Question

$A$ and $B$ are two ideal gases with following parameters:
$\begin{array}{cccc}\text{Gas}& \text{Molecules}& \text{Temperature}& \text{Kinetic energy}\\ A& N_A& -{123}^{\circ }\text{C}& E_A\\ B& N_B& {27}^{\circ }\text{C}& E_B\end{array}$
If $\frac{E_B}{E_A}$ = 2, then $\frac{N_B}{N_A}$ is equal to

• A. 4
• B. 2
• C. 3
• D. 1

Answer 1

The correct option is D 1

(KE) of $n$ moles $=\frac{3}{2}nRT$
$E_A=(KE{)}_A=\frac{3}{2}\left(\frac{N_A}{N_o}\right)R(150)---(1)$
where $N_o=$ Avogadro number, and
$E_B=(KE{)}_B=\frac{3}{2}\left(\frac{N_B}{N_o}\right)R(300)---(2)$
From (1) and (2)
$\frac{E_B}{E_A}=\frac{2N_B}{N_A}$
$\frac{E_B}{E_A}=2\left(Given\right)$
$\frac{E_B}{E_A}=\frac{2N_B}{N_A}=2$
$\therefore \frac{N_B}{N_A}=1$