Question

The molecular weight of ${{\text{O}}_{\text{2}}}$and ${{\text{N}}_{\text{2}}}$ are 32 and 28 respectively. At ${\text{1}}{{\text{5}}^{\text{o}}}{\text{C}}$, the pressure of 1 gm ${{\text{O}}_{\text{2}}}$ will be the same as that of 1gm ${{\text{N}}_{\text{2}}}$ in the same bottle at the temperature?
A.${\text{ - 2}}{{\text{1}}^{\text{0}}}{\text{C}}$
B.${\text{ - 1}}{{\text{3}}^{\text{0}}}{\text{C}}$
C.${\text{1}}{{\text{5}}^{\text{0}}}{\text{C}}$
D.56.4 V

Answer 1

Hint:According to Gay Lussac’s law, Pressure exerted is directly proportional to temperature. According to Boyle’s law, at a constant temperature, the volume of a given gas is inversely proportional to pressure ${V\propto }\dfrac{{\text{1}}}{{\text{P}}}$. According to Charle’s law, the volume is directly proportional to temperature at constant pressure ${V\propto T}$. Number of moles, ${{\text{n}}_{\text{1}}}{\text{ = }}\dfrac{{{\text{given mass of }}{{\text{O}}_{\text{2}}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
Here, we have to find out the temperature at which the pressure of ${{\text{O}}_{\text{2}}}$ and ${{\text{N}}_{\text{2}}}$, each of them taken 1 gram is the same.
In the question, the temperature of ${{\text{O}}_{\text{2}}}$ when it is equal to the pressure of ${{\text{N}}_{\text{2}}}$ is given
${\text{Temperature of }}{{\text{O}}_2}{\text{, }}{{\text{T}}_1}{\text{ = 1}}{{\text{5}}^{\text{0}}}{\text{C}}$
Now, convert temperature to Kelvin, so ${\text{Temperature of }}{{\text{O}}_{\text{2}}}{\text{ = 1}}{{\text{5}}^{\text{0}}}{\text{ + 273 = 298K}}$
We have to find the temperature of ${{\text{N}}_{\text{2}}}$ and can be taken as y
We know that according to Gay Lussac’s law, Pressure exerted is directly proportional to temperature. We also know that pressure is the same, Pressure is directly proportional to the number of moles.
${\text{P}}\alpha {\text{nT}}$
So, we have $\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}}}{{{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}}}$
Since in this particular question pressure of both of it are at the same pressure $\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = 1}}$
Therefore, ${\text{1 = }}\dfrac{{{{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}}}{{{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}}}$
Thus, ${{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}{\text{ = }}{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}$
Here, ${{\text{n}}_{\text{1}}}{\text{ = }}\dfrac{{{\text{given mass of }}{{\text{O}}_{\text{2}}}}}{{{\text{molar mass}}}}$
${{\text{n}}_{\text{1}}}{\text{ = }}\dfrac{1}{{32}}$ and ${{\text{n}}_{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{28}}}}$
We know ${{\text{T}}_{\text{1}}}$ is 298K, we need to find ${{\text{T}}_{\text{2}}}$, the temperature of ${{\text{N}}_{\text{2}}}$.
${{\text{n}}_{\text{1}}}{{\text{T}}_{\text{1}}}{\text{ = }}{{\text{n}}_{\text{2}}}{{\text{T}}_{\text{2}}}$
$\dfrac{{\text{1}}}{{{\text{32}}}}{\times 298 = }\dfrac{{\text{1}}}{{{\text{28}}}}{\times y}$
Therefore, ${\text{y = }}\dfrac{{{\text{28}}}}{{{\text{32}}}}{\times 298}$
${\text{y = 260}}{\text{.75K}}$
Now since the options are given in degree Celsius, ${\text{y = (260}}{\text{.75 - 273}}{{\text{)}}^0}C$
${\text{y = - 1}}{{\text{3}}^{\text{0}}}{\text{C}}$
Therefore, the correct option is (B) .

Note: To convert temperature in degree Celsius to Kelvin, added a 273 to it. To convert temperature in Kelvin into degree Celsius, subtracted a 273 from it. From these gas laws, the Ideal gas equation can be derived, PV= nRT. If the question is about a constant temperature instead of constant pressure as here, we need to consider Boyle’s law.