Question

What will be the relation between the $T_1$ of gas $1$ with $M_1=56$ and $T_2$ of gas $2$ with $M_2=44$ if the average speed of gas $1$ is equal to most probable speed of gas $2$ ?

• A. $T_1=T_2^2$
• B. $T_1=T_2$
• C. $T_1={\left(T_2\right)}^{1/2}$
• D. $T_1=1/T_2$

Answer 1

Answer: (B)

$c_{av}=\sqrt{\frac{8RT}{\pi M}};$
$c_{mp}=\sqrt{\frac{2RT}{M}}$
Given that, $\sqrt{\frac{8R{T}_1}{\pi M_1}}=\sqrt{\frac{2R{T}_2}{M_2}}$
$\Rightarrow \frac{8T_1}{\pi M_1}=\frac{2T_2}{M_2}$
$\frac{4T_1}{\pi \times 56}=\frac{T_2}{44}$
$\Rightarrow \frac{T_1}{T_2}=\frac{56\times \pi }{44\times 4}=1$
$\Rightarrow T_1=T_2$