GTF 101: Solution for question that i asked you to try during lecture on gases 2

1. The oxygen gas generated by decomposition of potassium chlorate is collected.  The volume of oxygen collected at 24ºC and at atmospheric pressure of 762 mmHg is 128 mL.  The pressure of water vapor at 24ºC is 22.4 mmHg.  Calculate the mass (in grams) of oxygen gas obtained.

Solution

From the Dalton’s law we know that: P _total = P O₂ + P H₂O; therefore, P O₂ = P _total - P H₂O

= 762 mmHg -22.4 mmHg

= 740 mmHg.

From the ideal equation we write PV = NRT = (m/ M) RT, where m and M are the mass of O₂ and molar mass of O₂, respectively.

By rearranging the equation we can derive to:  m = (PVM) / (RT)

= [(740 /760) atm (0.128 L) (32.00 g/mol)] / [(0.0821 L.atm/K.mol) (24+273.15 K)

= 0.164 g      

2.  Given that 3.50 moles of NH3 occupy 5.20 L at 47ºC, calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the Van der Waals equation

Solutions

(a)    (a) Using the ideal gas equation

We have the following data:  V = 5.20 L; T = (47+273.15 K) = 320.15K; n = 3.50 mol;

R = 0.0821 L.atm/K.mol

P = nRT / V, hence, P = [(3.50 mol) (0.0821 L.atm/K.mol) (320 K)] / 5.20 L

   = 17.7 atm

  

(b)   (b)  Using the van der Waals equation

Van der Waals constants for a and b for NH₃ are 4.17 atm.L²/ mol² and 0.0371 L / mol, respectively.

From the Van der Waals equation calculates:

  (an2 / V2) = (4.17 atm.L²/ mol²) (3.50 mol) ² / (5.20 L) ² = 1.89 atm

  (nb)  = (3.50 mol) (0.0371 L/mol) = 0.130 L

Substitute all these values in the van der Waals equation:

(P + 1.89 atm)  (5.20 L – 0.130 L) = (3.50 mol) (0.0821 L.atm/K.mol) (320.15 K)

P = 16.2 atm