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)

$T_1 = T_2$

Answer 2

$c_{av}=\sqrt{\frac{8RT}{\pi M}} ;$
$c_{mp}=\sqrt{\frac{2RT}{M}}$
Given that, $\sqrt{\frac{8RT_{1}}{\pi M_{1}}}=\sqrt{\frac{2RT_{2}}{M_{2}}}$
$\Rightarrow \frac{8T_{1}}{\pi M_{1}}=\frac{2T_{2}}{M_{2}}$
$\frac{4T_{1}}{\pi\times56}=\frac{T_{2}}{44}$
$\Rightarrow \frac{T_{1}}{T_{2}}=\frac{56\times\pi}{44\times4}=1$
$\Rightarrow T_{1}=T_{2}$