Question

Two gas cylinders having same capacity have been filled with ${\text{44}}$ g of ${{\text{H}}_{\text{2}}}$and ${\text{44}}$ g of ${\text{C}}{{\text{O}}_{\text{2}}}$respectively. If the pressure in ${\text{C}}{{\text{O}}_{\text{2}}}$cylinder is $1$ atmosphere at particular temperature, the pressure in the hydrogen cylinder at the same temperature is:
A. $2$ atmosphere
B. $1$ atmosphere
C. $22$ atmosphere
D. $44$ atmosphere

Answer 1

Hint:To answer this question we should know the ideal gas law. According to which product of pressure and volume is equal to the product of the number of moles, gas constant, and temperature. We will write the ideal gas equation for both of the given gases. Then by comparing both equations we will determine the pressure of hydrogen gas.


Complete solution:
The formula of the ideal gas is as follows:
${\text{pV = nRT}}$
${\text{p}}$is the pressure
V is the volume
${\text{n}}$ is the number of moles of ideal gas
R is the gas constant
T is the temperature
We can write the above equation for hydrogen and carbon dioxide gas as,
${{\text{p}}_{{\text{He}}}}{{\text{V}}_{{\text{He}}}}{\text{ = }}{{\text{n}}_{{\text{He}}}}{\text{R}}{{\text{T}}_{{\text{He}}}}$…..$(1)$
${{\text{p}}_{{\text{C}}{{\text{O}}_2}}}{{\text{V}}_{{\text{C}}{{\text{O}}_2}}}{\text{ = }}{{\text{n}}_{{\text{C}}{{\text{O}}_2}}}{\text{R}}{{\text{T}}_{{\text{C}}{{\text{O}}_2}}}$…..$(2)$
It is given that both gases are at the same temperature. The volume of the cylinder is not given and both gases have the same heat capacity, so the volume of both the gases is also the same.
So, on substituting equation $(1)$equal to equation $(2)$,
\[\dfrac{{{{\text{p}}_{{\text{He}}}}{{\text{V}}_{{\text{He}}}}}}{{{{\text{n}}_{{\text{He}}}}{\text{R}}{{\text{T}}_{{\text{He}}}}}}{\text{ = }}\dfrac{{{{\text{p}}_{{\text{C}}{{\text{O}}_2}}}{{\text{V}}_{{\text{C}}{{\text{O}}_2}}}}}{{{{\text{n}}_{{\text{C}}{{\text{O}}_2}}}{\text{R}}{{\text{T}}_{{\text{C}}{{\text{O}}_2}}}}}\]
We can cancel the temperature, volume and gas constant from both side of the above equation so,
\[\dfrac{{{{\text{p}}_{{\text{He}}}}}}{{{{\text{n}}_{{\text{He}}}}}}{\text{ = }}\dfrac{{{{\text{p}}_{{\text{C}}{{\text{O}}_2}}}}}{{{{\text{n}}_{{\text{C}}{{\text{O}}_2}}}}}\]…..$(3)$
Now, we will determine the mole of each gas as follows:
${\text{mole = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
For hydrogen gas,
The molar mass of hydrogen gas is $2$gm/mol.
On substituting ${\text{44}}$ g for mass and $2$gm/mol for molar mass.
${\text{mole = }}\,\dfrac{{44}}{2}$
${\text{mole = }}\,22$
So, the moles of hydrogen gas is $22$.
For carbon dioxide gas,
The molar mass of carbon dioxide gas is $44$gm/mol.
On substituting ${\text{44}}$ g for mass and $44$gm/mol for molar mass.
${\text{mole = }}\,\dfrac{{44}}{2}$
${\text{mole = }}\,1$
So, the moles of carbon dioxide gas is$1$.
On substituting $22$ g for mole of ${{\text{H}}_{\text{2}}}$, $1$ for mole of ${\text{C}}{{\text{O}}_{\text{2}}}$ and $1$ atmosphere for the pressure of carbon dioxide in equation $(3)$,
\[\dfrac{{{{\text{p}}_{{\text{He}}}}}}{{{{\text{n}}_{{\text{He}}}}}}{\text{ = }}\dfrac{{{{\text{p}}_{{\text{C}}{{\text{O}}_2}}}}}{{{{\text{n}}_{{\text{C}}{{\text{O}}_2}}}}}\]
\[\dfrac{{{{\text{p}}_{{\text{He}}}}}}{{{\text{22}}}}{\text{ = }}\dfrac{1}{1}\]
\[{{\text{p}}_{{\text{He}}}}{\text{ = }}\dfrac{1}{1} \times 22\]
\[{{\text{p}}_{{\text{He}}}}{\text{ = }}22\] atmosphere
So, the pressure in the hydrogen cylinder at the same temperature is \[22\] atmosphere.

Therefore, option (C) \[22\] atmosphere correct.


Note:Here, we have to compare two different gases in the same condition, so we used the ideal gas law. If we have to compare the same gas at different conditions then we use the combined gas law. According to this law, if an ideal gas is present at a condition of temperature, pressure, and volume, and any one or two parameters of temperature, pressure, or volume get changed then we can determine the third parameter by putting the initial conditions of temperature pressure and volume equal to the final condition of temperature, pressure, and volume. The relation between temperature pressure and volume of an ideal gas according to combined gas law is as follows:
$\dfrac{{{{\text{p}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}\, = \,\dfrac{{{{\text{p}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$