Question

The temperature at which the rms speed of oxygen molecules is 75\% of the rms speed of nitrogen molecules at a temperature of \( 287^\circ C \) is: 

• A. \( 87^\circ C \)
• B. \( 127^\circ C \)
• C. \( 227^\circ C \)
• D. \( 360^\circ C \)

Hint:

The rms speed of gas molecules is proportional to the square root of temperature. Use the relation \( v_{\text{rms}} \propto \sqrt{T} \) to solve temperature-related problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Root Mean Square (rms) Speed Formula 
The rms speed of gas molecules is given by: \[ v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}. \] For two different gases at different temperatures: \[ \frac{v_{\text{rms, O}_2}}{v_{\text{rms, N}_2}} = \sqrt{\frac{T_{\text{O}_2}}{T_{\text{N}_2}}}. \] 

Step 2: Substituting the Given Values 
Given: \[ v_{\text{rms, O}_2} = 0.75 v_{\text{rms, N}_2} \] and \[ T_{\text{N}_2} = 287 + 273 = 560K. \] Using the formula: \[ 0.75 = \sqrt{\frac{T_{\text{O}_2}}{560}}. \] 

Step 3: Solving for \( T_{\text{O}_2} \) 
Squaring both sides: \[ 0.5625 = \frac{T_{\text{O}_2}}{560}. \] \[ T_{\text{O}_2} = 560 \times 0.5625 = 315K. \] Converting to Celsius: \[ T_{\text{O}_2} = 315 - 273 = 87^\circ C. \] 

Step 4: Conclusion 
Thus, the required temperature is: \[ \boxed{87^\circ C}. \]