Question

What will be the pressure exerted by a mixture of $3.2g$ of methane and $4.4g$ of carbon dioxide contained in a $9d{{m}^{3}}$ flask at $27{}^\circ C$?

Answer 1

Recall a formula that incorporates number of moles, volume, pressure, and temperature. Calculate the total number of moles present in the flask to get the number of molecules that will exert pressure.

Complete step by step solution:
The formula that includes all the parameters that are given in the question is the ideal gas equation. The equation is as follows:
$PV=nRT$
Where, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of the substance, $R$ is the universal gas constant, and $T$ is the temperature given
We have all the information required except the number of moles of substance present in the container. The total number of moles will be obtained when the number of moles of each individual substance is added. The formula to calculate the number of moles is as follows:
$\text{no}\text{. of moles = }\dfrac{\text{given weight}}{\text{molecular weight}}$
We have the weight of both the substances given to us but we need to calculate the molecular weight. The molecular weight is given by the addition of the atomic weights of all the atoms present in the molecule. So, the molecular weights will be:
- Molecular weight of $C{{H}_{4}}$

& \text{Molecular weight of }C{{H}_{4}}=(1\times \text{atomic weight of }C)+(4\times \text{atomic weight of }H) \\\ & \text{Molecular weight of }C{{H}_{4}}=(1\times 12)+(4\times 1) \\\ & \text{Molecular weight of }C{{H}_{4}}=16 \\\ \end{aligned}$$ \- Molecular weight of $C{{O}_{2}}$ $$\begin{aligned} & \text{Molecular weight of }C{{O}_{2}}=(1\times \text{atomic weight of }C)+(2\times \text{atomic weight of }O) \\\ & \text{Molecular weight of }C{{O}_{2}}=(1\times 12)+(2\times 16) \\\ & \text{Molecular weight of }C{{O}_{2}}=44 \\\ \end{aligned}$$ Now, we will calculate the number of moles of each substance present. \- Moles of methane $$\begin{aligned} & \text{No}\text{. of moles of }C{{H}_{4}}\text{ = }\dfrac{3.2}{16} \\\ & \text{No}\text{. of moles of }C{{H}_{4}}\text{ = }0.2 \\\ \end{aligned}$$ \- Moles of carbon dioxide $$\begin{aligned} & \text{No}\text{. of moles of }C{{O}_{2}}=\dfrac{4.4}{44} \\\ & \text{No}\text{. of moles of }C{{O}_{2}}=0.1 \\\ \end{aligned}$$ \- Total number of moles present $$\begin{aligned} & \text{Total no}\text{. of moles = no}\text{.of moles of }C{{H}_{4}}+\text{no}\text{. of moles of }C{{O}_{2}} \\\ & \text{Total no}\text{. of moles = }0.2+0.1 \\\ & \text{Total no}\text{. of moles = }0.3 \\\ \end{aligned}$$ So, now we have the values: V = $9d{{m}^{3}}=9\times {{10}^{-3}}{{m}^{3}}$ R = $8.314J{{K}^{-1}}mo{{l}^{-1}}$ n = $0.3mol$ T = $27{}^\circ C=300K$ Now, we will put these values in the ideal gas equation and solve for pressure. $$\begin{aligned} & PV=nRT \\\ & P=\dfrac{nRT}{V} \\\ & P=\dfrac{0.3\times 8.314\times 300}{9\times {{10}^{-3}}} \\\ & P=83.14\times {{10}^{3}} \\\ & P=8.314\times {{10}^{4}}Pa \\\ \end{aligned}$$ **Hence, the pressure exerted by both the gases is $8.314\times {{10}^{4}}Pa$** **Additional information:** We can solve this problem without converting all the units to the SI units too. But make sure the value for R that you are taking is appropriate for the units of all the other values. **Note:** The pressure is directly proportional to the number of molecules that are present in the container, an estimate of how many molecules are present in the container will be given by the total number of moles present in the container.