5 Steps to a 5 AP Chemistry

and pressure have changed, so this is really a Gay-Lussac’s law problem. From Gay-Lussac’s

law you know that if you increase the temperature, the pressure should increase if the amount

and volume are constant. This means that when you calculate the new pressure, it should be

greater than 2.50 atm; if it is less, you’ve made an error. Also, remember that the tempera-

tures must be expressed in kelvin. 20°C =293 K (K =°C +273) and 80°C =353 K.

We will be solving for P 2 , so we will take the combined gas law and rearrange for P 2 :

(T 2 P 1 V 1 )/(T 1 V 2 ) =P 2

Substituting in the values:

(353 K)(2.50 atm)(5.0 L)/(293 K)(5.0 L) =P 2

3.0 atm =P 2

The new pressure is greater than the original pressure, making the answer a reasonable

one. Note that all the units canceled except atm, which is the unit that you wanted.

Let’s look at a situation in which two conditions change. Suppose a balloon has a

volume at sea level of 10.0 L at 760.0 torr and 20°C (293 K). The balloon is released and

rises to an altitude where the pressure is 450.0 torr and the temperature is − 10 °C (263 K).

You want to calculate the new volume of the balloon. You know that you have to express

the temperature in K in the calculations. It is perfectly fine to leave the pressures in torr. It

really doesn’t matter what pressure and volume units you use, as long as they are consistent

in the problem. The pressure is decreasing, so that should cause the volume to increase

(Boyle’s law). The temperature is decreasing, so that should cause the volume to decrease

(Charles’s law). Here you have two competing factors, so it is difficult to predict the end

result. You’ll simply have to do the calculations and see.

Using the combined gas equation, solve for the new volume (V 2 ):

(P 1 V 1 )/T 1 =(P 2 V 2 )/T 2

(P 1 V 1 T 2 )/(P 2 T 1 ) =V 2

Now substitute the known quantities into the equation. (You could substitute the

knowns into the combined gas equation first, and then solve for the volume. Do it

whichever way is easier for you.)

(760.0 torr)(10.0 L)(263 K)/(450.0 torr)(293 K) =V 2

15.2 L =V 2

Note that the units canceled, leaving the desired volume unit of liters. Overall, the

volume did increase, so in this case the pressure decrease had a greater effect than the tem-

perature decrease. This seems reasonable, looking at the numbers. There is a relatively small

change in the Kelvin temperature (293 K versus 263 K) compared to a much larger change

in the pressure (760.0 torr versus 450.0 torr).

Volume–Amount Relationship: Avogadro’s Law

In all the gas law problems so far, the amount of gas has been constant. But what if the

amount changes? That is where Avogadro’s law comes into play.

If a container is kept at constant pressure and temperature, and you increase the number

of gas particles in that container, the volume will have to increase in order to keep the pres-

sure constant. This means that there is a direct relationship between the volume and the

number of moles of gas (n). This is Avogadro’s lawand mathematically it looks like this:

V/n=ka or V 1 /n 1 =V 2 /n 2

108  Step 4. Review the Knowledge You Need to Score High

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