Question

The molecular weights of $O_2$ and $N_2$ are $32$ and $28$ respectively. At ${15}^0$C, the pressure of $1$gm $O_2$ will be the same as that of $1$ gm $N_2$ in the same bottle at the temperature:

• A. $-{21}^0$C
• B. $-{13}^0$C
• C. ${15}^0$C
• D. ${56.4}^0$V

Answer 1

Answer: (B)

$-{13}^0$C

Temperature of $O_2$ $=273+15=298K$

Temperature of $N_2$ $=x$

As $n$ $\propto$ $\frac{1}{T}$, as pressure is constant.

$⟹n_1{T}_1=n_2{T}_2⟹\left(\frac{1}{32}\right)\left(298\right)=\left(\frac{1}{28}\right)\left(x\right)⟹x=\frac{28}{32}\times 298=260.75K$

Temperature in ${}^{o}C=260.75-273=-{13}^{o}C$

Hence, option $B$ is correct.