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 \(K^{-1} mol^{-1}\))
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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.
To solve the problem of determining the mass of methane (\(CH_4\)) used in grams, given the usage of \(4.00 \times 10^3 \text{ m}^3\) under conditions of 1 atm and 300 K, we will use the ideal gas law and molecular calculations. Let's go through the steps:
Step 1: Use the Ideal Gas Law The ideal gas law is \(PV = nRT\), where:
\(P\) is the pressure in atm (1.0 atm).
\(V\) is the volume in liters (First, convert \(4.00 \times 10^3 \text{ m}^3\) to liters: \(4.00 \times 10^3 \times 1000 = 4.00 \times 10^6\) L).
\(n\) is the number of moles of gas.
\(R\) is the ideal gas constant (0.083 L atm K\(^{-1}\) mol\(^{-1}\)).
\(T\) is the temperature in Kelvin (300 K).
Plugging these values into the equation, we have: \(1.0 \times 4.00 \times 10^6 = n \times 0.083 \times 300\) Solve for \(n\): \(n = \frac{4.00 \times 10^6}{0.083 \times 300}\) \(n ≈ 1.608 \times 10^5 \text{ moles of } CH_4\)
Step 2: Convert Moles to Mass The molar mass of methane (\(CH_4\)) is 16 g/mol. Thus, the mass \(m\) in grams is given by: \(m = n \times \text{Molar Mass of } CH_4 = 1.608 \times 10^5 \times 16\) \(m ≈ 2.5728 \times 10^6 \text{ g}\)
Step 3: Compute \(x\) and Validate Here, the mass \(m\) is given as \(x \times 10^5\) g. Set up the equation: \(x \times 10^5 = 2.5728 \times 10^6\) Solve for \(x\): \(x = \frac{2.5728 \times 10^6}{10^5} = 25.728\) Round \(x\) to the nearest integer: \(x = 26\).
Verification: The computed value \(x = 26\) falls within the provided range of \(26,26\). Thus, our solution is verified to be correct. The value of \(x\) is 26.
