Question

At what temperature, the mean kinetic energy of $O_2$ will be the same as that of $H_2$ molecules at $-{73}^{\circ }C$?

• A. $-{173}^{o}C$
• B. $-{73}^{0}C$
• C. ${2927}^{0}C$
• D. ${527}^{0}C$

Answer 1

The correct option is B $-{73}^{0}C$

Mean kinetic energy $K.E=\frac{f}{2}nRT$

where $n=\frac{Mass}{Molarmass}$

So, $K.E_{O_2}=K.E_{H_2}$

Or $\frac{T_{O_2}}{M_{O_2}}=\frac{T_{H_2}}{M_{H_2}}$

Or $\frac{T_{O_2}}{32}=\frac{200}{2}$

$⟹T_{O_2}=3200K$

$⟹T_{O_2}=3200-273={2927}^{o}C$