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How many grams of methane gas (CH4) are in a 15.8 liter sample at 1.2 atmospheres and 27°C? Show all work used to solve this problem.
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\[pV = \frac{ m(CH4)RT }{ M(CH4) }\] Solve for m, so to get it alone: multiply by M(CH4), divide by RT \[\ m = \frac{ pVM(CH4) }{ RT }\] then convert to the needed units and plug in the known values: M(CH4) = 12+4 = 16 (roughly, Ar rounded) R(for p in atm and V in liters) = 0.082 T(must be in K) = 27 +273 = 300K Back to the equation: \[m = \frac{ 1.2*15.8*16 }{ 0.082*300 } = 12.332 grams \]
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