Library Guides: Chemistry Textbook: Gas Laws for Ideal Gas

Gas Laws for Ideal Gases

By the end of this section, you will be able to:

Identify the mathematical relationships between the various properties of gases
Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

Activity (Interactive Simulation): Exploring the relationships between gas properties

In the simulation below:
1. Click on "Ideal". Pump gas particles into the chamber by click-hold-and-dragging the handle of the pump four times. Does the pressure increase or decrease when the number of gas particles are increased?

2. Click on "Temperature (T)" under "Hold Constant" on the right. This will hold the temperature constant. Decrease the volume of the container (click-hold-and-drag the container handle on the left). Does the pressure increase or decrease when the volume of the container is decreased?

3. Click on "Pressure ↕V" under "Hold Constant" on the right. This will hold the pressure constant and allow the volume to vary. Heat the container by sliding the blue button underneath the container to the "Heat" setting. Does the gas volume increase or decrease when the gas is heated?

4. Pump more particles into the container (note that the pressure is still held constant; temperature is also kept constant as you are not heating or cooling). Does the volume increase or decrease when number of particles are increased?
(Answers are provided at the bottom of this webpage)

Pressure and Temperature: Amontons’s Law or Gay-Lussac’s Law

Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 6.2) and the pressure increases.

This figure includes three similar diagrams. In the first diagram to the left, a rigid spherical container of a gas to which a pressure gauge is attached at the top is placed in a large beaker of water, indicated in light blue, atop a hot plate. The needle on the pressure gauge points to the far left on the gauge. The diagram is labeled “low P” above and “hot plate off” below. The second similar diagram also has the rigid spherical container of gas placed in a large beaker from which light blue wavy line segments extend from the top of the liquid in the beaker. The beaker is situated on top of a slightly reddened circular area. The needle on the pressure gauge points straight up, or to the middle on the gauge. The diagram is labeled “medium P” above and “hot plate on medium” below. The third diagram also has the rigid spherical container of gas placed in a large beaker in which bubbles appear near the liquid surface and several wavy light blue line segments extend from the surface out of the beaker. The beaker is situated on top of a bright red circular area. The needle on the pressure gauge points to the far right on the gauge. The diagram is labeled “high P” above and “hot plate on high” below. — Figure 6.2 The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.

This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 6.3. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

This figure includes a table and a graph. The table has 3 columns and 7 rows. The first row is a header, which labels the columns “Temperature, degrees C,” “Temperature, K,” and “Pressure, kPa.” The first column contains the following values from top to bottom: negative 100, negative 50, 0, 50, 100, and 150. The second column contains the values, from top to bottom, 173, 223, 273, 323, 373, and 423. The third column contains the values 36.0, 46.4, 56.7, 67.1, 77.5, and 88.0. A graph appears to the right of the table. The horizontal axis is labeled “Temperature ( K ).” with markings and labels provided for multiples of 100 beginning at 0 and ending at 500. The vertical axis is labeled “Pressure ( kPa )” with markings and labels provided for multiples of 10, beginning at 0 and ending at 100. Six data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. A dashed line extends from the data point furthest to the left to the origin. The graph shows a positive linear trend. — Figure 6.3 For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at –273 °C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.

Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P-T relationship for gases is known as either Amontons’s law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:

P \propto T or P = constant \times T or P = k \times T

where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio $\frac{P}{T}$ is therefore constant (i.e., $\frac{P}{T} = k$ ). If the gas is initially in “Condition 1” (with P = P₁ and T = T₁), and then changes to “Condition 2” (with P = P₂ and T = T₂), we have that $\frac{P_{1}}{T_{1}} = k$ and $\frac{P_{2}}{T_{2}} = k,$ which reduces to $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} .$ This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)

EXAMPLE 6.3

Predicting Change in Pressure with Temperature

A can of hair spray is used until it is empty except for the propellant, isobutane gas.

(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?

(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

Solution: (a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)

(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P₁ and T₁ as the initial values, T₂ as the temperature where the pressure is unknown and P₂ as the unknown pressure, and converting °C to K, we have:

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} which means that \frac{360 kPa}{297 K} = \frac{P_{2}}{323 K}

Rearranging and solving gives: $P_{2} = \frac{360 kPa \times 323 K}{297 K} = 390 kPa$

Check Your Learning

A sample of nitrogen, N₂, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?

Answer: 400 torr

Volume and Temperature: Charles’s Law

If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.

LINK TO LEARNING

This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.

These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 6.4.

This figure includes a table and a graph. The table has 3 columns and 6 rows. The first row is a header, which labels the columns “Temperature, degrees C,” “Temperature, K,” and “Pressure, k P a.” The first column contains the values from top to bottom negative 100, negative 50, 0, 100, and 200. The second column contains the values from top to bottom 173, 223, 273, 373, and 473. The third column contains the values 14.10, 18.26, 22.40, 30.65, and 38.88. A graph appears to the right of the table. The horizontal axis is labeled “Temperature ( K ).” with markings and labels provided for multiples of 100 beginning at 0 and ending at 300. The vertical axis is labeled “Volume ( L )” with marking and labels provided for multiples of 10, beginning at 0 and ending at 30. Five data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. The graph shows a positive linear trend. — Figure 6.4 The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero.

The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’s law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.

Mathematically, this can be written as:

V α T or V = constant · T or V = k · T or V_{1} / T_{1} = V_{2} / T_{2}

with k being a proportionality constant that depends on the amount and pressure of the gas.

For a confined, constant pressure gas sample, $\frac{V}{T}$ is constant (i.e., the ratio = k), and as seen with the P-T relationship, this leads to another form of Charles’s law: $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} .$

EXAMPLE 6.4

Predicting Change in Volume with Temperature

A sample of carbon dioxide, CO₂, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?

Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking V₁ and T₁ as the initial values, T₂ as the temperature at which the volume is unknown and V₂ as the unknown volume, and converting °C into K we have:

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} which means that \frac{0.300 L}{283 K} = \frac{V_{2}}{303 K}

Rearranging and solving gives: $V_{2} = \frac{0.300 L \times 303 K}{283 K} = 0.321 L$

This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).

Check Your Learning:

A sample of oxygen, O₂, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?

Answer: 21.6 mL

EXAMPLE 6.5

Measuring Temperature with a Volume Change

Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm³ when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm³. Find the temperature of boiling ammonia on the kelvin and Celsius scales.

Solution: A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V₁ and T₁ as the initial values, T₂ as the temperature at which the volume is unknown and V₂ as the unknown volume, and converting °C into K we have:

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} which means that \frac{150.0 {cm}^{3}}{273.15 K} = \frac{131.7 {cm}^{3}}{T_{2}}

Rearrangement gives $T_{2} = \frac{131.7 {cm}^{3} \times 273.15 K}{150.0 {cm}^{3}} = 239.8 K$

Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.

Check Your Learning:

What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?

Answer: 635 mL

Volume and Pressure: Boyle’s Law

If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 6.5.

This figure contains a diagram and two graphs. The diagram shows a syringe labeled with a scale in m l or c c with multiples of 5 labeled beginning at 5 and ending at 30. The markings halfway between these measurements are also provided. Attached at the top of the syringe is a pressure gauge with a scale marked by fives from 40 on the left to 5 on the right. The gauge needle rests between 10 and 15, slightly closer to 15. The syringe plunger position indicates a volume measurement about halfway between 10 and 15 m l or c c. The first graph is labeled “V ( m L )” on the horizontal axis and “P ( p s i )” on the vertical axis. Points are labeled at 5, 10, 15, 20, and 25 m L with corresponding values of 39.0, 19.5, 13.0, 9.8, and 6.5 p s i. The points are connected with a smooth curve that is declining at a decreasing rate of change. The second graph is labeled “V ( m L )” on the horizontal axis and “1 divided by P ( p s i )” on the vertical axis. The horizontal axis is labeled at multiples of 5, beginning at zero and extending up to 35 m L. The vertical axis is labeled by multiples of 0.02, beginning at 0 and extending up to 0.18. Six points indicated by black dots on this graph are connected with a black line segment showing a positive linear trend. — Figure 6.5 When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of $\frac{1}{P}$ vs. V is linear.

Unlike the P-T and V-T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:

P α 1 / V or P = k · 1 / V or P · V = k or P_{1} V_{1} = P_{2} V_{2}

with k being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure $(\frac{1}{P})$ versus the volume (V), or the inverse of volume $(\frac{1}{V})$ versus the pressure (P). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (see Figure 6.6).

This diagram shows two graphs. In a, a graph is shown with volume on the horizontal axis and pressure on the vertical axis. A curved line is shown on the graph showing a decreasing trend with a decreasing rate of change. In b, a graph is shown with volume on the horizontal axis and one divided by pressure on the vertical axis. A line segment, beginning at the origin of the graph, shows a positive, linear trend. — Figure 6.6 The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a hyperbola, whereas (b) the graph of $(\frac{1}{P})$ vs. V is linear.

The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.

EXAMPLE 6.6

Volume of a Gas Sample

The sample of gas in Figure 6.5 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:

(a) the P-V graph in Figure 6.5

(b) the 1/P vs. V graph in Figure 6.5

Comment on the likely accuracy of each method.

(a) Estimating from the P-V graph gives a value for P somewhere around 27 psi.

(b) Estimating from the $\frac{1}{P}$ versus V graph give a value of about 26 psi.

(c) From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P₁V₁ = k and P₂V₂ = k which means that P₁V₁ = P₂V₂.

Using P₁ and V₁ as the known values 13.0 psi and 15.0 mL, P₂ as the pressure at which the volume is unknown, and V₂ as the unknown volume, we have:

P_{1} V_{1} = P_{2} V_{2} or 13.0 psi \times 15.0 mL = P_{2} \times 7.5 mL

Solving:

P_{2} = \frac{13.0 psi \times 15.0 mL}{7.5 mL} = 26 psi

It was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.

Check Your Learning:

The sample of gas in Figure 6.5 has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:

(a) the P-V graph in Figure 6.5

(b) the $\frac{1}{P}$ vs. V graph in Figure 6.5

Comment on the likely accuracy of each method.

Answer:

(a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow

Moles of Gas and Volume: Avogadro’s Law

The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.

In equation form, this is written as:

\begin{matrix} V \propto n & or & V = k \times n & or & \frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} \end{matrix}

Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T.

LINK TO LEARNING

Visit this interactive PhET simulation to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).

The Ideal Gas Law

To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:

Boyle’s law: PV = constant at constant T and n
Amontons’s law: $\frac{P}{T}$ = constant at constant V and n
Charles’s law: $\frac{V}{T}$ = constant at constant P and n
Avogadro’s law: $\frac{V}{n}$ = constant at constant P and T

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:

P V = n R T

where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol^–1 K^–1 and 8.314 kPa L mol^–1 K^–1.

Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.

EXAMPLE 6.7

Using the Ideal Gas Law

Methane, CH₄, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH₄. What is the volume of this much methane at 25 °C and 745 torr?

Solution:

We must rearrange PV = nRT to solve for V:

V = \frac{n R T}{P}

If we choose to use R = 0.08206 L atm mol^–1 K^–1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units:

n = 6 55 g {CH}_{4} \times \frac{1 mol}{16.043 {g CH}_{4}} = 40.8 mol

T = 25 °C + 273 = 298 K

P = 745 torr \times \frac{1 atm}{760 torr} = 0.980 atm

V = \frac{n R T}{P} = \frac{(40.8 mol) (0.08206 L {atm mol}^{–1} K^{–1}) (298 K)}{0.980 atm} = 1.02 \times 10^{3} L

It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.

Check Your Learning:

Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.

Answer: 350 bar

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ using units of atm, L, and K. Both sets of conditions are equal to the product of n $\times$ R (where n = the number of moles of the gas and R is the ideal gas law constant).

EXAMPLE 6.8

Using the Combined Gas Law

When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 6.6). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

This photograph shows a scuba diver underwater with a tank on his or her back and bubbles ascending from the breathing apparatus. — Figure 6.7 Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} ⟶ \frac{(153 atm) (13.2 L)}{(300 K)} = \frac{(3.13 atm) (V_{2})}{(310 K)}

Solving for V₂:

V_{2} = \frac{(153 atm) (13.2 L) (310 K)}{(300 K) (3.13 atm)} = 667 L

(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)

Check Your Learning:

A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.

Answer: 0.193 L

Standard Conditions of Temperature and Pressure

We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa).¹ At STP, one mole of an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure 6.7).

This figure shows three balloons each filled with He, NH subscript 2, and O subscript 2, respectively. Beneath the first balloon is the label “4 g of He” Beneath the second balloon is the label, “17 g of NH subscript 2.” Beneath the third balloon is the label “32 g of O subscript 2.” Each balloon contains the same number of molecules of their respective gases. — Figure 6.7 Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of ~22.4 L at STP.

Key Equations:

PV = nRT

Footnotes

1 The IUPAC definition of standard pressure was changed from 1 atm to 1 bar (100 kPa) in 1982, but the prior definition remains in use by many literature resources and will be used in this text.

Answers to Activity

1. Pressure increases
2. Pressure increases
3. Volume increases
4. Volume increases