1. ORDER FO MAGNITUDE ESTIMATION; At one atmosphere of pressure, what percentage

Question

1. ORDER FO MAGNITUDE ESTIMATION; At one atmosphere of pressure, what percentage of the volume of nitrogen gas is taken up by the volume of the nitrogen molecules (N2)? Assume a nitrogen molecule is about =3*10^-10 m in diameter. State clearly all other assumptions that you make and show your calculations. Woudl you expect the van der Waals' correction of volume to be important?

2. A closed chamber with an internal volume of 0.004 m^3 contains neon gas. When the chmber is cooled with liquid nitrogen to a temperature of -196 degree C, the internal gas pressure is 55kPa. Find the following values:

-The gas temperature in unites of Kelvins

-The number of moles of neon gas in the chamber.

-The number of neon atoms in the chamber.

-The mass of gas in the chamber. ONe mole of neon atoms has a mass of 20.18 g.

3. Supposed all of the neon gas is transferred to a new chamber. The new chamber is heated to 500 degrees C and the internal gas pressure reaches 613 kPa. What is the volume of the new chamber?

Explanation / Answer

2. Given values are V (of chamber)= 0.004m3 , T(of chamber)= -196oC

                                P = 55kPa = 55,000 Pa , R = 8.314JK-1mol-1

    To find out : (a) The gas temperature in units of kelvin :    -196oC +273 = 77K

                         (b) The number of moles of neon gas (n) in the chamber : Using gas equation PV=nRT

                putting the given values    55000Pa X 0.004m3 = n X 8.314JK-1mol-1 X 77K

                                                            n = 55 X 4 / 8.314 X 77

                                                                = 0.3436 moles

                          (c) The number of neon atoms in the chamber : since 1 mole = 6.023 X 1023 atoms

                                                               so 0.3436 moles= 0.3436 X 6.023 X 1023 = 2.07 X 1023 atoms

                           (d) The mass of gas in the chamber : since n = mass (g) / mass of 1 mole(g)

                                                         hence total mass = 0.3436 moles X 20.18g = 6.93g

3. if all the gas is transferred to a new chamber the n will be the same. so putting the rest of new values in gas equation : PV= nRT we get   613000Pa x V = 0.3436moles X 8.314JK-1mol-1 x 773K

                                                  V = 3.60 x 10-3 m3

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