FOUNDATION CHEMISTRY 2 LAB REPORT

EXPERIMENT 1: DETERMINING THE ENTHALPY OF VAPORIZATION

TRILOKESH A/L RAJA MOGAN (1001128099)

FACULTY OF ENGINEERING, ARCHITECTURE &BUILT ENVIRONMENT

UCSI UNIVERSITY

May 28, 2011

APPENDIX B: FORMAT FOR TABLE OF CONTENT

TABLE OF CONTENT

PAGE NO.

1.0 INTRODUCTION 1

2.0 MATERIAL & METHODOLOGY 2

3.0 RESULTS AND DISCUSSION 4

4.0 CONCLUSION 10

5.0 LIMITATION OF EXPERIMENT 10

6.0 REFERENCE 11

1.0 INTRODUCTION

In this experiment, a sample of air will be trapped over water, in an inverted graduated

cylinder in a beaker. As the temperature of the apparatus is changed, the number of moles of

water vapor in the gas phase will vary according to the Clapeyron equation, while that of the

trapped air will remain constant. The number of moles of air in the water-air mixture will be

found by reducing the temperature of the whole apparatus to about 5 °C. At that temperature, the

vapor pressure of water is so small that the volume measured for the trapped gas corresponds

only to the air present. Using the data collected, the enthalpy of vaporization can then be

calculated from the slope of a graph of ln PH2O (the vapor pressure of water) versus 1/T.

The objective of this experiment is to measure the vapor pressure of water as a function

of temperature change, prepare a graph of ln(P) as a function of and calculate the enthalpy

of vaporization (∆ H vap) of water.

2.0 MATERIAL AND METHODOLOGY

2.1 EXPERIMENTAL MATERIALS

Water, ice

2.2 EXPERIMENTAL APPARATUS

Beaker, graduated cylinder, thermometer, ruler, electrical heater

2.3 EXPERIMENTAL PROCEDURES

1. A 10 ml graduated cylinder was obtained and was filled with approximately 9 ml of

water.

2. A 1000 ml beaker was filled with approximately three-fourths full with water.

3. The top of the graduated cylinder was covered with a finger and inverted into a 1000 ml

beaker. The finger was not released until the mouth of the graduated cylinder was under

the surface of the water in the beaker.

4. Water was then added into the 1000 ml beaker to cover the graduated cylinder.

5. A ruler was used to measure the difference between the height of the water in the

graduated cylinder and height of the water in the beaker. This would provide a slight

adjustment for the pressure that the water exerts on the air in the graduated cylinder.

6. The barometric pressure was recorded in mmHg.

7. The 1000 ml beaker was heated on the electrical heater until the water was about 80 ºC. If

the air in the cylinder expanded past the scale on the graduated cylinder, the experiment

would have to be restarted using a smaller amount of air.

8. The temperature and the volume were recorded.

9. The beaker was cooled until the temperature reached 50ºC and the temperature and

volume was recorded every 5ºC.

10. Once the beaker reached 50 ºC, the beaker was cooled rapidly to about 0ºC by adding ice.

The gas volume and temperature was recorded at this low temperature. If the vapor

pressure was too low at this temperature it is assumed that all of the gas in the graduated

cylinder is air.

3.0 RESULTS AND DISCUSSION

3.1 CALCULATION AND ANALYSIS

Atmospheric Pressure: 760 mm Hg

Difference in height, h: 13.8-10 = 3.8 cm

Temperature (°C) Volume (mL)

80 8.0

75 7.6

70 6.4

65 5.2

60 4.8

55 4.6

50 4.5

45 4.4

40 4.3

35 4.2

30 4.1

25 4.0

20 3.9

15 3.8

10 3.7

5 3.6

Pressure exerted by water = h mm water x (1.00 mm Hg / 13.6 mm water) = 38 mm water x (1.00 mm Hg / 13.6 mm water)

= 2.79 mm Hg

PT = Patm + h mm water x (1.00 mm Hg / 13.6 mm water)

= 760 mm Hg + 2.79 mm Hg = 762.79 mm Hg Moles of trapped air

PTV = nairRT

P = total pressure in cylinderV = volume of air when T = 0n= number of moles of trapped airR =8.314kPa / 62.4 mm HgT = lowest temperature

PTV = nairRT

nair = PTV / RT = (762.79)(0.001) / (62.4)(276) = 0.76279 / 17222.4 = 0.0000443 mol

Total pressure in cylinder,Pτ 762.79 mm Hg

Moles of trapped air, n 0.0000443 mole

Corrected volume (mL)

Temperature(K)

Pair (atm) Pwater (atm) 1/T (1/K) ln Pwater

7.8 353 125.10 637.69 0.00283 6.467.4 348 129.99 632.80 0.00287 6.456.2 343 152.93 602.86 0.00292 6.405.0 338 186.87 575.92 0.00296 6.364.6 333 200.11 562.68 0.00300 6.334.4 328 209.21 553.58 0.00305 6.324.2 323 212.59 550.20 0.00309 6.311.8 280 430.00 332.79 0.00357 5.81

Partial pressure (PairV = nair RT)

For temperature 353 K and volume 7.8ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) ( 353K) / (0.0078)

=125.10mmHg

For temperature 348 K and volume 7.4ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) ( 348K) / (0.0074)

=129.99mmHg

For temperature 343 K and volume 6.2ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) ( 343K) / (0.0062) =152.93mmHg

For temperature 338 K and volume 5.0ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) (338K) / (0.0050) =186.87 mmHg

For temperature 333 K and volume 4.6ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) (333K) / ( 0.0046) =200.11mmHg

For temperature 328 K and volume 4.4ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) (333K) / ( 0.0044) =209.21 mmHg

For temperature 323 K and volume 4.2ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) (323K) / ( 0.0042) =212.59 mm Hg

For temperature 280K and volume 1.8ml

PairV = nair RTPair = (0.0000443) ( 62.4 mm Hg) (280K) /(0.0018) = 430.00 mmHgVapour pressure of water (PT = 762.79 mm Hg)

Pwater = PT – Pair

For temperature 353 K and volume 7.8mlPwater = PT – Pair = 762.79 mm Hg – 125.10mmHg =637.69 mmHg

For temperature 348 K and volume 7.4mlPwater = PT – Pair = 762.79 mm Hg –129.99mmHg = 632.80 mmHg

For temperature 343 K and volume 6.2mlPwater = PT – Pair = 762.79 mm Hg –152.93mmHg = 602.86 mmHg

For temperature 338 K and volume 5.0mlPwater = PT – Pair = 762.79 mm Hg –186.87 mmHg = 575.92 mmHg

For temperature 333 K and volume 4.6mlPwater = PT – Pair = 762.79 mm Hg –200.11mm Hg = 562.68 mm Hg

For temperature 328 K and volume 4.4mlPwater = PT – Pair = 762.79 mm Hg –209.21 mmHg = 553.58 mm Hg

For temperature 323 K and volume 4.2mlPwater = PT – Pair = 762.79 mm Hg –212.59 mm Hg = 550.20 mm Hg

For temperature 280 K and volume 1.8mlPwater = PT – Pair = 762.79 mm Hg – 430.00 mm Hg

= 332.79 mm Hg

A graph of ln Pwater (mmHg) against 1/T was plotted

ln P versus 1/T

y = -639.1x + 8.2674

6.286.3

6.326.346.366.386.4

6.426.446.466.48

0.0028 0.00285 0.0029 0.00295 0.003 0.00305 0.0031 0.00315

1/T

ln P

Series1

Linear (Series1)

Linear (Series1)

Regression Equation y= -639.1x + 8.2674

Hvap Hvap/R = - 639.1

Hvap = 639.1 * R

= 5313.47 J/mol

Percent error (5313.47 – 40.65) / 40.65 * 100

= 1.297 %

3.2 DISCUSSION

Vapor pressure is the pressure exerted on the inside of a container due to the vapors that

escape from a liquid. Some particles in a liquid have enough energy to go into the gas phase.

Since the particles are in gas phase, they exert a pressure on the inside of the container, but they

can also go back into liquid phase so equilibrium is reached where a constant pressure due to the

vapors is assumed. At higher temperatures this pressure changes with temperature because, the

liquid has more energy, and likewise there is a greater probability of the gas escaping, and the

escaped gas has more energy so it exerts a greater force on the container. Therefore, the vapor

pressure is directly proportional to temperature. From the graph, the ∆H vap value was obtained by

calculating the slope.

4.0 CONCLUSION

Based on the graph obtained, the enthalpy of vaporization of water can be calculated by using the

regression equation. The slope that is obtained is multiplied by the R constant to give the

∆ H vap. For this experiment, the ∆ H vap is 5313.47 J/mol. The percent error compared to the

theoretical value is 1.297 %.

5.0 LIMITATION OF EXPERIMENT

5.1 LIMITATION OF TECHNIQUE

One of the limitation in this experiment is the fact that an accurate value that corresponds to

the theoretical value of ∆ H vap cannot be obtained. This is due to the fact that heat is a kind of

energy that varies at different conditions. Therefore, the actual value is difficult to obtain and this

leads to a percent of error.

5.2 ERRORS THAT MAY HAVE OCCURRED

A few errors may have occurred during the experiment. Firstly, when the water was being

brought to boil on the electrical heater, it may have spilled and this may have changed the

pressure the water exerted on the air and could have skewed the volume measurements. Next,

when taking readings from the thermometer, parallax error could have occurred causing the

results to be slightly deviated.

5.3 IMPROVEMENT AND SAFETY PRECAUTIONS

To improve the results of the experiment, it is vital to follow all the correct procedures of the

experiment A few improvements can be done to reduce errors in the. Firstly, we can make sure

the thermometer is placed at a suitable position, not too high or too low, to obtain the accurate

change in temperature. If possible the experiment can be repeated to obtain an average on

calculations.

There are some safety precautions that we must follow throughout the experiment. Firstly, the

safety goggles should be worn throughout the experiment to avoid unwanted eye injuries. .

Furthermore, extra care has to be taken to make sure the electrical heater is handled carefully to

avoid burns.

6.0 REFERENCE

1. http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Enthalpy-of-Fusion-and-

Enthalpy-of-Vaporization-842.html

2. Thomas M. Moffett Jr., Plattsburgh State University, 2003 Edited 2010

faculty.plattsburgh.edu/tom.moffett/che112/enthalpiesofwater.pdf

accesed on 25 Feb 2011.

3. Kyle Miller November 19, 2006 Vapor Pressure and Enthalpy of Vaporization of Water

1st Edition