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, $\sqrt{\frac{8RT_{1}}{\pi M_{1}}}=\sqrt{\frac{2RT_{2}}{M_{2}}}$
$\Rightarrow \frac{8T_{1}}{\pi M_{1}}=\frac{2T_{2}}{M_{2}}$
$\frac{4T_{1}}{\pi\times56}=\frac{T_{2}}{44}$
$\Rightarrow \frac{T_{1}}{T_{2}}=\frac{56\times\pi}{44\times4}=1$
$\Rightarrow T_{1}=T_{2}$