## A Quick Introduction to Gases

These notes were written to help you quickly get to the point where you can work gas law problems on the quizzes and exams in this course.  They are not a detailed development of the topic.

The problems which will appear on exams will involve 4 "empirical" gas laws and one additional law known as the ideal gas law.

The empirical gas laws are as follows:

 Name of Gas Law Equation Parameters That Must Be Constant Boyle's Law P1V1 = P2V2 Temperature, Number of moles Charles' Law V1/ T1 = V2 / T2 Pressure, Number of moles Amonton's Law P1 / T1 = P2 / T2 Volume, Number of moles Combined Gas Law P1V1 / T1 = P2V2 / T2 Number of moles

The problems which you will encounter with these laws involve having a gas under an initial set of conditions (state 1) and then having the gas undergo some process which changes its state of existence to another set of conditions (state 2).  All the gas parameters will be specified except one, and the one missing can be solved for algebraically.  Generally, I ask for one of the state 2 parameters.

### Boyle's Law Sample Problems

#### Boyle's Law Example 1

A gas was confined in a cylinder fitted with a movable piston.  At 22.6 oC, the gas occupied a volume of 4.573 L under a pressure of 1.152 atm.  The gas was isothermally compressed, reducing its volume to 2.963 L.  What pressure was exerted by the comperssed gas?

#### Solution to Boyle's Law Example 1

We recognize that this is a Boyle's Law problem because the temperature remains constant (we are told the process is "isothermal", which means constant temperature).  Presumably, there is also no change in the number of moles of gas, since no amount of gas was ever mentioned in the problem.  Although the temperature is given, it is not needed, since it does not appear in Boyle's Law.

Before the process (compression), the gas volume is 4.573 L, so this is V1, and the gas pressure is 1.152 atm, so this is P1.  After the process (compression), the gas volume is 2.963 L, so this is V2.  The variable left unspecified is the pressure after the process, (P2) so this is the variable we must solve for.

The equation P1V1 = P2V2 can be solved for P2 to get

P2 = P1 ( V1 / V2 ) = 1.152 atm ( 4.573 L / 2.963 L ) = 1.778 atm     <=====   ANSWER

Notice that the "new" pressure is just the "old" pressure multiplied by a ratio of the two volumes.  A pattern like this exists in all problems involving the empirical gas laws.  It is possible to solve such problems by inspection, without actually having to set up the algebra, once you get used to it.  Notice in this case, for example, it makes sense that the larger volume is in the numerator.  Since the gas is being compressed, the final pressure should be larger than the initial one.  Therefore, the original pressure has to be multiplied by a number greater than one, to create a number larger than we started with.  The only other possible ratio we could write for the two volumes is ( 2.963 L / 4.573 L ), but this ratio would be less than one, and when multiplied by the origianl pressure, would yield a smaller pressure than we started with.  This would not make sense, in light of the fact that the gas is being compressed, thus we could have selected the proper ratio without ever setting up the algebra.

#### Boyle's Law Example 2

A gas was confined in a cylinder fitted with a movable piston.  At 19.6 oC, the gas occupied a volume of 7.336 L under a pressure of 3.196 atm.  The gas was isothermally compressed until its pressure reached 5.500 atm.  What volume was occupied by the compressed gas?

#### Solution to Boyle's Law Example 2

We recognize this as a Boyle's Law problem because the temperature remains constant ("isothermal") and there is apparently no change in the number of moles of gas, since moles (or amount of gas in any units, for that matter), is not mentioned in the problem.  The temperature is specified in the problem, but not needed, since temperature does not appear in Boyle's Law.

Reading through the problem, we can make the following variable assignments:

V1 = 7.336 L     P1 = 3.196 atm

V2 = ?                P2 = 5.500 atm

The Boyle's Law equation P1V1 = P2V2 can be solved for V2 to give

V2 = V1 ( P1 / P2 ) = 7.336 L ( 3.196 atm / 5.500 atm ) = 4.263 L    <=====   ANSWER

Notice that the "new" volume is just the "old" volume multiplied by a ratio of the pressures.  It makes sense that the smaller pressure is on top in the ratio.  This makes the ratio less than one, so when multiplied by the original volume, is gives us a smaller volume than we started with.  Since the gas is being compressed, it makes sense that the final volume should be smaller than the orignial.

#### Boyle's Law Example 3

A gas was confined in a cylinder fitted with a movable piston.  At 21.3 oC, the gas occupied a volume of 1.994 L under a pressure of 4.357 atm.  The pressure on the gas was then reduced to 3.264 atm, and the gas was allowed to expand isothermally to a new volume.  What was the volume of the expanded gas?

#### Solution to Boyle's Law Example 3

We recognize this as a Boyle's Law problem because the temperature does not change ("isothermal") and the number of moles apparently does not change.  The temperature has been given, but is not needed in the calculation.  Reading through the problem, we can make the following variable assignments:

V1 = 1.994 L     P1 = 4.357 atm

V2 = ?               P2 = 3.264 atm

The Boyle's Law equation, P1V1 = P2V2 can be solved for P2 to give

V2 = V1 ( P1 / P2 ) = 1.994 L ( 4.357atm / 3.264 atm ) = 2.662 L    <=====   ANSWER

It makes sense that the larger pressure ends up on top in the ratio.  The gas is expanding, so the final volume should be bigger than the initial one.  To obtain a larger volume than we started with, we must multiply by a number bigger than one.

#### Boyle's Law Example 4

A gas was confined in a cylinder fitted with a movable piston.  At 27.4 oC, the gas occupied a volume of 3.431 L under a pressure of 5.363 atm.  If you would like the gas volume to increase to 5.000 L, what pressure should you put on the gas?  Assume no change in temperature or number of moles of gas.

#### Solution to Boyle's Law Example 4

Lets see if we can solve this one without doing the algegra.  The unknown variable is a pressure.  The pressure we seek should be obtainable by multiplying the original pressure by a ratio of the volumes, since the pressure change is related to the volume change.  Recall that temperature and number of moles are constant in Boyle's Law problems.  We want the gas to expand, since we are trying to go from a volume of 3.431 L to a volume of 5.000 L.  In order to let the gas expand, the pressure on it must be reduced.  Thus, the final pressure should be smaller than the original.  This means the volume ratio should be constructed so as to be less than one -- that is, the smaller volume should be on top.  Therefore the calculation is

5.363 atm ( 3.431 L / 5.000 L ) = 3.680 atm    <=====  ANSWER

So, the pressure on the gas must be reduced from its original value to 5.363 atm down to a new, lower pressure of only 3.680 atm.  At this pressure, the gas volume will be 5.000 L, as desired.

#### Boyle's Law Example 5

A gas was confined in a cylinder fitted with a movable piston.  At 15.6 oC, the gas occupied a volume of 3.384 L under a pressure of 4.476 atm.  If you want to increase the gas pressure to 7.603 atm (at constant temperature), to what volume should the gas be compressed?  Assume no change in the number of moles of gas.

#### Solution to Boyle's Law Example 5

Lets again set up the calculation intuitively, without doing the algebra.  We want to increase the gas pressure, which means it must be compressed into a smaller volume.  The new volume can be calculated by multiplying the original volume by a ratio of the two pressures.  Since the new volume should be smaller than the original, the pressure ratio must be less than one.  This means the smaller pressure must be on top.  The calculation is therefore

3.384 L ( 4.476 atm / 7.603 atm ) = 1.992 L    <=====  ANSWER

Thus, the piston must be pushed in until the gas volume is reduced from its original value of 3.384 L to a new value of only 1.992 L.

### Charles' Law Sample Problems

#### Charles' Law Example 1

A gas was confined in a cylinder fitted with a movable piston.  At 22.6 oC, the gas occupied a volume of 3.254 L under a pressure of 1.200 atm.  The gas was isobarically heated to a temperature of 100.0 oC, causing it to expand.  What was the volume of the heated gas?

#### Solution to Charles' Law Example 1

We recognize this as a Charles' Law problem because the pressure does not change ("isobaric") and presumably, there is no change in the number of  moles of gas.  The pressure is given in the problem, but is not needed, because it does not appear in Charles' Law.  Reading through the problem, we can make the following variable assignments:

T1 = 22.6 oC = 295.75 K  (Note: Kelvin = Celsius + 273.15)
V1 = 3.254 L

T2 = 100.0 oC = 373.15 K
V2 = ?

The Charles' Law equation, V1 / T1 = V2 / T2 can be solved for V2 to give

V2 = V1 ( T2 / T1 ) = 3.254 L ( 373.15 K / 295.75 K ) = 4.106 L    <=====  ANSWER

Notice that a ratio pattern exists here, just as it did in the Boyle's Law problems.  The pressure and number of moles remain constant, so changes in volume are related only to changes in temperature.  The "new" volume is found by taking the "old" volume and multiplying it by the appropriate temperature ratio.  Since the gas is expanding, the final volume must be larger than the initial volume.  Therefore, the temperature ratio must be greater than one.  This means the larger temperature must be on top, as the algebra has shown us.  Always remember that temperature units MUST be Kelvin when doing gas calculations.

#### Charles' Law Example 2

A gas was confined in a cylinder fitted with a movable piston.  At 12.3 oC, the gas occupied a volume of 4.736 L under a pressure of 2.113 atm.  The gas was isobarically heated until its volume had increased to 5.000 L.  What was the temperature of the heated gas?  Experss your answer in oC.

#### Solution to Charles' Law Example 2

We recognize this as a Charles' Law problem because the pressure does not change ("isobaric") and presumably the number of moles of gas does not change.  The pressure is given in the problem, but is not needed, because it does not appear in Charles' Law.  Reading through the problem, we can make the following variable assignments:

T1 = 12.3 oC = 285.45 K   (Note: Kelvin = Celsius + 273.15)
V1 = 4.736 L

T2 = ?
V2 = 5.000 L

The Charles' Law equation, V1 / T1 = V2 / T2 can be solved for T2 to give

T2 = T1 ( V2 / V1 ) = 285.45 K ( 5.000 L / 4.736 L ) = 301.36 K = 28.2 o<===   ANSWER

Notice that the algebra dictates that we put the larger volume on top.  This makes sense, because it creates a ratio larger than one, so when multiplied by the original temperature, we get a higher temperature than we started with.  Notice also that even though our starting temperature was in oC and we wanted our final answer in oC, we still had to do the calculation in K.  We add 273.15 to the Celsius temperature to get Kelvin temperature, obtain the final temperature in Kelvin, and then subtract 273.15 to convert the temperature back to Celsius.

#### Charles' Law Example 3

A gas was confined in a cylinder fitted with a movable piston.  At 6.8 oC, the gas occupied a volume of 6.225 L under a pressure of 3.219 atm.  The gas was isobarically heated to a temperature of 110.7 oC.  What was the volume of the hot gas?

#### Solution to Charles' Law Example 3

This time, let's solve it intuitively, instead of going through the algebra.  Since the gas is being heated, it should expand, so the volume of the hot gas should be larger than the original volume.  This means the original volume needs to be multiplied by a ratio that is larger than one.  Since it is only the change in temperature that is changing the volume (pressure and number of moles remain constant), we must multiply the volume given in the problem by a temperature ratio with the larger temperature on top.  We must remember to add 273.15 to each temperature and thus enter them as Kelvin temperatures in the calculation.

6.225 L ( 383.85 K / 279.95 K ) = 8.535 L    <=====   ANSWER

Thus, when heated to 110.7 oC, the gas volume will be 8.535 L

#### Charles' Law Example 4

A gas was confined in a cylinder fitted with a movable piston.  At 15.4 oC, the gas occupied a volume of 2.573 L under a pressure of 1.646 atm.  To what temperature should the gas be heated to increase its volume ot 3.100 L?  Assume no change in pressure or number of moles of gas?

#### Solution to Charles' Law Example 4

Let's again solve the problem intuitively.  The change in temperature should be related to the change in volume, so we should be able to get the "new" temperature by multiplying the "old" temperature by the appropriate volume ratio.  Since the gas is being heated, the temperature we get for our answer needs to be higher than than the temperature we started with.  This means that in the volume ratio, the larger volume needs to be on top.

288.55 K ( 3.100 L / 2.573 L ) = 347.65 K = 74.5 oC    <=====   ANSWER

Notice that the calculation was done in Kelvin, even though the starting temperature was given in Celsius, and the final answer was to be expressed in Celsius.  The Kelvin answer is converted to Celsius by subtracting 273.15.

### Amonton's Law Sample Problems

#### Amonton's Law Example 1

A gas was contained in a rigid steel tank having a volume of 7.528 L.  At 21.4 oC, the gas exerted a pressure of 5.173 atm.  What pressure will the gas exert if it is heated to a temperature of 89.6 oC?

#### Solution to Amonton's Law Example 1

We recognize this as an Amonton's Law problem because the volume does not change and presumably, neither does the number of moles.  The problem does not explicitly say the volume does not change, but it refers to a rigid steel tank, and we can interpret that as meaning the volume is constant.  Moles are not mentioned, so we will assume they remain constant.  The volume is given in the problem, but is not needed, since it does not appear in Amonton's Law.

Reading through the problem, we can make the following variable assignments:

T1 = 21.4 oC = 294.55 K   (Note: Kelvin = Celsius + 273.15)
P1 = 5.173 atm

T2 = 89.6 oC = 362.75 K
P2 = ?

The Amonton's Law equation P1 / T1 = P2 / T2 can be solved for P2 to give

P2 = P1 ( T2 / T1 ) = 5.173 atm ( 362.75 K / 294.55 K ) = 6.371 atm    <=====   ANSWER

Notice that a ratio pattern again exists.  Since the volume and number of moles are not changing, the pressure change is related only to a temperature change.  The "new" pressure can be calculated by multiplying the "old" pressure by the appropriate temperature ratio.  Since the gas is being heated, the final pressure must be higher than the initial pressure.  This means the temperature ratio must be larger than one, so the larger temperature must be on top.  This is exactly what the algebra led us to do.  Notice that the calculation had to be done in Kelvin temperature units.

#### Amonton's Law Example 2

A gas confined in a 4.473 L steel tank exerted a pressure of 2.157 atm at 22.6 o.  In a warmer environment, the pressure inside the tank was found to be 2.306 atm.  What was the temperature in the warmer environment?

#### Solution to Amonton's Law Example 2

We recognize this as an Amonton's Law Problem because the volume does not change (we can assume a steel tank has a constant volume) and there is no mention of moles, so presumably there is no change in the number of moles of gas. The volume is given in the problem, but is not needed since it does not appear in Amonton's Law.  Reading through the problem, we can make the following variable assignments:

P1 = 2.157 atm
T1 = 22.6 oC = 295.75 K   (Note: Kelvin = Celsius + 273.15)

P2 = 2.306 atm
T2 = ?

The Amonton's Law equation P1 / T1 = P2 / T2 can be solved for T2 to give

T2 = T1 ( P2 / P1 ) = 295.75 K ( 2.306 atm / 2.157 atm ) = 316.18 K = 43.0 o<===   ANSWER

Notice that the algebra directs us to put the larger pressure on top in the pressure ratio.  This makes the ratio larger than one, which makes sense, because if the pressure of the gas has increased, as it has in this problem, the final temperature must be higher than the initial temperature.  As usual, gas calculations must be done in Kelvin, but we can convert our answer to Celsius if those are the units we want to express it in.

#### Amonton's Law Example 3

A sample of a gas having a volume of 3.470 L under a pressure of 2.507 atm at 21.1 oC was isochorically heated to a temperature of 62.8 oC.  What pressure will the gas exert under these conditions?

#### Solution to Amonton's Law Example 3

We recognize this as an Amonton's Law problem because the volume and number of moles are not changing.  Here, we are not told about a steel tank, but rather, that the gas is isochorically heated.  The term isochoric refers to constant volume.  No mention is made of moles, so we shall assume the number of moles remains constant.  Let's see if we can solve it intuitively this time.  If a gas is confined to a constant volume and heated, its pressure should increase.  The "new" pressure can be obtained by multiplying the "old" pressure by the appropriate temperature ratio.  Since the "new" pressure must be larger than the "old" pressure, the temperature ratio must be larger than one.  This means the larger temperature must be on top.  Of course, the temperatures need to be entered in Kelvin units.  Therefore, the calculation is

2.507 atm ( 335.95 K / 294.25 K ) = 2.862 atm

#### Amonton's Law Example 4

A steel tank at 17.3 oC contains a gas under a pressure of 4.973 atm.  For safety reasons, it is desired that the pressure not exceed 8.000 atm.  What is the maximum temperature (expressed in oC) to which the gas in the tank can be heated?  Assume no change in the number of moles of gas during the heating process.

#### Solution to Amonton's Law Example 4

Since this is a steel tank the volume can be assumed to be constant.  We are also told that we should assume no change in the number of moles.  The volume of the tank is not given in this problem, but we don't need it.  Let's again solve the problem intuitively, that is, without the algebra.  The highest pressure that can be tolerated is 8.000 atm, so we need to know at what temperature the pressure reaches this magnitude.  This temperature is clearly higher than the origianl temperature, because pressure increases with temperature.  So we must multiply the original temperature by a pressure ratio that is larger than one -- that is, it must have the higher pressure on top.  As usual, we must carry out the calculation in Kelvin units.  We can then convert our answer to Celcius for reporting it.

290.45 K ( 8.000 atm / 4.973 atm ) = 467.24 K = 194.1 oC     <=====   ANSWER

Thus, for this tank, the temperature should not be allowed to exceed 194.1 oC, because then the pressure would exceed the maximum safe working pressure of 8.000 atm.

### Combined Gas Law Sample Problems

#### Combined Gas Law Example 1

A gas was confined in a cylinder fitted with a movable piston.  At 12.8 oC, the gas occupied a volume of 8.175 L under a pressure of 1.824 atm.  The gas was simultaneously heated and compressed, so that its volume was 6.713 L and its temperature was 72.8 oC.  What pressure was exerted by the hot compressed gas?

#### Solution to Combined Gas Law Example 1

Notice that none of the individual laws (Boyle's, Charles' or Amonton's) can be used to solve this problem, because pressure, volume, and temperature of the gas are all changing.  The only thing that remains constant is the number of moles, since that was not mentioned in the problem.  Reading through the problem, we can make the following assignments:

T1 = 12.8 oC = 285.95 K   (Note: Kelvin = Celsius + 273.15)
V1 = 8.175 L
P1 = 1.824 atm

T2 = 72.8 oC = 345.95 K
V2 = 6.713 L
P2 = ?

The combined gas law equation P1V1 / T1 = P2V2 / T2 can be solved for P2 to give

P2 = P1 ( V1 / V2 ) ( T2 / T1 ) = 1.824 atm ( 8.175 L / 6.713 L ) ( 345.95 K / 285.95 K )

= 2.687 atm     <=====   ANSWER

The ratio patterns exist here too, but there are two ratios to consider instead of only one.  Since pressure, volume, and temperature are all changing, the change in pressure is affected both by volume changes and temperature changes.  We can analyze each of these effects separately to deduce where the variables should be placed in the ratios.  In this problem, the gas has been compressed into a smaller volume.  The effect of reducing the gas volume is to increase its pressure, so the volume ratio must be larger than one.  Thus the larger volume should be on top, as the algebra has directed us to do.  Looking at the temperatures, heating a gas should increase its pressure, so the temperature ratio should also be larger than one.  Again, the algebra has directed us to put the larger temperature on top, as we would expect.

#### Combined Gas Law Example 2

A gas was confined in a cylinder fitted with a movable piston.  At 18.4 oC, the gas occupied a volume of 6.312 L and exerted a pressure of 2.417 atm.  The gas was simultaneously heated and compressed, so that its pressure was 4.161 atm and its temperature was 59.7 oC.  What volume was occupied by the hot compressed gas.

#### Solution to Combined Gas Law Example 2

Again, only the combined gas law can be used to solve the problem, because volume, pressure and temperature are all changing.  Only the number of moles remains constant.  Reading through the problem, we can make the following variable assignments:

T1 = 18.4 oC = 291.55 K   (Note: Kelvin = Celsius + 273.15)
V1 = 6.312 L
P1 = 2.417 atm

T2 = 59.7 oC = 332.85 K
V2 = ?
P2 = 4.161 atm

The combined gas law equation P1V1 / T1 = P2V2 / T2 can be solved for V2 to give

V2 = V1 ( P1 / P2 ) ( T2 / T1 ) = 6.312 L ( 2.417 atm / 4.161 atm ) ( 332.85 K / 291.55 K )

= 4.186 L     <=====   ANSWER

The change in volume is affected both by a change in pressure and by a change in temperature.  We can consider each of these effects separately.  The pressure on the gas is being increased.  This factor alone would tend to shrink the gas, giving it a smaller volume.  Therefore, the pressure ratio must be less than one.  Looking at the calculation above, we see that that's exactly what the algebra has led us to do.  We had to put the smaller pressure on top.  The gas is also being heated.  This factor alone would tend to make the gas expand, giving it a larger volume.  Therefore the temperature ratio must be larger than one.  Indeed, the algebra has led us to put the larger temperature on top, giving a temperature ratio larger than one.  The pressure change and temperature change work against each other in this problem, but the effect of increasing pressure "wins", because the final volume of the gas is less than the initial volume.  That is, the direction of volume change is that predicted on the basis of the pressure change.

#### Combined Gas Law Example 3

A gas was confined in a cylinder fitted with a movable piston.  At 17.3 oC, the gas occupied a volume of 7.515 L under a pressure of 1.406 atm.  The gas was simultaneously heated and compressed, so that its volume was 5.221 L and its pressure was 2.515 atm.  To what temperature (expressed in oC) was the gas heated?

#### Solution to Combined Gas Law Example 3

Temperature, pressure and volume are all changing.  Only the number of moles of gas remains constant.  Therefore, only the combined gas law can be used to solve this problem. Reading through the text of the problem, we can make the following variable assignments:

T1 = 17.3 oC = 290.45 K   (Note: Kelvin = Celsius + 273.15)
V1 = 7.515 L
P1 = 1.406 atm

T2 = ?
V2 = 5.221 L
P2 = 2.515 atm

The combined gas law equation, P1V1 / T1 = P2V2 / T2 can be solved for T2 to give

T2 = T1 ( P2 / P1 ) ( V2 / V1 ) = 290.45 K ( 2.515 atm / 1.406 atm ) ( 5.221 L / 7.515 )

=  360.95 K = 87.8 oC     <=====   ANSWER

The "new" temperature is obtained by multiplying the "old" temperature by the appropriate ratios.  Let's look at these ratios and see if they make sense.  The pressure of the gas increases in this problem.  Ignoring the volume change for the moment, the increasing pressure implies increasing temperature.  On this basis, the "new" temperature must be higher than the "old" temperature, so the pressure ratio must be larger than one.  Looking at the algebraic solution, we see that it is.  Ignoring the pressure changes and looking at the voume changes, we see that the volume is decreasing in this problem.  If we pretend pressure is not changing, then the only way to shrink the gas is to cool it.  Thus, the decreasing volume implies cooling of the gas, or a final temperature that is lower than the initial temperature.  With this in mind, the volume ratio must be smaller than one.  Looking at the algebra, we see that it is.

Notice how we are able to isolate the effects of each variable and deal with them one at a time.  In the above example, we temporarily pretended we were cooling the gas, even though we knew we were in fact, heating it.  In this problem, the increasing pressure on the gas (which tends to shrink it) was overcoming the effects of temperature, which tends to expand it.  It was because of the increased pressure on the gas that it became smaller, in spite of being heated.  However, in analyzing the requirement for the volume ratio, we ignore the pressure change, and determine how the temperature would have to change to make the volume get smaller IF there was no pressure change.  Since temperature would have to become smaller, we construct a volume ratio that would tend to give us a smaller temperature than we started with.  This requires a volume ratio less than one.

#### Combined Gas Law Example 4

A gas was confined in a cylinder fitted with a movable piston.  At 101.4 oC, the gas occupied a volume of 9.926 L under a pressure of 2.487 atm.  The gas was cooled to a temperature of 23.8 oC, and compressed to a volume of 5.134 L.  What pressure did the gas exert under these conditions?

#### Solution to Gas Law Example 4

Let's try to solve this one intuitively.  The "new" pressure should be equal to the "old" pressure multiplied by the appropriate ratios.  The parameters that are causing the pressure to change are the volume change and the temperature change, so we must multiply by the appropriate ratios of these quantities.  The volume of the gas is getting smaller.  Forcing the gas into a smaller volume (ignoring the temperature change) should cause the pressure to increase.  Thus the volume ratio needs to be larger than one.  We should put the larger volume on top.  The gas is being cooled.  This fact alone (ignoring the volume change) should reduce the pressure of the gas.  So the temperature ratio should be less than one.  We should put the smaller temperature on top.  Therefore, our calculation is

2.487 atm ( 9.926 L / 5.134 L ) ( 296.95 K / 374.55 K ) = 3.812 atm   <=====   ANSWER

The reduction in volume and the cooling work against each other.  The reduction in volume has a greater effect than the cooling, because the pressure increased, as would be predicted for a compression.  The cooling offsets some of the pressure increase, however.  The pressure would have been higher if we had the same volume reduction without the simultaneous cooling.  Try it for yourself and see.  Do a Boyle's Law calculation for the isothermal compression of a gas from 9.926 L to 5.134 L, using 2.484 atm as the original pressure.  If done correctly, you should obtain 4.808 atm as the final pressure, as compared to only 3.812 atm when the gas is cooled as described in the above problem.

### Ideal Gas Law Sample Problems

#### Ideal Gas Law Example 1

What pressure is exerted by 9.768 g of CO2 in a 3.774 L vessel at 26.5 oC?

#### Solution to Ideal Gas Law Example 1

In this problem, unlike the ones worked so far, we don't have a "before and after" type problem.  We have a single state of the gas, and all of the parameters have been specified but one.  Notice also, that a specific gas has been given (CO2), rather than leaving the gas unspecified, as has been done in the empirical gas laws.  When a specific gas is given, and/or you have only a single state, this is a good indication that the problem calls for the Ideal Gas Law.

The equation of the Ideal Gas Law is PV=nRT, where R = 0.08205783 L atm / K mol.  We can solve this equation for the pressure to get P = nRT / V.  However, we can't use this just yet, because we don't know the number of moles of CO2.  We do, however, know the number of grams, and these can be converted to moles.

9.768 g CO2  ( 1 mol CO2 / 44.0098 g CO2 ) = 0.221950566 mol CO2

Now we can calculate the pressure:

P = nRT / V = (0.221950566 mol ) (0.08205783 L atm / K mol ) ( 299.65 K ) / 3.774 L

P = 1.446 atm     <=====   ANSWER

#### Ideal Gas Law Example 2

A steel tank contains 14.73 g of Cl2 gas under a pressure of 4.825 atm at 22.7 oC.  What is the volume of the tank?

#### Solution to Ideal Gas Law Example 2

This is a one-state problem involving a specific gas; the Ideal Gas Law should be used to solve it.  The equation of the Ideal Gas Law is PV=nRT.  Solving this for volume gives V= nRT / P.  However, we need to know the number of moles of Cl2 to solve it.  We can get this from the mass of Cl2.

14.73 g Cl2  ( 1 mol Cl2 / 70.9054 g Cl2 ) = 0.207741582 mol Cl2

Now we can calculate the volume:

V = nRT / P = ( 0.207741582 mol ) ( 0.08205783 L atm / K mol ) ( 295.85 K ) / 4.825 atm

V = 1.045 L     <=====   ANSWER

#### Ideal Gas Law Example 3

A steel reaction vessel having a volume of 8.761 L contains 4.822 g of H2 gas under a pressure of 7.016 atm.  What is the temperature of the H2 gas?

#### Solution to Ideal Gas Law Example 3

We have a specified gas and only one state (no "before and after") so we must use the Ideal Gas Law to solve it.  For the Ideal Gas Law, we must know the number of moles of gas:

4.822 g H2  ( 1 mol H2 / 2.01588 g H2 ) = 2.392007461 mol H2

Then, solving the Ideal Gas Law, PV=nRT for the temperature gives

T = PV / nR = ( 7.016 atm ) (8.761 L ) / ( 2.392007461 mol ) ( 0.08205783 L atm / K mol )

T = 313.16 K = 40.0 oC     <=====   ANSWER