Ideal Gas Law Problems

1)                  How many molecules are there in 985 mL of nitrogen at 0.0° C and    1.00 x 10^-6 mm Hg?

2)                  Calculate the mass of 15.0 L of NH₃ at 27° C and 900. mm Hg.

3)                  An empty flask has a mass of 47.392 g and 47.816 g when filled with acetone vapor at 100.° C and 745 mm Hg.  If the volume of the flask is 247.3 mL, what is the molar mass of the acetone?

4)                  Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg.

5)                  6.3 mg of a boron hydride is contained in a flask of 385 mL at 25.0° C and a             pressure of 11 torr.

(a)               Determine the molar mass of the hydride.

(b)              Which of the following hydrides is contained in the flask, BH₃, B₂H₆,             or B₄H₁₀?

6)                  A volume of 26.5 mL of nitrogen gas was collected in a tube at a temperature           of 17° C and a pressure of 737 mm Hg.  The next day the volume of the        nitrogen was 27.1 mL with the barometer still reading 737 mm Hg.  What was the temperature on the second day?

Solutions

1)                  P = 1.00 x 10^-6 mm Hg                     T = 0.0° C + 273 = 273 K

V  = 985 mL                            R = 0.0821 L·atm/mol·K

            PV = nRT        

            n = PV/RT       

            n = 1.00 x 10^-6 mm x 1 atm/760 mm x 985 mL x 1 L/10³ mL/

      (0.0821 L·atm/mol·K x 273 K) = 5.78 x 10^-11 moles N₂

n_molecules = 5.78 x 10^-11 moles N₂ x 6.02 x 10²³ N₂ molecules/1 mol N₂ 

                              = 3.48 x 10¹³ N₂ molecules  

2)                  P = 900. mm Hg                                T = 27° C + 273 = 300 K

V  = 15.0 L                                          R = 0.0821 L·atm/mol·K

            PV = nRT        

            n = PV/RT       

            n = 900. mm x 1 atm/760 mm x 15.0 L/(0.0821 L·atm/mol·K x 300 K) =

n = 0.721 moles NH₃ x 17.04 g NH₃/1 mol NH₃ = 12.3 g NH₃ 

3)                  P = 745 mm Hg                                 T = 100.° C + 273 = 373 K

V  = 247.3 mL                                    R = 0.0821 L·atm/mol·K

            m_vapor = 47.392 g – 47.816 g = 0.424 g

            PV = nRT        

             n = m/MM

            PV = mRT/MM

            MM = mRT/PV          

            MM = 0.424 g x 0.0821 L·atm/mol·K x 373 K/(745 mm x 1 atm/760 mm 

                        x 247.3 mL x 1 L/10³ mL) = 53.6 g/mol

4)                  P = 671 mm Hg                                 T = 47° C + 273 = 320. K

V  = 478 mL                            R = 0.0821 L·atm/mol·K

            PV = nRT        

             n = m/MM

            D = m/V = P x MM/R x T

            D = 671 mm x 1 atm/760 mm x 83.80 g/mol/(0.0821 L·atm/mol·K x 320. K)

    = 2.82 g/L

5)                  P = 11 torr                                         T = 25.0° C + 273 = 298 K

V  = 385 mL                            R = 0.0821 L·atm/mol·K

m  = 6.3 mg

            PV = nRT        

n    = m/MM

            PV = mRT/MM

            MM = 6.3 mg x 1 g/10³ mg x 0.0821 L·atm/mol·K x 298 K)/

                         (11 torr x 1 mm/1 torr x 1 atm/760 mm x 385 mL x 1 L/10³ mL)

        = 27.7 g/mol

B₂H₆ because its molar mass is 27.7 g.

6)                  P₁ = 737 mm Hg                                P₂ = 737 mm Hg

            V₁ = 26.5 mL                                      V₂ = 27.1 mL

            T₁ = 17° C + 273 = 290. K                 T₂ = ?

            P₁V₁ = nRT₁ 

            P₂V₂ = nRT₂ 

P₁V₁/ P₂V₂ = nRT₁/ nRT₂

V₁/V₂ = T₁/T₂   (Charles’s Law)

T₂ = V₂/V₁ x T₁ 

T₂ = 27.1 mL/26.5 mL x 290. K = 297 K = 24° C