A sample of butane gas, C4H10, was slowly heated at a constant pressure of 0.80 bar. The volume of the gas was measured at a series of different temperatures and a plot of volume vs. temperature was constructed. The slope of the line was 0.0208 L·K-1. What was the mass of the sample of butane? (Assume that the temperature is 25°C.)
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The ideal gas law:
PV = nRT
V = (nR/P)(T)
If you plot V vs T, the slope of the line (0.0208) is equal to the coefficient of T, which is nR/P.
P = gas pressure in atm = 0.80 bar x (1 atm / 1.013 bar) = 0.79 atm
0.0208 = nR / P
n = (0.0208)(P)/R = (0.0208)(0.79)/(0.0821) = 0.20 moles of C4H10
0.20 moles C4H10 x (58.1 g C4H10 / 1 mole C4H10) = 12 g C4H10
PV = nRT
V = (nR/P)(T)
If you plot V vs T, the slope of the line (0.0208) is equal to the coefficient of T, which is nR/P.
P = gas pressure in atm = 0.80 bar x (1 atm / 1.013 bar) = 0.79 atm
0.0208 = nR / P
n = (0.0208)(P)/R = (0.0208)(0.79)/(0.0821) = 0.20 moles of C4H10
0.20 moles C4H10 x (58.1 g C4H10 / 1 mole C4H10) = 12 g C4H10