Question

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

Answer 1

Hint: Molecular weight is determined by the atomic mass of the element in the periodic table. When the temperature increases, the molecules of gas that have more kinetic energy strikes the surface of the container with more force.
-Volume and temperature are more effective factors of the gas component.

Complete step by step answer:
The molecular weight of ${O_2}$ is 32 and ${N_2}$ is 28. Here both gases contain the same weight of 1 gm. ${O_2}$ gas strike in the bottle at the temperature of ${15^ \circ }C$ .
We will strike ${N_2}$ in the same bottle in which bottle ${O_2}$ gas strike.
Here we find out the temperature of 1 gm of ${N_2}$ to be filled in the same bottle by using an equation.
An equation given below,
 ${n_1}{T_1} = {n_2}{T_2}$
We put the given values into the above equation.
Temperature of ${O_2}$ = 273+25=298 K
Temperature of ${N_2}$ = (?)
As per $P \propto nT$ , pressure is the same.
So, ${n_1}{T_1} = {n_2}{T_2}$
Here, ${n_1} = \dfrac{1}{{32}}$
            ${n_2} = \dfrac{1}{{28}}$
   \[
  \dfrac{1}{{32}} \times 298 = \dfrac{1}{{28}} \times {T_2} \\
  {T_2} = \dfrac{{28}}{{32}} \times 298 = 260.75K \\
 \]
Temperature in $^ \circ C$ =260.75-273
                                     $ = - {13^ \circ }C$

Hence, the correct option is (B).

Note:
The property of gases changes by factors of temperature, pressure and volume. Robert boyle found the formula for ideal gases. The formula is PV=nRT. He derived PV=constant. Product of the pressure of gas and the volume of gas is constant. We find the temperature with help of molecular weight by using this formula.