CHU 201 Lab 6 Rev 1 C12-17 4. Calculations 4.1.1 Subtrnet M, from M, to give the
ID: 704430 • Letter: C
Question
CHU 201 Lab 6 Rev 1 C12-17 4. Calculations 4.1.1 Subtrnet M, from M, to give the mass of Co2 in the full bag for each trial 4.1.2 Using a molecular weight for Co2 of 44.01 grams/mole, calculate the number of moles for each trial. the bag in 1.5 in K, the volume in L. from 1.11 and the barometric pressure in atm (1.12) for each trial. Calculate R from the ideal gas equation using the pumber of moles in 4.1.2, the temperature of water (°C), the Record the following data for each butane trial in table 2: the temperature of the mass of butan ml.. 4.2.1 e used (M, Me, the barometric pressure and the volume of butane observed in From table 5 in the Laboratory Handbook (Appendix B) determine the vapor pressure of water at the temperature of the water bath (WT) and subtract it from the barometric pressure (be sure to use consistent units!). Enter this value in the table as corrected pressure and convert this pressure 4.2.2 4.2.2 Calculate the molecular weight (molar mass) using the ideal gas law using compatible units and enter into the table (see Chang Example 5.9). Determine the mean value of the molecular weight for the three trials. Determine the percent difference between the mean value for the molecular weight of butane and the MW for pure butane of 58 g/mol. % difference-ss.aveMW x 100% 58 4.3.1 Calculate the PV product for each of the data points obtained 4.3.2 Determine the mean value (average) for the PV product values. 4.33 Determine the factor by which you adjust the voltume in 3,3 and 3,4 and calculate the füctor by which the pressure changes.Explanation / Answer
Pressure of butane
PV=nRT
P = nRT/ V
1. Given,
V = 35.8
mass = 0.082g
T = 24 (celcius)
moles = grams/molarmass = 0.082/58.12 = 1.41*10^-3
P = nRT / V
= 1.41*10^-3 * 8.314 * 297.15 / 35.8 = 0.0973
Pressure of butane in atm = 9.6027634e-7
2. Given,
V = 35.8
mass = 0.083g
T = 24 (celcius)
moles = grams/molarmass = 0.083/58.12 = 1.428*10^-3
P = nRT / V
= 1.428*10^-3 * 8.314 * 297.15 / 35.8 = 0.0985
Pressure of butane in atm = 9.7211942e-7
3. Given,
V = 35.4
mass = 0.081g
T = 24 (celcius)
moles = grams/molarmass = 0.081/58.12 = 1.39*10^-3
P = nRT / V
= 1.39*10^-3 * 8.314 * 297.15 / 35.4 = 0.0970
Pressure of butane in atm = 9.573156e-7
Converting volume in grams into litre:
Multiply your mole value by the molar volume constant, 22.4L.
35.8g = >
moles = grams/ molar mass = 35.8 / 58.12 = 0.6159
Volume in litre = 0.6159 * 22.4 = 13.797L
35.4g = >
moles = grams/ molar mass = 35.4 / 58.12 = 0.609
Volume in litre = 0.6159 * 22.4 = 13.643L
Water temperature in Kelvin
25 degree Celcius = 298.15K
“partial pressure” refers to the pressure that each gas in a gas mixture exerts against its surroundings, such as a sample flask, a diver’s air tank, or the boundary of an atmosphere. You can calculate the pressure of each gas in a mixture if you know how much of it there is, what volume it takes up, and its temperature. You can then add these partial pressures together to find the total pressure of the gas mixture, or, you can find the total pressure first and then find the partial pressures.
Calculating partial pressure
1. Define the partial pressure equation for the gases you're working with. ...
2. Convert the temperature to degrees Kelvin. ...
3. Find the number of moles of each gas present in the sample. ...
4. Plug in the values for the moles, volume, and temperature. ...
5. Plug in the value for the constant R.
Boyle’s law – K = PV
Pascal(Pa) – 1atm = 101,325 PA
Charle’s law – K = V/T
Molar mass (?*R) = PV/T
Partial pressure of water at 25degree Celcius = 23.8