The root mean square speed of hydrogen is `sqrt(5)` times than that of nitrogen. If T is the temperature of the gas, then :

Options

A. `T_(H_(2))=T_(N_(2)`
B. `T_(H_(2))gtT_(N_(2)`
C. `T_(H_(2))ltT_(N_(2)`
D. `T_(H_(2))=sqrt(7)T_(N_(2)`

Explanation

To solve the problem, we need to analyze the relationship between the root mean square (RMS) speed of hydrogen (H₂) and nitrogen (N₂) gases, given that the RMS speed of hydrogen is \(\sqrt{5}\) times that of nitrogen.

### Step-by-Step Solution:

1. **Understand the Formula for RMS Speed**:
The root mean square speed (\(V_{rms}\)) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature of the gas,
- \(M\) is the molar mass of the gas.

2. **Set Up the Equation for Both Gases**:
Let \(V_{rms, H_2}\) be the RMS speed of hydrogen and \(V_{rms, N_2}\) be the RMS speed of nitrogen. According to the problem:
\[ V_{rms, H_2} = \sqrt{5} \cdot V_{rms, N_2} \]

3. **Express RMS Speeds Using the Formula**:
Using the formula for RMS speed, we can express the speeds for both gases:
\[ V_{rms, H_2} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} \]
\[ V_{rms, N_2} = \sqrt{\frac{3RT_{N_2}}{M_{N_2}}} \]

4. **Substituting into the Equation**:
Substituting these expressions into the equation from step 2 gives:
\[ \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{5} \cdot \sqrt{\frac{3RT_{N_2}}{M_{N_2}}} \]

5. **Square Both Sides**:
Squaring both sides to eliminate the square roots results in:
\[ \frac{3RT_{H_2}}{M_{H_2}} = 5 \cdot \frac{3RT_{N_2}}{M_{N_2}} \]

6. **Cancel Out Common Terms**:
Since \(R\) and \(3\) are common on both sides, we can cancel them out:
\[ \frac{T_{H_2}}{M_{H_2}} = 5 \cdot \frac{T_{N_2}}{M_{N_2}} \]

7. **Substituting Molar Masses**:
The molar mass of hydrogen (\(M_{H_2}\)) is 2 g/mol and that of nitrogen (\(M_{N_2}\)) is 28 g/mol:
\[ \frac{T_{H_2}}{2} = 5 \cdot \frac{T_{N_2}}{28} \]

8. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 28T_{H_2} = 10T_{N_2} \]

9. **Rearranging the Equation**:
Rearranging this equation gives:
\[ \frac{T_{H_2}}{T_{N_2}} = \frac{10}{28} = \frac{5}{14} \]

10. **Conclusion**:
This means that the temperature of hydrogen is less than that of nitrogen, specifically:
\[ T_{N_2} > T_{H_2} \]