Question

Flask $A$ contains $H_2$ gas and $B$ contains $C{H}_4$ gas have equal mass and volume at 300 $K$ and 600 $K$ respectively. Which flask will have a greater number of molecules?

(Assuming ideal behaviour for both the gases.)

• A. $A$
• B. $B$
• C. Both $A$ and $B$ are equal
• D. None

Answer 1

The correct option is A $A$

Let $n_A$ be the number of molecules of $H_2$ and $n_B$ be the number of molecules of $C{H}_4$.

We know that,

$n_A=\frac{m}{2};n_B=\frac{m}{16}$

where $m$ is the mass of the gas of two flasks $A$ and $B$.

$\therefore \frac{n_A}{n_B}=\frac{m}{2}/\frac{m}{16}=8$

Now, the number of molecules in these flasks are given as :

$N_A=\frac{m}{2}\times N,N_B=\frac{m}{16}\times N$

where $N$ is Avogadro's constant.

$\therefore \frac{N_A}{N_B}=8$

$\Rightarrow N_A>N_B$

Hence, option $\left(A\right)$ is correct.