Animated Solution for Chemistry - States of Matter: A home owner uses $4.00\times 10^3{\text{ m}}^3$ of methane (${\text{CH}}_4$) gas, (assume ${\text{CH}}_4$ is an ideal gas) in a Year to heat his home. Under the pressure of $1.0\text{ atm}$ and $300\text{ K}$, mass of gas used is $x\times 10^5\text{ g}$. The value of $x$ is ...... . (Nearest integer) (Given, $R=0.083{\text{ L atm K}}^{-1}{\text{mol}}^{-1}$)

Imagine a massive house that needs heating for an entire year. The homeowner uses a huge volume of methane gas to keep the cold at bay. Let's visualize this gas and note down the physical conditions given to us. We are told that the volume of the gas is $V=4.00\times 10^3{\text{ m}}^3$. The pressure is a standard $p=1.0\text{ atm}$, and the temperature is a comfortable $T=300\text{ K}$.

We need to find the total mass of this gas. Since methane is behaving as an ideal gas here, we can rely on our trusty master equation that connects pressure, volume, temperature, and the amount of gas. Yes, the Ideal Gas Equation!

The Ideal Gas Equation is given by $pV=nRT$. We know that the number of moles $n$ is equal to the given mass $w$ divided by the molar mass $M$. Substituting this into our equation gives us $pV=\frac{w}{M}RT$. Rearranging for the mass $w$, we get $w=\frac{pVM}{RT}$.

Before we plug in the numbers, we must be very careful with units. This is where many students make a silly mistake! The gas constant $R$ is given as $0.083{\text{ L atm K}}^{-1}{\text{mol}}^{-1}$. Because $R$ uses liters, our volume must also be in liters. We know that $1{\text{ m}}^3$ is equal to $1000\text{ L}$. Therefore, our volume $V=4.00\times 10^3{\text{ m}}^3$ becomes $4\times 10^6\text{ L}$.

Now, let's put all the pieces together. The pressure is $1.0\text{ atm}$, the volume is $4\times 10^6\text{ L}$, and the molar mass of methane (${\text{CH}}_4$) is $16\text{ g/mol}$. We divide this by $R$, which is $0.083$, and the temperature, $300\text{ K}$.

Let's simplify the denominator first. Multiplying $0.083$ by $300$ gives us $24.9$. In the numerator, $4$ times $16$ is $64$. So we have $64\times 10^6$ divided by $24.9$.

Dividing $64$ by $24.9$ gives approximately $2.57$. So the mass is $2.57\times 10^6\text{ g}$. The question asks for the answer in the format of $x\times 10^5\text{ g}$. By shifting the decimal point, we get $25.7\times 10^5\text{ g}$. Rounding to the nearest integer, we find that $x=26$!

Always remember to check the units of $R$ before substituting your values. If the gas wasn't ideal, we would have to use the Van der Waals equation, which accounts for intermolecular forces and the volume of the gas molecules themselves.