Sam studies a mixture of helium and argon of one liter at 25 degrees celsius and

Question

Sam studies a mixture of helium and argon of one liter at 25 degrees celsius and 1 atm pressure in a cylinder. He brings the cylinder in contact with liquid nitrogen (66K) at which a small amount of a white substance deposits on the walls of the cylinder and the volume of the gas reduced to 220mL. The pressure remains constant. Neglecting the volume of the solid formed, what was the partial pressure of the helium at 25 degrees celsius? What is the white powder?

BP of helium is 4.2K and that of argon is 87.45K. Melting point of argon is 90.15K.

Explanation / Answer

From the given data,

Volume of gaseous mixture at 25 °C and at 1 atm = 1 litre.

Since the Melting point of argon is 90.15K, argon at 66K exists as a solid.

Thus, the white powder is solidified argon.

From the data, volume decreased by solidification of argon = 220 mL

Thus, the volume of argon at 25 °C and at 1 atm = 220 mL =0.22 L

           the volume of helium at 25 °C and at 1 atm = 780 mL = 0.78 L

By definition, partial pressure of helium = total pressure x mole fraction of helium gas.

Now, according to Avagadro’s relation, 1 mole of any gas occupy 22.4 litres of volume at 25°C, 1 atm.

Thus, number of moles of helium = (volume of helium in litres/ 22.4)x 1 = 0.03482 moles

                number of moles of argon = (volume of argon in litres/ 22.4)x 1 = 0.009821 moles

mole fraction of helium = n/ (n+N) = 0.03482/ (0.03482+0.009821) = 0.78

partial pressure of helium = 1 x 0.78 = 0.78 atm

Alternative method:

Number of moles of Helium, n = PV/RT = 1x0.78/0.0821x298 = 0.03188

Number of moles of Argon, n = PV/RT = 1x0.22/0.0821x298 = 0.008992

mole fraction of helium = n/ (n+N) = 0.03188/ (0.03188+0.008992) = 0.78

partial pressure helium = 1 x 0.78 = 0.78 atm

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