Chemistry 11 Practice Quiz for Gas Laws Solutions

 

 

1)      We placed the sealed bag into room temp water and found the volume.  When placed in ice water we noticed that the volume was less.  When we placed it in progressively hotter water the volume grew and grew.

2)      This is a Dalton’s Law of partial pressures question.  According to Dalton’s Law the pressure of a gas mixture is made up by adding the partial pressures due to each gas.  The partial pressure of each gas is proportional to the number of a particles of each gas.  Since there are equal portions of the three gases in this question they will each make up one third of the total pressure.  Since the total pressure is  2.4 atm, each gas must have 0.80 atm partial pressure.  Bonus: By doubling the amount of helium the mixture will be one half helium and one quarter each of nitrogen and oxygen.  So the helium partial pressure will be one half of 2.4 atm or 1.2 atm.  The oxygen and nitrogen will each have partial pressure of 0.60 atm.

3)      STP is 0°C and 1.0 atm.  Volume is given as 1.5 L.  All the units must be changed to those of R (see R and conversions, below).

 

                                                0°C = 273 K

1.0 atm. = 1.01×105 Pa

1.5 L = 0.0015 m3

 

            The question asks for n so rearrange PV=nRT to solve for n:

                                                 n = PV/RT

 

Now put the data in correct units into the equation:

 

                                                n = 1.01×105 Pa×0.0015 m3/8.31 Pa·m3/mol·K×273 K

                                                   =  6.7×10-2 moles

 

4)      T1 =  20ºC  = 293 K, P1 = 1.6 atm., T2 =  100ºC = 373 K and we are being asked to find P2.      V and n will be constant here so use no subscripts:

 

P1V =  nRT1

P2V =  nRT2

 

So

 

P1/P2 =T1 /T2

 

 Solving for P2:

P2 = P1T2 /T1

 

                                                            = 1.6 atm.× 373K/293K

= 2.0 atm.

 

 

 

5)      Do the following conversions:

a)      650 Torr to Pa

650Torr[1.01×105Pa/760Torr] = 8.64×104Pa

b)      2.26 × 1024units to moles

2.26 × 1024units[1 mole/6.02× 1023units] = 3.75 moles

c)      95 mL to m3 = 0.095 m3

95mL[1L/1000mL] = 0.095L

0.095L[1m3 /1000L] = 0.000095m3 = 9.5×10-5 m3

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