Question 35: Thermodynamics & KTG

Question

The rms velocity of the gas molecule at $327^{\circ} \mathrm{C}$ is same as the rms velocity of the oxygen molecules at $27^{\circ} \mathrm{C}$. If the molecular weight of oxygen is 32 then the molecular weight of the given gas molecule is:

Accepted Answer

First, convert temperatures to kelvin:
TÃ¢ÂÂ = 327 ÃÂ°C + 273 = 600 K
TÃ¢ÂÂ = 27 ÃÂ°C + 273 = 300 K
For any gas,
$$v_{\rm rms} = \sqrt{\frac{3RT}{M}}\,, $$
where M is the molar mass.
Since the rms speeds are equal,
$$\sqrt{\frac{3R\,T_1}{M_{\rm gas}}} \;=\;\sqrt{\frac{3R\,T_2}{M_{\rm O_2}}}$$
Ã¢ÂÂ
$$\frac{T_1}{M_{\rm gas}} \;=\;\frac{T_2}{M_{\rm O_2}}$$
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$$M_{\rm gas} = M_{\rm O_2}\,\frac{T_1}{T_2} = 32\;\frac{600}{300} = 64.$$
Answer: 64.

Answer

32

Answer

64

Answer

96

Answer

128

Thermodynamics & KTG

The rms velocity of the gas molecule at 327^{\circ} \mathrm{C} is same as the rms velocity of the oxygen molecules at 27^{\circ} \mathrm{C}. If the molecular weight of oxygen is 32 then the molecular weight of the given gas molecule is:

Choose the correct answer: