ N o w, T_{N} & = & 300^{\circ} C + 273 K \\ = & 573 K \\ \Rightarrow & T_{H} = (\frac{573}{28}) \times 2 = 40 \cdot 92 K (a s, M_{N} = 28, M_{H} = 2) \end{array}$
Thus the temperature of hydrogen gas is $40 \cdot 92 K$ .

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