Question

A home owner uses \(4.00 \times 10^3 \text{ m}^3\) of methane (\(CH_4\)) gas, (assume \(CH_4\) is an ideal gas) in a year to heat his home. Under the pressure of 1.0 atm and 300 K, mass of gas used is \(x \times 10^5\) g. The value of \(x\) is ________. (Nearest integer) (Given R = 0.083 L atm \(K^{-1} mol^{-1}\))

Hint:

Be careful with units: \(1 \text{ m}^3\) is 1000 liters. Using the wrong volume unit is the most common mistake in gas law problems.

Answer 1

Correct Answer: 26

Step 1: Understanding the Concept:
Methane behaves as an ideal gas, so we can use the ideal gas equation to find the number of moles present in the given volume at a specific pressure and temperature.
The total mass of the gas is the product of the number of moles and the molar mass of methane.
Step 2: Key Formula or Approach:
1. Ideal Gas Law: \(PV = nRT \Rightarrow n = \frac{PV}{RT}\).
2. Mass: \(m = n \times M_{CH_4}\).
Step 3: Detailed Explanation:
1. Convert Units:
Volume \(V = 4.00 \times 10^3 \text{ m}^3\).
Since \(1 \text{ m}^3 = 1000 \text{ L}\), \(V = 4.00 \times 10^6 \text{ L}\).
Molar mass of \(CH_4\) (\(M\)) = \(12 + (4 \times 1) = 16 \text{ g/mol}\).

2. Calculate Moles (\(n\)):
\[ n = \frac{1.0 \text{ atm} \times 4.00 \times 10^6 \text{ L}}{0.083 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}} \]
\[ n = \frac{4.00 \times 10^6}{24.9} \approx 160642.57 \text{ mol} \]

3. Calculate Mass (\(m\)):
\[ m = 160642.57 \text{ mol} \times 16 \text{ g/mol} \approx 2570281.12 \text{ g} \]
\[ m = 25.7 \times 10^5 \text{ g} \]
Rounding to the nearest integer, \(x = 26\).
Step 4: Final Answer:
The value of \(x\) is 26.