Question

Find the total pressure exerted by \[1.6\] g of methane and \[2.2\] gram of \[{\text{C}}{{\text{O}}_{\text{2}}}\] contained in a 4 liter flask at ${\text{2}}{{\text{7}}^{\text{0}}}{\text{C}}$.

Answer 1

Hint: In this question, we will use ideal gas equation ${{\text{P}}_{\text{T}}}{\text{V = }}{{\text{n}}_{\text{T}}}{\text{RT}}$, total pressure, volume and temperature are given to us in question, R is universal gas constant $\left( {{\text{8}}{\text{.314 atm L}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)$. To find the total number of moles, add the number of moles of methane and carbon dioxide.

Formula used: ${{\text{P}}_{\text{T}}}{\text{V = }}{{\text{n}}_{\text{T}}}{\text{RT}}$ where,
${{\text{P}}_{\text{T}}}$- Total pressure exerted
${{\text{n}}_{\text{T}}}$ - Total number of moles
V - Volume of system
R - Universal gas constant
T – Temperature (in Kelvin)

Complete step by step answer:
We have been given a mixture of two gases methane $\left( {{\text{C}}{{\text{H}}_{\text{4}}}} \right)$ and carbon dioxide $\left( {{\text{C}}{{\text{O}}_{\text{2}}}} \right)$.
It is given that:
Mass of ${\text{C}}{{\text{H}}_{\text{4}}}$\[ = \] $1.6$ g
Mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ \[ = \] $2.2$ g
Volume of container = 4 L
To convert degree Celsius to Kelvin, we can use the conversion:${\text{T(in kelvin) = }}{{\text{t}}^{\text{0}}}{\text{C + 273}}$.
Temperature = ${27^ \circ }{\text{C = 300K}}$
As we know that,
Molecular mass of ${\text{C}}{{\text{H}}_{\text{4}}}$${{ = 12 + 1 \times 4 = 16 g mo}}{{\text{l}}^{ - 1}}$
Molecular mass of ${\text{C}}{{\text{O}}_{\text{2}}}$${{ = 12 + 16 \times 2 = 44 g mo}}{{\text{l}}^{{\text{ - 1}}}}$
No. of moles of ${\text{C}}{{\text{H}}_{\text{4}}}$$ = \dfrac{{{\text{mass of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Molecular mass of C}}{{\text{H}}_{\text{4}}}}}$
Mass of methane is given in the question and molecular mass has been calculated by us.
No. of moles of ${\text{C}}{{\text{H}}_{\text{4}}}$$ = \dfrac{{{\text{1}}{\text{.6}}}}{{{\text{16}}}}$
Dividing we will get:
No. of moles of ${\text{C}}{{\text{H}}_{\text{4}}}$ = $0.1$ moles
No. of moles of ${\text{C}}{{\text{O}}_{\text{2}}}$$ = \dfrac{{{\text{mass of C}}{{\text{O}}_{\text{2}}}}}{{{\text{Molecular mass of C}}{{\text{O}}_2}}}$
Similarly,
No. of moles of ${\text{C}}{{\text{O}}_{\text{2}}}$$ = \dfrac{{2.2}}{{{\text{44}}}}$
Dividing the above equations,
 No. of moles of ${\text{C}}{{\text{O}}_{\text{2}}}$${\text{ = 0}}{\text{.05}}$ moles
${{\text{n}}_{\text{T}}}$ is the total no. of moles which can be obtained by adding the no. of moles of individual components present in the mixture.
Hence,
$
  {{\text{n}}_{\text{T}}}{\text{ = }}{{\text{n}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}{\text{ + }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} \\
  {{ = \;0}}{\text{.1 + 0}}{\text{.05}} \\
  {\text{ = 0}}{\text{.15 moles}} \\
$
Use ideal gas equation ${{\text{P}}_{\text{T}}}{\text{V = }}{{\text{n}}_{\text{T}}}{\text{RT}}$ and rearrange the equation by moving V to the denominator on right hand side.
Now, substituting the values, that are given and the one we calculated, in the above equation we will get,
 ${{\text{P}}_{\text{T}}} = \dfrac{{{\text{0}}{{.15 \times 0}}{{.0821 \times 300}}}}{{\text{4}}}$
Solving this will give, ${{\text{P}}_{\text{T}}} = \dfrac{{3.6945}}{{\text{4}}}$.
After dividing, the pressure will come out to be
${{\text{P}}_{\text{T}}}{\text{ = 0}}{\text{.9236 atm}}$
Hence, the total pressure exerted by the system is ${\text{0}}{\text{.9236 atm}}$.

Note:
We always use temperature only in Kelvin and not in degree Celsius. So, we need to make sure we convert the unit of the temperature. The above question is based on Dalton’s law of partial pressure that is total pressure of a mixture of gases is equal to the sum of partial pressure of each component in mixture ${{\text{P}}_{\text{T}}}{\text{ = }}{{\text{P}}_{\text{1}}}{\text{ + }}{{\text{P}}_{\text{2}}}{\text{ + }}...................{\text{ + }}{{\text{P}}_{\text{n}}}$. That is why in the ideal gas equation, we used a total number of moles. For a binary mixture, mixture involving 2 components (as in our question) we can calculate total no. of moles = ${{\text{n}}_{\text{T}}}{\text{ = }}{{\text{n}}_{\text{A}}}{\text{ + }}{{\text{n}}_{\text{B}}}$. We can use unit of R in which unit we want to obtain the value of pressure:
${\text{R = 8}}{\text{.314 J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{R = 8}}{\text{.314 Pa L}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
In terms of bar and atmosphere we can write it as,
 ${\text{R = 0}}{\text{.0821 atm L}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{R = 0}}{\text{.083 bar L}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$