Question

At what temperature the molecules of nitrogen will have the same $rms$ velocity as the molecules of oxygen at ${127}^{o}C$

• A. ${457}^{o}C$
• B. ${273}^{o}C$
• C. ${350}^{o}C$
• D. ${77}^{o}C$

Answer 1

The correct option is D ${77}^{o}C$

$rms$ velocity of gas is $v_{rms}=1.73{\left(\frac{RT}{M}\right)}^{\frac{1}{2}}$

$M=$ Molecular mass

For oxygen, $M=16\times 2$

$T={127}^{o}C=127+273=400K$

For nitrogen, $M=17\times 2,T=$?

$\Rightarrow 1.73{\left(\frac{RT}{M}\right)}_{O_2}^{\frac{1}{2}}=1.73{\left(\frac{RT}{M}\right)}_{N_2}^{\frac{1}{2}}$

$\Rightarrow {\left(\frac{T}{M}\right)}_{O_2}^{\frac{1}{2}}={\left(\frac{T}{M}\right)}_{N_2}^{\frac{1}{2}}$

$\Rightarrow \sqrt{T_{N_2}}=\sqrt{\frac{M_{N_2}}{M_{O_2}}{T}_{O_2}}$

$=\sqrt{\frac{28}{32}\times 400}$

$=\sqrt{\frac{7}{8}\times 400}$

$\Rightarrow T_{N_2}=\frac{7}{8}\times 400=350K$

$T_{N_2}=350K-273K={77}^{o}C$

Hence, the correct option is $\text{D}$