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Chemistry 1°

Lab 14: Determination of R: The Gas-Law Constant

OBJECTIVES

• To understand how real gases obey the ideal-gas law
• To determine the ideal-gas-law constant, R

INTRODUCTION

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.

Pre-Lab Discussion

1. Under what conditions of temperature and pressure would you expect gases to obey the ideal-gas equation?
2. Gases obey the ideal-gas law at high temperatures and low pressures.

3. Calculate the value of R in L-atm/mol-K by assuming that an ideal gas occupies 22.4 L/mol at STP.

4. R= = .0821

5. Why do you equalize the water levels in the flask and the beaker?
6. 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.

7. Why does the vapor pressure of water contribute to the total pressure in the flask?
8. 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.

9. What is the value of an error analysis?
10. An error analysis allows you to judge the reliability of your data and gives an indication of the potential sources of error.

11. Suggest reasons why real gases might deviate from the ideal-gas law on the molecular level.
12. 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.

13. At present, automobile batteries are sealed. When lead storage batteries discharge, they produce hydrogen. Suppose the void volume in the battery is 100 mL at 1 atm of pressure and 25ºC. What would be the pressure increase if 0.05 g H2 were produced by the discharge of the battery? Does this present a problem? Do you know why sealed lead storage batteries were not used in the past?
14. PV = nRT

(298 K)

P = = = 6 atm

15. Why is the corrective term to the volume subtracted and not added to the volume in the van der Waals equation?
16. 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.

17. A sample of pure gas at 20ºC and 670 mm Hg occupied a volume of 562 cm3. How many moles of gas does this represent? (HINT: Use the value of R that you found in question 2.)
18. (562 cm^3)(.001 L) = .562 L (670 mm Hg)(1.00 atm) = .88 atm

1 cm^3 760 mm Hg

PV=nRT

(.88 atm)(.562 L) = n (.082 L-atm/mol-K) ( 293K )

n = .02058 mol = .021 mol

19. A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100ºC and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?
20. 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

21. Which gas would you expect to behave more like an ideal gas, Ne or HBr? Why?

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.

MATERIALS

Balance

Bunsen burner and hose

Test tube

250-mL beaker

Rubber stoppers (2)

Pinch clamp

Clamp

Barometer

Glass tubing with 60-degree bends (2) and straight pieces (2)

Rubber tubing

Thermometer

Ring stand

Plug of fiberglass

KClO3

MnO2

SET-UP

Take safety precautions and put on goggles and lab apron. The set-up should look like the following:

PROCEDURES

1. Measure out about 0.02g of MnO2 and 0.3g of KClO3.
2. Add MnO2 and KClO3 in a test tube and weigh to the nearest .001g. Record the mass of the test tube with the chemicals in the Results section.
3. Add fiberglass into the test tube very close to the rubber stopper.
4. Insert a rubber stopper attached to the tube B into the test tube.
5. Attach tube B to the flask but make sure it does not extend below the water level in the flask.
6. Fill glass tube A and the rubber tubing with water by loosening the pinch clamp and attaching a rubber bulb to and applying pressure through tube B.
7. Close the clamp when the tube is filled.
8. Mix the solids in the test tube by rotating the tube. Make sure none of the mixture is lost and does not touch the rubber stopper.
9. Fill the beaker with water half way. Insert tube A into the beaker.
10. Open the pinch clamp and then lift the beaker until the levels of water in the flask and beaker are identical. [By equalizing the levels, you produce atmospheric pressure in the flask and test tube.]
11. Close the clamp, discard the water in the beaker and dry the beaker.
12. Put tube A in the beaker again and open the pinch clamp. The flow should soon stop.
13. If the flow does not stop, then the apparatus is not air tight and you should start again.
14. Gently heat the lower part of the test tube (where the chemicals are). when the rate of gas evolution slows considerably, increase the rate of heating, and heat until no more oxygen is evolved. Water from the flask should flow into the beaker.
15. Wait until the apparatus cools to room temperature and then pour the water into the graduated cylinder. Record the volume of water in the Results section
16. Remove the test tube from the apparatus and weigh the test tube with its contents. Record this under "Mass of test tube + contents after reaction" in the Results section.
17. The difference in this mass and the mass before the reaction is the mass of O2 produced.
18. Record the barometric pressure in the Results section.
19. Get the room temperature. Record it in the Results section.
20. Use a table to get the vapor pressure of water at the room temperature and record it in the Results section.
21. Do not dispose of the chemicals in a waste paper basket or a sink. Consult your instructor as to where and how to dispose the contents.
22. Calculate the gas-law constant, R using the data and ideal gas law equation.
23. RESULTS

1. Mass of test tube + KClO3 + MnO2 ____42.993 g_______
2. Mass of test tube + contents after reaction ___ 42.874 g_______
3. Mass of oxygen produced _____0.119 g_______
6. Mass of Water _____ 95 g_______
7. Temperature of Water ___21 C° or 294 K___
8. Density for water ______.99792 g/L___
9. Volume of water____.095 L______ = volume of O2 gas ___.095 L_____
10. Barometric pressure _760.26 mmHg (1.00 atm)
11. Vapor pressure of water _ 18.6 mmHg (.0245 atm)
12. Pressure of O2 gas (show calculations) _741.7 mmHg (.976 atm)__
13. 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

(760 mmHg)

14. Gas-law constant, R, from ideal gas law (show calculations) __.085 L-atm/mol-K
15. 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

16. R from the van der Waals equation (show calculations) ___.085 L-atm/mol-K
17. n = (.119 g)/(32.00 g/mol)

= .00372 mol

(P+(n2a/V2))(V-nb)=nRT

(.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

18. Accepted value of R__.08206 L-atm/mol-K___ (source of R value) __text book___
19. Uncertainty in R (show calculations) .0848 ! .0005 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

Maximum:

R = (741.8 mmHg/760. MmHg)(.0951 L)(32.00 g/mol)

(.1189 g)(293K)

R = .0853 L-atm/mol-K

Minimum:

R = (741.6 mmHg/760. MmHg)(.0949 L)(32.00 g/mol)

(.1191 g)(295K)

R = .0843 L-atm/mol-K

Average:

.0853 L-atm/mol-K + .0843 L-atm/mol-K

2

Difference:

.0853 L-atm/mol-K - .0843 L-atm/mol-K = .0005

2

R = .0848 ! .0005 L-atm/mol-K

CONCLUSIONS

Questions

1. Does your value of R agree with the accepted value within your uncertainty limits?
2. No, our R does not agree with the accepted value within your uncertainty limits but it is close.

3. Discuss possible sources of error in the experiment; indicate the ones that you feel are most important.
4. 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.

5. Which gas would you expect to deviate more from ideality, H2 or HBr? Explain your answer.
6. 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.

7. How does the solubility of oxygen in water affect the value of R you determined? Explain your answer.
8. 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

9. a. Use the van der Waals equation to calculate the pressure exerted by 1.000 mol of Cl2 in 22.41L at 0.0ºC. The van der Waals constants for Cl2 are a = 6.49 L2 atm/mol2 and b = 0.0562 L/mol.

( 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.

1. How much potassium chlorate is needed to produce 20.0 mL of oxygen gas at 670 mm Hg and 20ºC?
2. 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

760 mm Hg

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

3 mol O2

3. If oxygen gas were collected over water at 20ºC and the total pressure of the wet gas were 670 mm Hg, what would be the partial pressure of oxygen?
4. 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

760. mm Hg

5. An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen. If 23.0 g of this gas were found to occupy 5.6 L at STP, what are the empirical and molecular formulas for this oxide of nitrogen?
6. 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

7. The gauge pressure in the automobile tire reads 32 pounds per square inch (psi) in the winter at 32ºF. The gauge reads the difference between the tire pressure and the atmospheric pressure (14.7 psi). In other words, the tire pressure is the gauge reading plus 14.7 psi. If the same tire were used in the summer at 110ºF and no air had leaked from the tire, what would be the tire gauge reading in the summer? (HINT: Recall that ºC=5/9(ºF – 32).)

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