Question

The molecular weights of ${O_2}$ and ${N_2}$are $32$ and $28$ respectively. At $15^\circ C$, the pressure of $1gm$ ${O_2}$ will be the same as that of $1gm$ ${N_2}$in the same bottle at the temperature:
A. $- 21^\circ C$
B. $13^\circ C$
C. $15^\circ C$
D. $56.4^\circ C$

Answer 1

We know that the ideal gas equation is given by $\Rightarrow PV = nRT$, here $P =$ pressure of the gas , $V =$volume of the gas , $n =$ number of moles $R =$ideal gas constant , $T =$ temperature of the gas in Kelvin. The experiment is conducted in the same bottle so the volume will be the same in both the cases.

Complete step by step answer:
As in the question we are given that the molecular weights of ${O_2}$ and ${N_2}$ are $32$ and $28$ respectively. And $15^\circ C$, the pressure of $1gm$ ${O_2}$ will be the same as that of $1gm$ ${N_2}$ in the same bottle. Let the pressure, volume, number of moles and temperature of ${O_2}$ be ${P_1},{V_1},{n_1},{T_1}$ respectively. Similarly let the pressure, volume, number of moles and temperature of${N_2}$be ${P_2},{V_2},{n_2},{T_2}$. From the question we know that ${V_1} = {V_2}$.The pressures are also equal so ${P_1} = {P_2}$. So we can write ${P_1}{V_1} = {P_2}{V_2}$.The value ${T_1} = (273 + 15)K = 288K$.
The ideal gas equation of ${O_2}$ can be written as -
$\Rightarrow {P_1}{V_1} = {n_1}R{T_1} \\\ \Rightarrow {P_1}{V_1} = \dfrac{1}{{32}}R(288) \\\$
The ideal gas equation of ${N_2}$can be written as
$\Rightarrow {P_2}{V_2} = {n_2}R{T_2} \\\ \Rightarrow {P_2}{V_2} = \dfrac{1}{{28}}R({T_2}) \\\$
We know that ${P_1}{V_1} = {P_2}{V_2}$, so
${P_1}{V_1} = {P_2}{V_2} \\\ \Rightarrow \dfrac{{288R}}{{32}} = \dfrac{{{T_2}R}}{{28}} \\\ \Rightarrow {T_2} = \dfrac{{28 \times 288}}{{32}} \\\ \therefore {T_2} = 252K = - 21^\circ C \\\$

From the above explanation and calculation it is clear to us that the correct answer of the given question is option : A. $- 21^\circ C$.

Note:
${O_2}$ and ${N_2}$are the example of diatomic gases because these molecules are made up of two atoms . ${O_2}$ and ${N_2}$ constitute the major portion of the atmosphere. Nitrogen constitutes $78.1\%$ of the atmosphere and oxygen constitutes about $20.946\%$ of the atmosphere.