Question

The density of uncompressed air is 1.293 kg/m³, and one mole of air (containing Avogadro's number of molecules) has a mass of 28.94 g.

a) Use the density and the volume of the uncompressed air 45 cc to determine the mass of the air in the syringe, m_air, in grams. (You will have to do some unit conversions.)

m_air =  ______ g

b)Use the mass of one mole of air and Avogadro's number to determine N, the number of molecules in the syringe.

C) Now suppose the volume of the syringe is 38.8 cc. You measure the pressure as 1.28 atm, and the temperature as 19.9 Celsius. Assume: the diameter of the plunger on the syringe is 1.66 cm.

Find
- N, the number of molecules of gas in the syringe: __________ molecules

Answer 1

a) density = 1.293 kg/m³ = 1.293*10^-3 g/ml

Thus, mass of 45 cc of air = volume*density = 5.8185*10^-2 g

b) moles of air = mass/molar mass = 5.8185*10^-2/28.94 = 2.01*10^-3

Thus, number of air molecules = moles*Avagadro's Number = (2.01*10^-3)*(6.022*10²³) = 1.21*10²¹ molecules

c) Applying Ideal Gas Equation i.e. P*V = n*R*T we get

moles of air, n = (P*V)/(R*T) = (1.28*0.0388)/(0.0821*292.9) = 2.065*10^-3 moles

Thus, molecules of air in the syringe = moles*Avagadro's Number = 1.244*10²¹ molecules