Ideal Gas Experiments
PV = and where:
P Pressure of the gas.
V Volume of the gas.
n Number of moles of the gas.
R The ideal gas constant.
T Temperature of the gas.
The above formula is used while determining the ideal gas constant R, as in the experiment where carbon dioxide was used to determine R.
Since n = mass of the gas / relative molecular mass of the same gas, the combined gas equation can also be used to determine the molecular weight of a gas, just as in the experiment of determining the molecular weight of butane gas.
From n = PV/RT, the molecular weight of a gas can be calculated since:
Molecular Weight (Relative Molecular Mass) = Mass of the gas /n.
It is important to note that no gas will behave in an ideal manner under normal circumstances. Because of this, most gases are real (in fact all), meaning they do not obey gas laws unless subjected to certain conditions such as high pressure. In this paper, sample equations are discussed which are from experiments conducted to determine the ideal gas constant R, the molecular weight of the gas, and to investigate the relationship between pressure and volume of a gas.
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CHM201 General Chemistry and Laboratory I
Laboratory 6 Ideal Gas Experiments
(Due July 10, 2013)
Name M2
Group 5
(A lab quiz will be worth 10 points.)
1. Determination of the Ideal Gas Constant "R" using Carbon Dioxide(20 points)
Mass of beaker (optional) 

Mass of plastic bag (Error! Reference source not found.) 

Temperature 
21.5 0C 
The volume of plastic bag 
3750ml 
Barometric pressure 
774.8mmHg 
Trial 
1 
2 
3 
4 (wet) 
5 (wet) 
Mf in grams 
120.257 
110.38 
100.424 
69.35 
59.43 
Mi in grams 
126.80 
120.24 
110.38 
74.33 
69.35 
R in LmmHgK1mol1 
66.30045 
43.99634 
43.5721 
87.16837 
43.75993 
Rave in LmmHgK1mol1 
51.28963 
65.46415 

% Error 
17.7572% 
4.97161% 
2. Determining the Molecular Weight of a Gas. (20 points)
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Room temperature TR 
25 0C 
Water temperature  Tw 
28 0C 
Barometric Pressure Paytm 
1.019atm 
Vapor Pressure of Water Peter 
0.037368atm 
The pressure of Butane Pbutane (Paytm P? water) 
0.981632atm 
Trial 
1 
2 
3 
Total (Error! Reference source not found.) 
Mf in grams 
11.263 
11.180 
11.101 

MI in grams 
11.387 
11.263 
11.180 

Mbutane in grams 
0.124 
0.083 
0.079 
Mave = 0.095333 
Vbutane in ml 
36.351 
36.125 
36.198 
Vave = 36.22467 
Molar Mass 
65.71144g 
3. Pressure/Volume Measurements. (20 points)
V 
20 cm3 
16 cm3 
9 cm3 
6 cm3 
4 cm3 
P mmHg 
369.6 
461.6 
784.1 
1102.7 
1519.8 
PV 
7392 
7385.6 
7056.9 
6616.2 
6079.2 
V 
5 cm3 
10 cm3 
20 cm3 
P mmHg 
768.7 
400.7 
208.6 
PV 
3843.5 
4007 
4172 
V 
20 cm3 
10 cm3 
5 cm3 
P mmHg 
332.7 
644.3 
1168.7 
PV 
6654 
6443 
5843.5 
1. How can you account for the fairly large percent error in the gas constant determination? There are both experimental and scientific factors involved. Include both.
Experimentally, it is not easy to measure the exact mass of gas as some will automatically escape to the environment. This will lead to some errors encountered while taking readings. This fraction of a gas that escapes to the environment contains numerous molecules. The random movement of molecules within an ideal gas gives the ideal gas its unique properties when obeying gas laws.
Some of the instruments used while experimenting are not exactly accurate. Some have inaccurate calibrations that might result in significant errors when used. For instance, an electronic balance might be more accurate than the common beam balance when measuring mass. Measuring cylinders are always more accurate compared to beakers when measuring volumes. The small difference within the measurements results in various errors observed.
Poor judgment while taking readings is also a common source of experimental errors. Individuals tend to assume and approximate the last digit of their measurements while taking readings. It is important to note that even 1cm3 of gas has the capability of changing the behavior of the gas to some extent. Readings should be taken several times and the average calculated to reduce this error.
Scientifically, carbon dioxide is not an ideal gas. Its molecules are not identical hence cannot give a valid result when used to determine the gas constant. Its molecules move randomly but do not obey Newton's law of motion. The particles of an ideal gas occupy negligible volume compared to the volume of the gas itself. The molecules of carbon dioxide interact with each other while those of an ideal gas react totally independent of all others (Clugston and Flemming, 2000). The changes in kinetic energy within the molecules of carbon dioxide could have led to the great percentage error obtained.
An ideal gas is considered as a point mass, that is, its molecules are extremely small thus its mass is almost zero. Because of this, there is no change in the kinetic energy of its molecules since it has no rest mass. The collision between molecules of an ideal gas is said to be elastic (there is neither attractive nor repulsive energy involved) and takes a negligible amount of time.
The pressure exerted by carbon dioxide during the experiment was also not strong enough compared to the pressure exerted by an ideal gas. The molecules of an ideal gas do not have attractive forces that would hold them back when they collide at an impact. They collide with less energy but exert greater pressure than real gases. For the ideal gas constant to be determined accurately, the pressure exerted by the gas must be strong enough.
The ability of carbon dioxide to liquefy at very low temperatures and very high pressure could have also resulted in the high percentage error recorded. When a gas liquefies, it no longer possesses the properties of a gas. The molecular interactions within the liquid form of the gas increase and greater changes in energy occur within the molecules. The molecules exert a very small amount of pressure that cannot be used to determine the gas constant. The gas is said to have acquired the properties of a liquid. In the liquid state, it occupies a definite volume and the movement of molecules becomes dependent on each other. The collisions become more inelastic and the molecules become more compacted together by strong intermolecular forces.
2. If some water remained on the lighter at the time of the second weighing (after collecting the gas), how would this affect your determination of molecular weight? Be specific.
The molecular weight or molecular mass of a compound or molecule simply refers to the total mass of the elements that make up the compound or molecule. It determines the stoichiometry nature of a compound. The remains of water on the lighter may have been due to the subjection of butane to a high temperature and low pressure. This means that butane was literally combusting to produce some water molecules. The mass of butane would be greatly reduced. This will eventually lead to a decrease in the molecular weight of butane when determined using the equation of state.
The calculation of the molecular weight of gas entirely depends on the three gas laws. They require that the mass of gas be fixed. Variation of the mass of gas leads to unreliable results. The change in mass of butane is symbolized by the remains of water on the lighter. When the mass of butane continuously changes, it no longer obeys the gas laws. The equation of state (PV = NRT) cannot be used to determine its mass as this will be blatantly ineffective. If we made the effect of assuming while calculating the molecular weight of butane, the result will not be even close to the actual molecular weight of butane.
3. Which of the individual gas laws is confirmed by the results of part 3?
Part 3 involves the relationship between pressure and the volume of gas. This relationship is basically known as the Boyles law (Clugston and Flemming, 2000). Boyles law states that, for a fixed mass of an ideal gas at a constant temperature, the pressure of the ideal gas is inversely proportional to its volume. This is expressed mathematically as:
P ? 1/V
PV = C
Where:
P Pressure.
V Volume.
C Constant.
The equation can also be stated as that the product of pressure and volume of a fixed mass of an ideal gas is a constant providing that its temperature is kept constant.
Thus, the law can also be expressed as:
P1V1 = P2V2
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Where:
P1 Initial pressure of the gas.
V1 Initial volume of the gas.
P2 Final pressure of the gas.
V2 Final volume of the gas.
The equation above shows that, as the pressure increases, the volume of the gas decreases in proportion. Conversely, as the pressure of an ideal gas decreases, its volume also increases proportionally.
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