Tutorial – Combining the Gas Laws and the Universal Gas Law

(Ideal Gas Law)    (Std 4c & 4h)


Read through this tutorial very carefully.  It has almost everything that needs to be done in the most difficult of problems and goes through the logic of why each step needs to be done in detail.  I hope this helps !!!!


Up until Ch 12.5, you used separate laws to do each type of problem as follows:


            Boyle’s              P1V1 = P2V2       meaning   Volume varies inversely with Pressure


                                       V1        V2

            Charles’ Law     ------  =  ------     meaning  Volume varies directly with Temperature

                                      T1         T2


            Avogadro’s Law


                                     V1        V2

                                    ------  =  ------      meaning  Volume varies directly with n (number of moles)

                                     n1         n2


            If we combine Boyle’s Law, Charles’ Law and Avogadro’s Law formulas, we come up with a Combined Gas Law that can be used to solve any one of the three !!!  Here’s what it looks like:


                                     P1V1      P2V2

                                    ------  =  ------     

                                    T1n1       T2n2


            Now lets see how to do the examples I gave you before (Boyle & Charles) using this new tool:


              Example (Boyle’s Law):

                        If you have P1 = 235 kPa  and  P2 = 9,500 Torr and  V1 = 925 ml, what will  V2  be?


You must first change  235 kPa to Torr  ( P1 ) so that P1 and  P2 are in the same

units of measure and we have to change kPa to Pa before we can change to atm.

    .                   1000 Pa = 1 kPa        and       101,325 Pa = 1 atm         and         1 atm = 760 Torr

So, converting would go like this:


                        235 kPa         1000 Pa             1 atm            760 Torr

                        ---------   x   ----------   x  -------------  x  ----------     =   1763 Torr  =  P1

 1                   kPa           101,325 Pa          1 atm


                        Now we have everything we need to use the new formula to solve this Boyle’s Law

                        problem:                                P1V1            P2V2

                                                                    ----------   =  ----------     

                                                                      T1n1             T2n2

First we look for the variables that do not change.  In this case temperature (T) and moles of gas (n)does do not change.  So we remove them from the formula and this is what’s  left:


                             P1  V1      =        P2 V2


                        1,763 Torr    x   925 ml    =      9,500 Torr    x    V2


                        Divide both sides by 1 Torr to get rid of that unit of measure and also divide both

sides by 9,500 to get that number to the left side of the equation.  Now the equation

looks like this:


1,763 x 925 ml

-----------------   =     V2           Answer      V2   =   171.7 ml



            Compare this logic to what we did before in the Boyle’s Law tutorial.


         Example (Charles Law):

                        We start with temperature at 25 °C and an end with a temperature of 58 °C.  The

                        volume starts at 75 ml.  What is the final volume?



                        T1 =  25 °C        T2 =  58 °C       V1 = 75 ml      and    V2 =  unknown


                        We cannot use Charles Law until the temperatures at changed to Kelvin.  So, be sure

                        you know how to use p. 44 of the textbook.  It tells you how to convert any

                        temperature unit of measure to any other temperature unit of measure.  Using the first

                         formula in the chart, we can do the following conversions before we use Charles Law:


                        25 °C  +   273 °C  =  298 °K = T1

                        58 °C  +   273 °C  =  331 °K = T2


                        We use the Combined Gas Law as follows:

                                     P1V1      P2V2

                                    ------  =  ------     

                                    T1n1       T2n2

                        What are the two variables that do not change?   The first is number of moles (n).  So

                        we remove n from the formula, which leaves:

                                     P1V1      P2V2

                                    ------  =  ------     

                                      T1          T2

The other variable that does not change is pressure.  Taking that out leaves the familiar Charles Law formula:

                                       V1         V2

                                    ------  =  ------     

                                      T1          T2


                        Now we can proceed as we did in the last tutorial on Charles Law, plug in the numbers:


                                      75 ml             T2

                                    --------   =   ---------

                                    298 °K         331 °K


                        The unit of measure  °K  crosses out on both side of the = sign.  The unit of measure

                        left is ml.    To finish off the math, multiply 75 ml x 331 and then divide your answer

                        by 298:                         Answer        83.3 ml  =  T2



 Now let’s look the the new type of problem, where only one variable is unknown, and no other variables change, we neec the Universal Gas Law:      


                                    PV = nRT


                           0.08206 atm L         

            R =   ------------------------

                            Moles °K



            What mass of helium gas is needed to pressurize a 100.0 L tank to 255 atm at 25 °C ?



            First, we know that we can use PV = nRT because we are being asked to find one out of the 4 variables – n (moles of Helium).  Note they didn’t ask for moles, but grams.  That means we have to find moles first and then convert to grams.  If you are having trouble with that type of conversion, see my tutorial:  

                        Converting grams to moles to atoms or PDF   Std 3


            What are we given?

                        V = 100.0 L – which is already in the correct unit of measure

                        P = 255 atm – which is already in the correct unit of measure

                        T = 25 °C  - which must be converted to Kelvin

                                                25 + 273 = 298 °K


            Now we can put the numbers into the formula:

                        (225 atm)(100.0 L)  =  n (0.08206)(298 °K)


                        (225 atm)(100.0 L)                                  22,500

                        --------------------  =  n                 n  =  --------    =   920 mol He

                        (0.08206)(298 °K)                                 24.45


            The problem wants to know the mass of the He.   So, we must convert to grams as follows:

                        1 mol He = 4.0 grams


                        920 mol He        4.0 grams

                        ------------  x  ------------ =  3,680 grams                 This is the answer you

                               1               1 mol He                                             were asked to find.