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 of the flask will have greater total kinetic energy?
Assuming ideal behavior for both the gases.

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

Answer 1

The correct option is A $A$

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

Given,

$T_A=300K$

$T_B=600K$

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

where $N$ is Avogadro's constant.

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

Also, as we know that,
$TotalKE=KE\times Numberofmolecules$

$\therefore Total(KE{)}_A=\frac{3}{2}k{T}_A{N}_A$

$\therefore Total(KE{)}_B=\frac{3}{2}k{T}_B{N}_B$

Hence,

$\frac{Total(KE{)}_A}{Total(KE{)}_B}=\frac{T_A{N}_A}{T_B{N}_B}=\frac{300}{600}\times 8=4$

$\therefore Total(KE{)}_A>Total(KE{)}_B$

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