Animated Solution for Chemistry - States of Matter: The pressure exerted by a non-reactive gaseous mixture of $6.4\text{ g}$ of methane and $8.8\text{ g}$ of carbon dioxide in a $10\text{ L}$ vessel at $27^{\circ }\text{C}$ is ...... $\text{kPa}$ (Round off to the nearest integer) (Assume gases are ideal, $R=8.314{\text{ J mol}}^{-1}{\text{ K}}^{-1}$ Atomic mass, $\text{C}:12.0\text{ u}$, $\text{H}:1.0\text{ u}$, $\text{O}:16.0\text{ u}$)

Imagine you are looking at a sealed container with a volume of $10\text{ L}$. Inside this vessel, we have a mixture of two gases: $6.4\text{ g}$ of methane ($C{H}_4$) and $8.8\text{ g}$ of carbon dioxide ($C{O}_2$). The entire system is maintained at a pleasant ambient temperature of $27^{\circ }\text{C}$. Our mission is to find the total pressure exerted by this gaseous mixture on the walls of the container.

Before we dive into the math, we must acknowledge a crucial piece of information given in the problem: the gases are non-reactive. This means they will happily coexist without forming any new chemical compounds. Because they don't react and we are assuming ideal behavior, we can treat the mixture as a single ideal gas where the identity of the particles doesn't matter—only their total number matters.

According to Dalton's Law of Partial Pressures, the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Mathematically, this implies that the total pressure is directly proportional to the total number of moles in the container.

To use this master equation, our first objective is to find $n_{\text{total}}$, which is simply the sum of the moles of methane and the moles of carbon dioxide.

Let's break down the mixture component by component. First, we tackle methane ($C{H}_4$). The molar mass of methane is the sum of the atomic mass of one carbon atom and four hydrogen atoms:

Now, we find the number of moles by dividing the given mass by the molar mass:

Next, we do the same for carbon dioxide ($C{O}_2$). Its molar mass is:

Calculating its moles:

Adding them together gives us the total number of gas particles (in moles) bouncing around in our vessel:

Now we are ready to substitute our values into the ideal gas equation. But wait, there is a brilliant shortcut we can use here to avoid messy unit conversions!

The universal gas constant $R$ is given as $8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

Did you know that $1\text{ Joule}$ is exactly equal to $1\text{ kPa}\cdot \text{L}$?

Let's prove it quickly: $1\text{ J}=1\text{ N m}=1{\text{ Pa m}}^3$. Since $1{\text{ m}}^3=1000\text{ L}$ and $1\text{ Pa}=10^{-3}\text{ kPa}$, multiplying them gives exactly $1\text{ kPa L}$.

This means we can rewrite the gas constant as:

By using this specific form of $R$, we can plug in our volume directly in Liters, and our resulting pressure will automatically pop out in kilo-Pascals (kPa)! Also, don't forget to convert the temperature to Kelvin: $T=27+273=300\text{ K}$.

Let's substitute everything into our rearranged ideal gas equation:

Let's simplify the numbers. The $0.6$ multiplied by $300$ gives $180$. Dividing that by the volume $10$ leaves us with $18$. Now we just multiply:

The question asks us to round off to the nearest integer. Looking at the decimal $.652$, we round up:

And there we have it! By understanding the physical setup, leveraging Dalton's Law, and using a clever unit shortcut, we arrived at the perfect answer.