Question

If pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ (real gas) in a container is given by
${\text{P}}\,{\text{ = }}\,\dfrac{{{\text{RT}}}}{{{\text{2V - b}}}} \dfrac{{\text{a}}}{{{\text{4}}{{\text{b}}^{\text{2}}}}}$ , then mass of the gas in container is:
A. $11$ g
B. $22$g
C. $33$g
D. $44$g

Answer 1

We will write the van der Waals equation of pressure for the real gas. Then arrange the equation to get the equation similar to the given equation so we can compare both of the equations. Comparing both of the equations determine the value of the number of moles. Then by using the mole formula determine the mass of carbon dioxide.

Complete step by step solution: The van der Waals equation for real gas is as follows:
${\text{P}}\,{\text{ + }}\dfrac{{{\text{a}}\,{{\text{n}}^2}}}{{{{\text{V}}^{\text{2}}}}}\left( {{\text{V - nb}}} \right) = \,{\text{nRT}}$
Where,
P is the pressure.
a and b are constant.
V is the volume.
R is the gas constant.
T is the temperature.
n is the number of moles.
Rearrange the van der Waals equation for pressure as follows:
${\text{P}}\, = \,\dfrac{{{\text{nRT}}}}{{{\text{V - nb}}}} - \dfrac{{{\text{a}}\,{{\text{n}}^2}}}{{{{\text{V}}^{\text{2}}}}}$
Divide the first part with n and the second part with ${{\text{n}}^2}$to make the van der Waals equation similar to the given equation.
${\text{P}}\, = \,\dfrac{{{\text{nRT/n}}}}{{{\text{V/n - nb/n}}}} - \dfrac{{{\text{a}}\,{{\text{n}}^2}/{{\text{n}}^2}}}{{{{\text{V}}^{\text{2}}}{\text{/}}{{\text{n}}^2}}}$
$\Rightarrow {\text{P}}\, = \,\dfrac{{{\text{RT}}}}{{{\text{V/n}} - {\text{b}}}} - \dfrac{{{\text{a}}\,}}{{{{\text{V}}^{\text{2}}}{\text{/}}{{\text{n}}^2}}}$…..$(1)$
The given equation is as follows:
${\text{P}}\,{\text{ = }}\,\dfrac{{{\text{RT}}}}{{{\text{2V - b}}}} - \dfrac{{\text{a}}}{{{\text{4}}{{\text{b}}^{\text{2}}}}}$…..$(2)$
By comparing the equation $(1)$and$(2)$,
${\text{V/n}} = \,2{\text{V}}$
$\Rightarrow {\text{n}}\, = \,\dfrac{{\text{V}}}{{2\,{\text{V}}}}$
$\Rightarrow {\text{n}}\, = \,\dfrac{1}{{2\,}}$
Use the mole formula to determine the amount of carbon dioxide as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
The molar mass of carbon dioxide is $44\,{\text{g/mol}}$.
Substitute $44\,{\text{g/mol}}$ for molar mass and $1/2$for moles.
$\dfrac{1}{2}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{44}}\,{\text{g}}\,/{\text{mol}}}}$
$\Rightarrow {\text{mass}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}\,{{ \times 44}}\,$
$\Rightarrow {\text{mass}}\,{\text{ = }}\,{\text{22}}\,{\text{g}}$
So, if pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ (real gas) in a container is given by ${\text{P}}\,{\text{ = }}\,\dfrac{{{\text{RT}}}}{{{\text{2V - b}}}} - \dfrac{{\text{a}}}{{{\text{4}}{{\text{b}}^{\text{2}}}}}$ , then mass of the gas in container is ${\text{22}}$gram.

Therefore, option (B) $22$g, is correct.

Note: The comparison of the second part also gives the same value for n. Van der Waals constant b represents the correction in volume. The unit of b is the same as the unit of volume. So, b and V can cancel out each other. So,
${{\text{V}}^{\text{2}}}{\text{/}}{{\text{n}}^2}\, = \,4\,{{\text{b}}^2}$
${{\text{n}}^2}\, = \,{{\text{V}}^{\text{2}}}{\text{/}}\,4\,{{\text{b}}^2}$
${{\text{n}}^2}\, = \,1/4$
${\text{n}}\,{\text{ = }}\,{\text{1/2}}$
Real gas disobeys the ideal gas law. In real gas attraction forces work so, the actual volume is less than the volume determined by the ideal gas law. So, a constant b is subtracted from volume. Pressures applied by the real gas are larger than the pressure applied by real gases, so some amount of pressure in terms of constant ‘a’ is added in ideal gas pressure. These corrections give the pressure-volume equation for real gases. The unit of constant ‘a’ is the unit of pressure.