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Ideal gases are gases that have no attractive or repulsive forces between molecules.

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- 1. Determining the molar volume of Carbon DioxideAuthor: Dr. Robert D. Craig, Ph.D.Purpose of the experiment:Determine the molar volume of an ideal gas,By measuring the mass and volume of carbon dioxide gas sample
- 2. Determining the molar volume of Carbon DioxideBackground informationIdeal gases are gases that have no attractive or repulsive forces between molecules.
- 3. Determing the molar volume of Carbon DioxideAvogadro’s law states that equal volumes of ideal gases, under identical temperature and pressure Conditions, contain equal numbers of molecules. Thus, at the same temperature and pressure, one mole of any ideal gas has the same volume as one mole of any other ideal gas.
- 4. Determing the molar volume of Carbon Dioxide• This volume is called the molar volume (Vm) of ideal gases. Understandard condition (STP) ,which are one atmosphere (at,) of 760 mm Hg pressure and 273 kelvin (K), the molar volume of any ideal gas is 22.4 L
- 5. • The ideal gas law (equation 1) expresses the relationship between pressure ( P) volume (V) , number of moles present (n), and temperature (T in kelvins) for any ideal gas sample
- 6. .
- 7. ,• R is the gas law constant. So long as T, P and V are expressed in kelvins , atmospheres , and liters, respectively, R equals 8.21 x 10-2 L atm. Mol. K
- 8. .• If we measure the volume of a known mass of an ideal gas under one set of condition (to those labeled “2”) we can use equation 1 to write
- 9. .N = P1 V1/ RT1 and N = P2 V2/R T2
- 10. .• Because the mass of gas does not change , n remains constant. Therefore, we can combine the two equations to form equation 2 on the next page. As part of this process, we can cancel R on each side , because R is a constant
- 11. Pg 138• As part of this process, we can cancel R on each side , because R is a constant
- 12. .• In order for equation 2 to be valid, we must express the temperature in Kelvins. We can convert any temperature from degrees Celsius to kelvins by adding 273
- 13. .• In equation 2, we can express the pressure in any of several units: inches or centimeters of mercury, torr, or atmospheres. Similarly, we can express the volume in either milliliters or liters
- 14. .• We can use equation2 and an ideal gas sample to determine the molar volume of that gas under standard conditions.
- 15. • For example, suppose a 0.250 g sample of carbon dioxide (CO 2) gas has a volume of 0.145 L at 754 torr and 27 oC .
- 16. .• To determine the molar volume of this sample at STP we begin by solving equation 2 for V stp, where condition 2 is STP• V stp = P1 V 1 T stp / T 1 P stp
- 17. .Next we substitute experimental values into equation 3Vstp = (754 torr ) ().145 L ) (273 K)/• (27 + 273) K ( 760 torr)
- 18. . Vstp = (754 torr ) ().145 L ) (273 K)/ (27 + 273) K ( 760 torr)• = 0.131L
- 19. .• The resulting volume is not 22.4 L, as you might have expected, because the sample mass was only 0.250 g/mol. However, we can use this result to calculate the molar volume of CO2 at STP, using equation 4
- 20. .Vm , L/mol = ( V stp, in L ) ( M.WT of sample) (mass of sample, g)= (0.131 L/0.250 g CO 2) ( 44.01 g CO 2/ 1 mol CO 2) =23.1 L/ mol
- 21. .We can find the percent error for this result relative to the theoretical ideal gas molar volume of 22.4 L/mol using equation 5Percent error, %= [ (experimental value of Vm, L/mol) -22.4 L/mol) x ( 100 %)•
- 22. .For the molar volume determined abovePercent error , % = ( 23. 1 L/mol – 22.4 L/ mol)/ 22.4 L/ mol=3%
- 23. • In this experiment, you will calculate the molar volume of CO 2 gas ( in L/mol) by determining the mass of a known volume of CO 2 gas at laboratory temperature and pressure.•
- 24. • You will determine the CO 2 mass by first weighing t a flask filled with air, and tghen reweighing the same flask filed with CO2 gas. You will fill the flask with CO2 gas by allowing a piece of solid CO 2 (dry ice) to vaporize in the flask.
- 25. • The vaporizing CO 2 will force all the air from the flask, because CO2 gas is considerably more dense than air. The flask becomes filled with CO2 gas, which we can assume is at laboratory temperature and pressure.
- 26. .The gas volume is the same in both cases; it is the volume of the flask.The mass you will determine from the first weighing of the flask (m1) is the mass of the flaske (mf) plus the mass of air contained in the flask (ma) , as shown in equation 6 m1 = mf + ma
- 27. .m1 = m F + maThe mass you will determine from the second weighing of the flask (m2) is the mass of the flask (mf) plus the mass of CO2 gas contained in the flask(mCO2) (equation 7)m2 = mf+ mCO2
- 28. .We can calculate the difference between the second and first masses using equation 8.m2 - m1 = ( mF + mCO2) - (mF+ ma )= mCO 2 - maWe can calculate the mass of CO2 gas usingEquation 9, which is a rearrangement of Equation 8.
- 29. mCO2 = m2– m1+ maEquation 8.mCO2 = m2– m1+ maAs equation 9 shows, the mass of CO2 gas in the flask is equal to the difference between the two weighings plus the mass of air contained In the flask
- 30. .We can calculate the mass of air, ma, by multiplying the density of air at laboratory temperature and pressure by the volume of air contained in the flask (equation 10)ma = (density of air, g/mL )(volume of flask, mL)
- 31. .The CRCYou can obtain the literature value for the density of air under your lab conditions in the CRC*based on you data, you will calculate the molar volume of CO2 gas** please also calculate the percent error in your analysis
- 32. http://itl.chem.ufl.edu/2045_s99/lect ures/lec_bc.html

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