Lab 14:Determination of R: The Gas-Law Constant
The ideal-gaw law equation, PV=nRT, is obeyed by most gases at room temperature and atmospheric pressure. However, there are small deviations from this and consequently, the van der Waals equation, (P+(n2a/V2))(V-nb)=nRT, is used because it takes these deviations into account. It is also applicable over a wider range of temperatures and pressures. In the equation, a and b are gas constants. nb is the correction for the finite volume of the molecules and the (n2a/V2) term accounts for the intermolecular attractions.
The gas constant is known as R and is records in terms of L-atm/mol-K. By using the van der Waals equation and ideal-gas law for an enclosed sample of oxygen, R can be determined. KClO3 reacts with MnO2 to form KCl and O2. Then, by using Dalton’s law of partial pressures, the pressure of the gas O2 is found and it can then be implemented in the ideal-gas law and van der Waals equation to find R.
Gases obey the ideal-gas law at high temperatures and low pressures.
Equalizing the water levels equalizes the pressures and ensures that the total pressure in the flask is atmospheric and does not contain a contribution from the pressure due to the height of the water column.
Since gaseous and liquid water are in dynamic equilibrium, there will always be some water vapor above a sample of liquid water. Since the vapor pressure of water is reasonably high at ambient temperature, it makes a significant contribution to the total pressure.
An error analysis allows you to judge the reliability of your data and gives an indication of the potential sources of error.
The ideal-gas law assumes that there are no forces of attraction between the individual gaseous molecules. Whenever this isn’t so, real molecules will not obey the ideal-gas law. This would e expected to occur at very high pressures and at very low temperatures where molecules are so close to one another that they necessarily interact. The ideal-gas law also assumes that the gas particles have no volume. At high pressures their volume may become appreciable relative to the volume of the container.
PV = nRT
The term is subtracted to count for the difference between ideal and real gas particles. Small deviations from the ideal-gas law are observed because real-gas molecules are finite in size and exhibit mutual attractive forces. The term subtracted takes into account these two causes for deviation.
(562 cm^3)(.001 L) = .562 L (670 mm Hg)(1.00 atm) = .88 atm
(.88 atm)(.562 L) = n (.082 L-atm/mol-K) ( 293K )
n = .02058 mol = .021 mol
PV = (mRT) / M CH = 12.01g + 1.01g = 13.02 g
D = m/V = PM/RT
D = ( 1.00 atm) ( 13.02g) / ( .08206 L-atm/mol-K) ( 373 K)
D = .425 g/L
(2.550 g/L) / (.425 g/L) = 6.00
molecular formula: C6H6
You would expect Ne to behave more like an ideal gas because it has a smaller molar mass. This means that the intermolecular attractive forces are smaller than that of HBr and therefore, it is more like an ideal gas.
Bunsen burner and hose
Rubber stoppers (2)
Glass tubing with 60-degree bends (2) and straight pieces (2)
125-mL Erlenmeyer flask
100 mL Graduated cylinder
Plug of fiberglass
Take safety precautions and put on goggles and lab apron. The set-up should look like the following:
P(atmosphere) = P(O2) + P(water vapor)
760.26 mmHg = P(O2) + 18.6 mmHg
P(O2) = 741.7 mmHg
(741.7 mmHg)(1 atm) = .976 atm
PV = mRT/M
(.976 atm)(.095 L) = (.119 g)R(294 K)/(32.00 g/mol)
R = .0848 L-atm/mol-K
R = .085 L-atm/mol-K
n = (.119 g)/(32.00 g/mol)
= .00372 mol
(.976 atm) + (((.00372)^2 * 1.36 L2-atm/mol2)/(.095)^2 L))(.095 L – (.00372 mol)(.0318 L/mol) = (.00372 mol)R(294 K)
=.08485 = .085 L-atm/mol-K
R = PV/nT
P = 741.7 mmHg ! .1 mmHg
V = .095 L ! .0001 L
m = .119 g ! .0001 g
T = (21+273) °C ! 1 °C
R = (741.8 mmHg/760. MmHg)(.0951 L)(32.00 g/mol)
R = .0853 L-atm/mol-K
R = (741.6 mmHg/760. MmHg)(.0949 L)(32.00 g/mol)
R = .0843 L-atm/mol-K
.0853 L-atm/mol-K + .0843 L-atm/mol-K
.0853 L-atm/mol-K - .0843 L-atm/mol-K = .0005
R = .0848 ! .0005 L-atm/mol-K
No, our R does not agree with the accepted value within your uncertainty limits but it is close.
There are a lot of possible sources of error in the experiment, such as not being able to get equal pressures in the beaker and flask, not measuring the amount of water in the graduated cylinder accurately, leaving the water in the graduated cylinder out overnight (even though it was covered), a loss of significant figures during the calculations, the intermolecular forces between the molecules in the gas (even though the van der Waals equation accounts for some of it, it may not account for all), and measurement of barometric pressure could be inaccurate.
You would expect H2 to behave more like an ideal gas because it has a polar bond and it’s ability to polarize is greater because the molecular weight is higher than HBr therefore it is more likely to interact with the environment.
The oxygen had to have been not very water soluble in order for it to displace the water in the florence flask. The displaced water was then our volume reading for amount of oxygen produced, which was key in determining the value of R
( P + (an2/v2)) x ( V - nb) = nRT
[P + (6.49 L2 atm/mol2 )(1.000mol)2 /( 22.41 L)2 ) ] x ( 22.41 L - (1.000 mol)(0.0562 L/mol) =
(1.000mol) (.08206 L-atm/mol-K) ( 273.15K)
P = .9898 atm = .990 atm
b. Which factor is the major cause for deviation from ideal behavior, the volume of the Cl2 molecules or the attractive forces between them?
The attractive forces are a greater cause for deviation from the ideal behavior.
2KClO3 (s) (reacts with MnO2 (s) and heat) à 2KCl (s) + 3O2 (g)
KClO3 = 122.54 g/mol (670 mm Hg)(1.00 atm) = .88 atm
PV = nRT
(.88 atm) ( .0200 L ) = n ( .08206 L-atm/mol-K) ( 293 K)
n = 7.32 x 10^-4 mol O
(7.32 x 10 ^-4 mol O)(2 mol KClO3) = 4.9 x 10^ -4 mol KClO3
Total pressure = Pressure of O2 + Pressure of H2O
670 mm Hg = 17.5 mm Hg + P H2O
P(H2O) = 652.5 mm Hg = 653 mm Hg
(653 mm Hg)(1.00 atm) = .859 atm
PV = mRT / M
( 1.00 atm) ( 5.6 L) = (23.0g) ( .08206 L-atm/mol-K) ( 273.15 K) / M
M = 92.0603 g/mol = 92 g/mol
92 g/mol x .696 = 64 g/mol O (64 g/mol O) / (16.00 g/mol O) = 4.0 mol O
92 g/mol x .304 = 28 g/mol N (28 g/mol N) / (14.01 g/mol N) = 2.0 mol N
Empirical Formula: NO2
Molecular Formula: N2O4
Gauge = Tire Pressure - Atmospheric Pressure (14.7 psi)
Tire Pressure = Gauge + Atmospheric Pressure (14.7 psi)
Winter: TP = 32 psi + 14.7 psi = 46.7 psi = 47 psi
Summer: TP = x +14.7 psi
32 °F = 0 °C = 273 K
110 °F = (5/9)(78) °C = 43.3 °C = 316.5 K
P/T = P/T
(47 psi)/(273 K) = x/(316.5 K)
x = 54
54 = x + 14.7 psi
x = 39