The molecular weights of $O_{2}$ and $N_{2}$ are 32 and 28 respectively. At $15℃,$ the pressure of $1\ g\ O_{2}$ will be the same as that of $1\ g\ N_{2}$ in the same bottle at the temperature

Question

Dr. Abubacker · Accepted Answer

(a) For $1\ g$ gas $PV = rT = \left( \frac{R}{M} \right)T$
Since $P$ and $V$ are constant $\Rightarrow T \propto M \Rightarrow \frac{T_{N_{2}}}{T_{O_{2}}} = \frac{M_{N_{2}}}{M_{O_{2}}}$
$\Rightarrow \frac{T_{N_{2}}}{(273 + 15)} = \frac{28}{32} \Rightarrow T_{N_{2}} = 252K = - 21℃$

Q. The molecular weights of $O_{2}$ and $N_{2}$ are 32 and 28 respectively. At $15℃,$ the pressure of $1\ g\ O_{2}$ will be the same as that of $1\ g\ N_{2}$ in the same bottle at the temperature

Explanation

Reviewed by Dr. Abubacker

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