What is the determinate error of Experiment 5 - Driving Forces? The sample data

Question

What is the determinate error of Experiment 5 - Driving Forces? The sample data below is based on volumes found in an inverted graduated cylinder submerged in water with an air bubble in it. The data can be used to extrapolate the enthalpy of formation of water. What is the number of moles of air found in the cylinder at 5 degree C? Find the pressure of air, P_air, at 50 degree C (in bar). Find the pressure of water, P_H2O, at 50 degree C (in bar). What is the Clausius-Clapeyron equation and why is it useful? What does each term represent in a linear plot?

Explanation / Answer

The experiment is to determine the vapor pressure of water at different temperature.

The Experiment Overview:

        Two third of 1L or 500mL beaker is filled with hot water. Then 20 - 21mL of distilled water is added to 25mL graduated cylinder. Two holed rubber stopper is placed on the cylinder and quickly inverted into the beaker. Then beaker is heated and volume of gas which is trapped in graduated cylinder is measured at different temperature.

      2.   The number of moles of air present in cylinder at 5 degree celcuis :

                        The gas law calculations required temperatures in kelvin. So convert the temperature to kelvin.

                                  K = Temp. in celcius + 273.2

                                  K = 5 + 273.2

                                  K = 278.2

At temperature of 0 to 5 degree celcius the only gas in the inverted cylinder is dry air. So let us use ideal gas law to calculate the number of mole air at 278.2 Kelvin.

                                                      P V = nRT

                       Pressure (P) is the atmospheric pressure 0.982 atm

                  Volume (V) is Volume measured at 5 degree celcius minus 0.2

                                    V   = 4.4 - 0.2

                                   V    = 4.2mL

                 R = 0.0821 L atm mole-1K-1 universal gas constant

          on rearranging we have,

                                    n   = PV / RT

                                    n   = 0.982 atm x 0.0042 L / 0.0821 L atm mole-1 K-1 x 278.2 K

                                   nair = 1.805 x 10-4 moles

        3.   The pressure of air Pair at 50 degree celcius :

                                        Again using ideal gas law we have,

                                            P = nRT/V

                               n number of moles of air is constant so nair = 1.805 x 10-4 moles

                              V = 5.8 - 0.2

                               V = 5.6mL

                              T = 50 + 273.2K

                                      T = 323.2K

                    P = 1.805 x 10-4 moles   x    0.0821 L atm mole-1 K-1 x 323.2K   / 0.0056L

                         Pair = 0.855 atm

                     1 atm = 1.01325 bar

                    Pair (in bar) = 0.866 bar

          4.   The pressure of water PH2o :

                                 PH2O = Atmospheric pressure - Pressure of air

                                         = (0.982 - 0.855)atm

                                  PH2O = 0. 127 atm

                           PH2O(in bar) = 0.129 bar

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.