Question

The ratio of rms speeds of helium gas molecules at a temperature of \(127^\circ\text{C}\) and oxygen gas molecules at a temperature of \(527^\circ\text{C}\) is \[ (\text{Molar masses of helium and oxygen gases are }4\text{ and }32\text{ respectively}) \]

• A. \(\sqrt2:1\)
• B. \(4:1\)
• C. \(2:1\)
• D. \(\sqrt8:1\)

Hint:

Remember, \[ \boxed{ v_{\rm rms}=\sqrt{\frac{3RT}{M}} } \] or \[ \boxed{ v_{\rm rms}\propto\sqrt{\frac{T}{M}}. } \] Always convert the temperature into Kelvin before substitution.

Answer 1

The Correct Option is C

Step 1: Recall the formula for rms speed. The rms speed of an ideal gas is \[ v_{\rm rms}=\sqrt{\frac{3RT}{M}}, \] where \[ T=\text{absolute temperature} \] and \[ M=\text{molar mass}. \] Hence, \[ v_{\rm rms}\propto\sqrt{\frac{T}{M}}. \]

Step 2: Convert the temperatures into Kelvin. For helium, \[ T_1=127+273=400\,\text{K}. \] For oxygen, \[ T_2=527+273=800\,\text{K}. \] Also, \[ M_{\rm He}=4, \qquad M_{\rm O_2}=32. \]

Step 3: Calculate the ratio. \[ \frac{v_{\rm He}}{v_{\rm O_2}} = \sqrt{\frac{400/4}{800/32}} = \sqrt{\frac{100}{25}} = \sqrt4 = 2. \] Therefore, \[ \boxed{v_{\rm He}:v_{\rm O_2}=2:1.} \] Hence, the correct option is \(\boxed{(C)}\).