What Temperature Must a Gas Initially at 10 Ââ°c Be Brought to for the Pressure to Triple?

LEARNING OBJECTIVES

Past the end of this department, you will exist able to:

• Place the mathematical relationships between the various properties of gases
• Use the platonic gas constabulary, and related gas laws, to compute the values of various gas properties under specified weather

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could wing (Figure 1), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, force per unit area, volume, temperature, and amount of gas. Although their measurements were not precise by today'southward standards, they were able to determine the mathematical relationships between pairs of these variables (eastward.g., pressure and temperature, pressure level and volume) that agree for an ideal gas—a hypothetical construct that real gases approximate under certain weather condition. Eventually, these private laws were combined into a unmarried equation—the ideal gas law—that relates gas quantities for gases and is quite authentic for low pressures and moderate temperatures. We will consider the cardinal developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas police.

Figure 1. In 1783, the showtime (a) hydrogen-filled balloon flight, (b) manned hot air airship flying, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed information technology with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Pressure and Temperature: Amontons'southward Law

Imagine filling a rigid container attached to a force per unit area estimate with gas and so sealing the container so that no gas may escape. If the container is cooled, the gas inside as well gets colder and its force per unit area is observed to subtract. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain abiding. If nosotros heat the sphere, the gas inside gets hotter (Figure 2) and the pressure increases.

Figure 2. The effect of temperature on gas pressure: When the hot plate is off, the pressure level of the gas in the sphere is relatively low. As the gas is heated, the pressure level of the gas in the sphere increases.

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

Figure 3. For a abiding volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot exist 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 zip.

Guillaume Amontons was the start to empirically establish the relationship between the force per unit area and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P–T human relationship for gases is known as either Amontons's police force or Gay-Lussac's law . Nether either name, it states that the force per unit area of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held abiding. Mathematically, this tin can be written:

[latex]P\propto T\text{ or }P=\text{constant}\times T\text{ or }P=m\times T[/latex]

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 [latex]\frac{P}{T}[/latex] is therefore constant (i.east., [latex]\frac{P}{T}=k[/latex] ). If the gas is initially in "Condition 1" (with P = P ₁ and T = T ₁), and so changes to "Condition 2" (with P = P ₂ and T = T _two), nosotros have that [latex]\frac{{P}_{1}}{{T}_{1}}=k[/latex] and [latex]\frac{{P}_{2}}{{T}_{2}}=k,[/latex] which reduces to [latex]\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{two}}.[/latex] This equation is useful for pressure-temperature calculations for a bars gas at constant volume. Notation 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 zip ). (Also annotation that at that place are at least iii means we tin can describe how the pressure of a gas changes every bit its temperature changes: Nosotros tin can use a tabular array of values, a graph, or a mathematical equation.)

Example 1

Predicting Change in Pressure with Temperature

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

(a) On the can is the warning "Store but at temperatures below 120 °F (48.viii °C). Do not incinerate." Why?

(b) The gas in the tin is initially at 24 °C and 360 kPa, and the can has a book 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 tin can?

Solution

(a) The can contains an amount of isobutane gas at a abiding volume, so if the temperature is increased by heating, the pressure will increment proportionately. High temperature could atomic number 82 to loftier pressure level, causing the can to outburst. (Also, isobutane is combustible, then incineration could cause the can to explode.)

(b) We are looking for a pressure modify due to a temperature alter at constant volume, so we will use Amontons'southward/Gay-Lussac's police force. Taking P ₁ and T ₁ as the initial values, T _two as the temperature where the force per unit area is unknown and P ₂ as the unknown pressure, and converting °C to K, nosotros take:

[latex]\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}\phantom{\dominion{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\frac{360\text{kPa}}{297\text{Grand}}=\frac{{P}_{2}}{323\text{G}}[/latex]

Rearranging and solving gives: [latex]{P}_{2}=\frac{360\text{kPa}\times 323\cancel{\text{One thousand}}}{297\abolish{\text{K}}}=390\text{kPa}[/latex]

Check Your Learning

A sample of nitrogen, North₂, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it take if cooled to –73 °C while the volume remains constant?

Book and Temperature: Charles's Law

If we fill a balloon with air and seal information technology, the balloon contains a specific amount of air at atmospheric pressure, let's say 1 atm. If nosotros put the airship 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 airship very cold, it will shrink a great deal, and it expands over again when it warms up.

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

These examples of the upshot of temperature on the book of a given amount of a confined gas at abiding pressure are truthful 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 Effigy 4.

Effigy four. The volume and temperature are linearly related for one mole of methyl hydride gas at a constant pressure of ane atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 1000 considering marsh gas liquefies at this temperature; when extrapolated, it intersects the graph's origin, representing a temperature of absolute nil.

The human relationship between the volume and temperature of a given corporeality of gas at constant pressure level is known as Charles's police in recognition of the French scientist and airship flight pioneer Jacques Alexandre César Charles. Charles's law states that the volume of a given corporeality of gas is directly proportional to its temperature on the kelvin scale when the force per unit area is held abiding.

Mathematically, this can be written as:

[latex]V\propto T\text{or}V=\text{constant}\cdot T\text{or}V=one thousand\cdot T\text{or}{5}_{1}\text{/}{T}_{1}={V}_{2}\text{/}{T}_{2}[/latex]

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

For a bars, constant force per unit area gas sample, [latex]\frac{V}{T}[/latex] is constant (i.due east., the ratio = chiliad), and as seen with the V–T relationship, this leads to another class of Charles's law: [latex]\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{two}}.[/latex]

Case 2

Predicting Change in Volume with Temperature

A sample of carbon dioxide, CO₂, occupies 0.300 50 at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?

Solution

Considering we are looking for the volume modify caused by a temperature modify at constant pressure, this is a job for Charles's police force. Taking V ₁ and T ₁ equally the initial values, T _ii every bit the temperature at which the volume is unknown and Five ₂ as the unknown book, and converting °C into G nosotros have:

[latex]\frac{{V}_{one}}{{T}_{1}}=\frac{{5}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\dominion{0.4em}{0ex}}\frac{0.300\text{L}}{283\text{G}}=\frac{{V}_{2}}{303\text{Grand}}[/latex]

Rearranging and solving gives: [latex]{V}_{ii}=\frac{0.300\text{Fifty}\times \text{303}\cancel{\text{Yard}}}{283\cancel{\text{K}}}=0.321\text{L}[/latex]

This answer supports our expectation from Charles'due south police force, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its book (from 0.300 L to 0.321 L).

Check Your Learning

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

Instance 3

Measuring Temperature with a Volume Change

Temperature is sometimes measured with a gas thermometer past observing the modify in the volume of the gas every bit the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a book 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. Detect the temperature of boiling ammonia on the kelvin and Celsius scales.

Solution

A volume change caused by a temperature alter at abiding force per unit area ways we should use Charles's law. Taking 5 _ane and T ₁ as the initial values, T _ii as the temperature at which the book is unknown and V ₂ equally the unknown volume, and converting °C into M we have:

[latex]\frac{{V}_{1}}{{T}_{ane}}=\frac{{V}_{two}}{{T}_{2}}\text{ which ways that }\frac{150.0{\text{cm}}^{3}}{273.15\text{Thousand}}=\frac{131.7{\text{cm}}^{3}}{{T}_{2}}[/latex]

Rearrangement gives [latex]{T}_{2}=\frac{131.7{\abolish{\text{cm}}}^{3}\times 273.xv\text{K}}{150.0{\abolish{\text{cm}}}^{3}}=239.viii\text{1000}[/latex]

Subtracting 273.15 from 239.eight Yard, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.

Bank check Your Learning

What is the book of a sample of ethane at 467 K and 1.1 atm if information technology occupies 405 mL at 298 K and 1.1 atm?

Volume and Pressure: Boyle's Police

If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If nosotros slowly button in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its force per unit area increases; if we pull out the plunger, the volume increases and the pressure decreases. This case of the effect of book 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 subtract its pressure. In fact, if the volume increases by a certain gene, the pressure decreases by the aforementioned factor, and vice versa. Volume-force per unit area information for an air sample at room temperature are graphed in Figure v.

Effigy 5. When a gas occupies a smaller volume, it exerts a college pressure; when it occupies a larger book, it exerts a lower pressure (bold the corporeality of gas and the temperature practise not change). Since P and V are inversely proportional, a graph of 1/P vs. V is linear.

Unlike the P–T and V–T relationships, force per unit area and volume are not direct proportional to each other. Instead, P and Five exhibit inverse proportionality: Increasing the force per unit area results in a decrease of the book of the gas. Mathematically this can be written:

[latex]P\alpha 1\text{/}V\text{ or }P=k\cdot 1\text{/}Five\text{ or }P\cdot 5=k\text{ or }{P}_{1}{5}_{1}={P}_{2}{Five}_{two}[/latex]

Figure 6. The relationship between force per unit area and volume is inversely proportional. (a) The graph of P vs. V is a parabola, whereas (b) the graph of (one/P) vs. V is linear.

with k beingness a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure [latex]\left(\frac{1}{P}\correct)[/latex] versus the volume (Five), or the inverse of volume [latex]\left(\frac{1}{V}\right)[/latex] versus the pressure (V). Graphs with curved lines are hard to read accurately at depression or high values of the variables, and they are more difficult to utilise in plumbing fixtures theoretical equations and parameters to experimental data. For those reasons, scientists oft try to find a way to "linearize" their data. If nosotros plot P versus V, we obtain a hyperbola (come across Effigy 6).

The relationship between the book and pressure level of a given amount of gas at abiding temperature was beginning published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle's police force : The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure nether which it is measured.

Example 4

Volume of a Gas Sample

The sample of gas in Figure 5 has a volume of 15.0 mL at a pressure level of thirteen.0 psi. Determine the pressure level of the gas at a book of 7.5 mL, using:

(a) the P–5 graph in Figure 5

(b) the [latex]\frac{1}{P}[/latex] vs. 5 graph in Figure 5

(c) the Boyle'south law equation

Comment on the likely accuracy of each method.

Solution

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

(b) Estimating from the [latex]\frac{one}{P}[/latex] 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 abiding temperature is always equal to the aforementioned value. Therefore we accept P _ane V _one = k and P ₂ V _two = k which means that P ₁ Five _one = P ₂ 5 ₂.

Using P ₁ and V ₁ equally the known values 0.993 atm and ii.xl mL, P ₂ every bit the pressure level at which the volume is unknown, and V ₂ as the unknown volume, we have:

[latex]{P}_{1}{5}_{1}={P}_{2}{5}_{two}\text{or}13.0\text{psi}\times 15.0\text{mL}={P}_{two}\times 7.5\text{mL}[/latex]

Solving:

[latex]{V}_{2}=\frac{thirteen.0\text{psi}\times 15.0\cancel{\text{mL}}}{vii.5\cancel{\text{mL}}}=26\text{mL}[/latex]

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

Bank check Your Learning

The sample of gas in Figure five has a volume of 30.0 mL at a pressure of six.5 psi. Determine the volume of the gas at a pressure level of xi.0 mL, using:

(a) the P–V graph in Figure 5

(b) the [latex]\frac{i}{P}[/latex] vs. 5 graph in Figure 5

(c) the Boyle's police force equation

Comment on the likely accuracy of each method.

Answer:(a) most 17–18 mL; (b) ~xviii mL; (c) 17.7 mL; it was more hard to approximate well from the P–V graph, so (a) is likely more inaccurate than (b); the calculation will be every bit accurate as the equation and measurements permit

Chemistry in Action: Breathing and Boyle's Police

What exercise you practise about xx times per minute for your whole life, without suspension, and frequently without fifty-fifty being aware of it? The respond, of course, is respiration, or breathing. How does information technology work? It turns out that the gas laws apply here. Your lungs take in gas that your torso needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your breast cavity and making your lung book larger. The increase in volume leads to a decrease in force per unit area (Boyle's constabulary). This causes air to menstruum into the lungs (from high pressure to low pressure). When you breathe, the process reverses: Your diaphragm and rib muscles relax, your breast cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle's law again), and air flows out of the lungs (from high pressure level to depression pressure). You and so breathe in and out again, and again, repeating this Boyle'due south law cycle for the residuum of your life (Effigy 7).

Effigy 7. Breathing occurs considering expanding and contracting lung book creates pocket-sized force per unit area differences between your lungs and your environment, causing air to be drawn into and forced out of your lungs.

Moles of Gas and Volume: Avogadro'south Constabulary

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 police : For a confined gas, the volume (5) and number of moles (n) are direct proportional if the pressure and temperature both remain abiding.

In equation form, this is written as:

[latex]\begin{assortment}{ccccc}V\propto n& \text{or}& V=g\times northward& \text{or}& \frac{{V}_{1}}{{n}_{1}}=\frac{{V}_{2}}{{due north}_{2}}.\end{array}[/latex]

Mathematical relationships can besides be determined for the other variable pairs, such equally P versus n, and north versus T.

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

The Ideal Gas Law

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

• Boyle'south law: PV = constant at constant T and n
• Amontons'south police force: [latex]\frac{P}{T}[/latex] = constant at constant Five and north
• Charles'southward constabulary: [latex]\frac{5}{T}[/latex] = abiding at abiding P and due north
• Avogadro's police: [latex]\frac{5}{n}[/latex] = constant at constant P and T

Combining these iv laws yields the ideal gas law , a relation between the force per unit area, volume, temperature, and number of moles of a gas:

[latex]PV=nRT[/latex]

where P is the pressure of a gas, 5 is its volume, north 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 limited pressure, book, and temperature will determine the proper grade of the gas constant as required past dimensional assay, the most ordinarily encountered values being 0.08206 L atm mol^–one 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 constabulary (or the other gas laws) are said to exhibit ideal behavior or to gauge the traits of an ideal gas . An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to finer explain the gas laws as will be described in a later module of this affiliate. Although all the calculations presented in this module assume ideal behavior, this assumption is simply reasonable for gases under conditions of relatively low force per unit area and loftier temperature. In the last module of this chapter, a modified gas police force will be introduced that accounts for the not-ideal beliefs observed for many gases at relatively loftier pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P, 5, n, and T. Specifying any four of these terms volition permit use of the ideal gas law to calculate the fifth term every bit demonstrated in the post-obit instance exercises.

Instance 5

Using the Platonic Gas Police

Methane, CH₄, is beingness considered for use as an culling automotive fuel to supersede gasoline. 1 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 Five: [latex]5=\frac{nRT}{P}[/latex]

If we cull to apply R = 0.08206 Fifty atm mol^–i Yard^–1, so the corporeality must be in moles, temperature must be in kelvin, and force per unit area must be in atm.

Converting into the "correct" units:

[latex]northward=655\text{g}\cancel{{\text{CH}}_{4}}\times \frac{i\text{mol}}{xvi.043{\cancel{\text{one thousand CH}}}_{4}}=40.8\text{mol}[/latex]

[latex]T=25\text{\textdegree C}+273=298\text{1000}[/latex]

[latex]P=745\abolish{\text{torr}}\times \frac{1\text{atm}}{760\cancel{\text{torr}}}=0.980\text{atm}[/latex]

[latex]V=\frac{nRT}{P}=\frac{\left(40.viii\cancel{\text{mol}}\right)\left(0.08206\text{L}\abolish{{\text{atm mol}}^{-1}{\text{One thousand}}^{\text{-1}}}\right)\left(298\cancel{\text{Yard}}\correct)}{0.980\abolish{\text{atm}}}=1.02\times {10}^{3}\text{Fifty}[/latex]

Information technology would crave 1020 L (269 gal) of gaseous methane at nearly 1 atm of pressure to replace 1 gal of gasoline. Information technology requires a large container to agree enough methane at 1 atm to supercede several gallons of gasoline.

Cheque Your Learning

Summate the force per unit area in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-50 storage tank of a modern hydrogen-powered machine.

If the number of moles of an ideal gas are kept constant under two different sets of weather condition, a useful mathematical human relationship called the combined gas police force is obtained: [latex]\frac{{P}_{1}{V}_{one}}{{T}_{i}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}[/latex] using units of atm, L, and M. Both sets of conditions are equal to the production of n × R (where n = the number of moles of the gas and R is the platonic gas law abiding).

Example 6

Using the Combined Gas Law

Figure 8. Scuba divers use compressed air to breathe while underwater. (credit: modification of piece of work by Mark Goodchild)

When filled with air, a typical scuba tank with a volume of 13.2 Fifty has a pressure of 153 atm (Figure viii). 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 sea where the pressure is 3.13 atm?

Letting i represent the air in the scuba tank and two represent the air in the lungs, and noting that body temperature (the temperature the air volition be in the lungs) is 37 °C, we take:

[latex]\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{ii}{Five}_{2}}{{T}_{2}}\phantom{\dominion{0.4em}{0ex}}\rightarrow\phantom{\rule{0.4em}{0ex}}\frac{\left(153\text{atm}\right)\left(13.2\text{Fifty}\correct)}{\left(300\text{One thousand}\right)}=\frac{\left(3.13\text{atm}\correct)\left({V}_{2}\right)}{\left(310\text{1000}\right)}[/latex]

Solving for V _two:

[latex]{V}_{2}=\frac{\left(153\abolish{\text{atm}}\correct)\left(thirteen.2\text{L}\correct)\left(310\text{K}\right)}{\left(300\text{K}\right)\left(iii.13\cancel{\text{atm}}\correct)}=667\text{L}[/latex]

(Note: Be advised that this particular case is i in which the assumption of ideal gas beliefs is not very reasonable, since information technology involves gases at relatively high pressures and depression temperatures. Despite this limitation, the calculated volume can exist viewed every bit a good "ballpark" guess.)

Check Your Learning

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

The Interdependence between Sea Depth and Pressure in Scuba Diving

Figure 9. Scuba divers, whether at the Not bad Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the corporeality of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)

Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure nine) or in the Caribbean area, divers must understand how force per unit area affects a number of bug related to their comfort and safe.

Pressure level increases with ocean depth, and the pressure level changes virtually quickly as divers accomplish the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Virtually pressure level measurements are given in units of atmospheres, expressed equally "atmospheres absolute" or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in improver to 1 ATA of pressure from the atmosphere at sea level.

As a diver descends, the increase in force per unit area causes the torso's air pockets in the ears and lungs to shrink; on the ascension, the decrease in pressure causes these air pockets to aggrandize, potentially rupturing eardrums or bursting the lungs. Defined must therefore undergo equalization by calculation air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses past equalization techniques; the corollary is as well true on rising, divers must release air from the torso to maintain equalization.

Buoyancy, or the ability to command whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands considering of lower pressure according to Boyle's constabulary (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or adventure an uncontrolled rise that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or adventure an uncontrolled descent, facing much higher pressures near the ocean floor.

The pressure besides impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more than compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is two ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast every bit at the surface.

Standard Conditions of Temperature and Force per unit area

We accept 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). At STP, an ideal gas has a volume of about 22.iv Fifty—this is referred to as the standard molar volume (Figure 10).

Effigy 10. Since the number of moles in a given book of gas varies with pressure and temperature changes, chemists employ standard temperature and pressure level (273.xv K and i atm or 101.325 kPa) to report backdrop of gases.

Cardinal Concepts and Summary

The beliefs of gases tin be described by several laws based on experimental observations of their properties. The pressure of a given corporeality of gas is direct proportional to its absolute temperature, provided that the volume does not change (Amontons'due south law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's police force). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law). Under the same weather condition of temperature and pressure level, equal volumes of all gases contain the same number of molecules (Avogadro'south police force).

The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the force per unit area of the gas, Five is its volume, due north is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas abiding.

Chemistry Stop of Chapter Exercises

1. Sometimes leaving a cycle in the sun on a hot 24-hour interval will cause a blowout. Why?
2. Explicate how the book of the bubbles wearied by a scuba diver (Figure 8) change as they ascension to the surface, assuming that they remain intact.
3. One mode to state Boyle's law is "All other things being equal, the pressure of a gas is inversely proportional to its volume."
   1. What is the meaning of the term "inversely proportional?"
   2. What are the "other things" that must be equal?
4. An alternating way to country Avogadro's constabulary is "All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas."
   1. What is the meaning of the term "direct proportional?"
   2. What are the "other things" that must be equal?
5. How would the graph in Figure iv change if the number of moles of gas in the sample used to determine the curve were doubled?
6. How would the graph in Figure 5 alter if the number of moles of gas in the sample used to determine the curve were doubled?
7. In addition to the data found in Figure 5, what other information do we need to find the mass of the sample of air used to decide the graph?
8. Determine the volume of i mol of CH₄ gas at 150 K and ane atm, using Figure iv.
9. Decide the pressure of the gas in the syringe shown in Figure 5 when its book is 12.5 mL, using:
   1. the appropriate graph
   2. Boyle's law
10. A spray tin is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what volition be the force per unit area in the hot can?
11. What is the temperature of an 11.two-L sample of carbon monoxide, CO, at 744 torr if it occupies xiii.3 50 at 55 °C and 744 torr?
12. A 2.fifty-Fifty volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no modify in pressure.
13. A balloon inflated with 3 breaths of air has a book of 1.seven L. At the same temperature and pressure, what is the volume of the balloon if 5 more same-sized breaths are added to the balloon?
14. A weather airship contains 8.80 moles of helium at a pressure level of 0.992 atm and a temperature of 25 °C at ground level. What is the book of the balloon nether these conditions?
15. The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.viii g of nitrogen gas. What was the force per unit area in the bag in kPa?
16. How many moles of gaseous boron trifluoride, BF₃, are contained in a 4.3410-L bulb at 788.0 K if the force per unit area is one.220 atm? How many grams of BF₃?
17. Iodine, I₂, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I_two vapor at a pressure of 0.462 atm?
18. How many grams of gas are present in each of the following cases?
    1. 0.100 50 of CO_ii at 307 torr and 26 °C
    2. 8.75 L of C₂H₄, at 378.3 kPa and 483 K
    3. 221 mL of Ar at 0.23 torr and –54 °C
19. A high altitude airship is filled with 1.41 × x⁴ L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
20. A cylinder of medical oxygen has a book of 35.4 L, and contains O_two at a pressure of 151 atm and a temperature of 25 °C. What book of O₂ does this represent to at normal body conditions, that is, 1 atm and 37 °C?
21. A big scuba tank (Figure 8) with a book of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of two.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C?
22. A 20.0-L cylinder containing 11.34 kg of butane, C_fourH_ten, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the force per unit area in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.
23. While resting, the boilerplate 70-kg human male person consumes 14 Fifty of pure O₂ per hour at 25 °C and 100 kPa. How many moles of O₂ are consumed by a 70 kg man while resting for ane.0 h?
24. For a given amount of gas showing ideal behavior, draw labeled graphs of:
    1. the variation of P with V
    2. the variation of 5 with T
    3. the variation of P with T
    4. the variation of [latex]\frac{1}{P}[/latex] with V
25. A liter of methyl hydride gas, CH₄, at STP contains more than atoms of hydrogen than does a liter of pure hydrogen gas, H₂, at STP. Using Avogadro's constabulary every bit a starting point, explain why.
26. The effect of chlorofluorocarbons (such equally CCl_twoF₂) on the depletion of the ozone layer is well known. The use of substitutes, such as CH₃CH₂F(thousand), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:
    1. CCl₂F_two(g)
    2. CH_threeCH₂F(g)
27. As 1 g of the radioactive element radium decays over i year, it produces one.sixteen × ten¹⁸ alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the force per unit area in pascal of the helium gas produced if information technology occupies a book of 125 mL at a temperature of 25 °C?
28. A airship that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the superlative of Mount Crumpet in British Columbia. If the concluding volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mountain Crumpet?
29. If the temperature of a stock-still amount of a gas is doubled at constant volume, what happens to the pressure?
30. If the volume of a fixed corporeality of a gas is tripled at abiding temperature, what happens to the pressure?

Selected Answers

2. As the bubbles ascent, the pressure decreases, so their book increases as suggested by Boyle'due south constabulary.

four. (a) The number of particles in the gas increases as the volume increases. This relationship may be written every bit northward = constant × V. It is a straight human relationship.(b) The temperature and pressure must be kept constant.

half dozen. The curve would be farther to the right and in a higher place, but the aforementioned bones shape.

8. The figure shows the change of 1 mol of CH₄ gas equally a function of temperature. The graph shows that the volume is about 16.3 to 16.5 50.

10. The first thing to recognize about this problem is that the volume and moles of gas remain constant. Thus, we can use the combined gas law equation in the grade:

 [latex]\frac{{P}_{2}}{{T}_{2}}=\frac{{P}_{ane}}{{T1}_{}}[/latex]

[latex]{P}_{2}=\frac{{P}_{1}{T}_{2}}{{T}_{ane}}=1344\text{torr}\times \frac{475+273.15}{23+273.xv}=3.forty\times {10}^{3}\text{torr}[/latex]

12. Apply Charles'due south police to compute the volume of gas at the higher temperature:

V ₁ = 2.50 L

T ₁ = –193 °C = 77.fifteen One thousand

5 ₂ = ?

T _two = 100 °C = 373.15 K

[latex]\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{2}}[/latex]

 [latex]{V}_{2}=\frac{{Five}_{i}{T}_{two}}{{T}_{ane}}=\frac{2.fifty\text{L}\times 373.15\abolish{\text{One thousand}}}{77.fifteen\cancel{\text{K}}}=12.1\text{L}[/latex]

xiv.PV = nRT

[latex]V=\frac{nRT}{P}=\frac{8.80\cancel{\text{mol}}\times 0.08206\text{50}\abolish{\text{atm}}{\cancel{\text{mol}}}^{\text{-i}}\cancel{{\text{K}}^{\text{-ane}}}\times 298.xv\abolish{\text{G}}}{0.992\cancel{\text{atm}}}=217\text{L}[/latex]

16. [latex]n=\frac{PV}{RT}\frac{1.220\cancel{\text{atm}}\left(iv.3410\text{L}\right)}{\left(0.08206\text{L}\cancel{\text{atm}}\text{mol}{\text{-1}}^{}\cancel{{\text{K}}^{\text{-i}}}\correct)\left(788.0\abolish{\text{K}}\right)}=0.08190\text{mol}=8.190\times {x}^{\text{-2}}\text{mol}[/latex]

 [latex]northward\times \text{molar mass}=viii.190\times {10}^{\text{-2}}\cancel{\text{mol}}\times 67.8052\text{g}{\cancel{\text{mol}}}^{\text{-1}}=5.553\text{k}[/latex]

18. In each of these problems, we are given a volume, pressure level, and temperature. We can obtain moles from this information using the molar mass, m = nℳ, where ℳ is the molar mass:

[latex]P,Five,T\stackrel{northward=PV\text{/}RT}{\to }n,\stackrel{grand=n\left(\text{molar mass}\correct)}{\to }\text{grams}[/latex]

or nosotros can combine these equations to obtain:

[latex]\text{mass}=m=\frac{PV}{RT}\times [/latex]ℳ

(a)  [latex]\begin{array}{l}307\cancel{\text{torr}}\times \frac{1\text{atm}}{760\cancel{\text{torr}}}=0.4039\text{atm}26\textdegree C=299.1 K\\ \text{Mass}=m=\frac{0.4039\cancel{\text{atm}}\left(0.100\cancel{\text{L}}\right)}{0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{\text{-ane}}\cancel{{\text{K}}^{\text{-one}}}\left(299.1\abolish{\text{Yard}}\right)}\times 44.01\text{g}{\text{mol}}^{\text{-1}}=7.24\times {10}^{\text{-ii}}\text{g};\end{array}[/latex]

(b)[latex]\text{Mass}=1000=\frac{378.3\cancel{\text{kPa}}\left(8.75\cancel{\text{L}}\right)}{8.314\abolish{\text{Fifty}}\abolish{\text{kPa}}{\text{mol}}^{\text{-1}}\abolish{{\text{K}}^{\text{-1}}}\left(483\cancel{\text{M}}\correct)}\times 28.05376\text{grand}{\text{mol}}^{\text{-i}}=23.1\text{g;}[/latex]

(c)  [latex]\begin{array}{l}\\ 221\cancel{\text{mL}}\times \frac{one\text{L}}{1000\abolish{\text{mL}}}=0.221\text{50}-54\textdegree C+273.15=219.fifteen\text{Thou}\\ 0.23\abolish{\text{torr}}\times \frac{ane\text{atm}}{760\cancel{\text{torr}}}=three.03\times {10}^{\text{-four}}\text{atm}\\ \text{Mass}=m=\frac{iii.03\times {10}^{\text{-4}}\cancel{\text{atm}}\left(0.221\cancel{\text{L}}\correct)}{0.08206\cancel{\text{Fifty}}\cancel{\text{atm}}{\text{mol}}^{\text{-1}}\cancel{{\text{K}}^{\text{-1}}}\left(219.xv\cancel{\text{K}}\right)}\times 39.978\text{g}{\text{mol}}^{\text{-1}}=1.5\times {10}^{\text{-4}}\text{g}\finish{array}[/latex]

20.  [latex]\frac{{P}_{ii}}{{T}_{2}}=\frac{{P}_{one}}{{T}_{1}}[/latex]

T ₂ = 49.five + 273.fifteen = 322.65 Yard

[latex]{P}_{2}=\frac{{P}_{ane}{T}_{ii}}{{T}_{1}}=149.6\text{atm}\times \frac{322.65}{278.xv}=173.five\text{atm}[/latex]

22. Calculate the amount of butane in 20.0 L at 0.983 atm and 27°C. The original amount in the container does not matter. [latex]north=\frac{PV}{RT}=\frac{0.983\abolish{\text{atm}}\times 20.0\cancel{\text{Fifty}}}{0.08206\cancel{\text{50}}\cancel{\text{atm}}{\text{mol}}^{\text{-ane}}\abolish{{\text{K}}^{\text{-1}}}\left(300.1\cancel{\text{K}}\right)}=0.798\text{mol}[/latex] Mass of butane = 0.798 mol × 58.1234 grand/mol = 46.four g

24. For a gas exhibiting ideal behavior:

26. (a) Make up one's mind the tooth mass of CCl₂F₂ then summate the moles of CCl₂F₂(1000) present. Use the ideal gas law PV = nRT to calculate the book of CCl₂F₂(g):

 [latex]\text{10.0 thou}{\text{CCl}}_{2}{\text{F}}_{two}\times \frac{1\text{mol}{\text{CC1}}_{two}{\text{F}}_{2}}{120.91\text{g}{\text{CCl}}_{2}{\text{F}}_{2}}=0.0827\text{mol}{\text{CCl}}_{ii}{\text{F}}_{2}[/latex]

PV = nRT, where northward = # mol CCl_twoF_two

[latex]ane\text{atm}\times 5=0.0827\text{mol}\times \frac{0.0821\text{L atm}}{\text{mol K}}\times 273\text{Thou}=1.85\text{L}{\text{CCl}}_{2}{\text{F}}_{ii};[/latex]

(b) [latex]10.0\text{1000}{\text{CH}}_{3}{\text{CH}}_{ii}\text{F}\times \frac{i\text{mol}{\text{CH}}_{iii}{\text{CH}}_{2}\text{F}}{48.07{\text{g CH}}_{3}{\text{CH}}_{2}\text{F}}=0.208\text{mol}{\text{CH}}_{3}{\text{CH}}_{2}\text{F}[/latex]

PV = nRT, with northward = # mol CH₃CH_iiF

1 atm × Five = 0.208 mol × 0.0821 L atm/mol K × 273 Yard = four.66 L CH₃ CH_ii F

28. Identify the variables in the problem and determine that the combined gas police [latex]\frac{{P}_{1}{V}_{1}}{{T}_{one}}=\frac{{P}_{ii}{V}_{ii}}{{T}_{2}}[/latex] is the necessary equation to utilize to solve the trouble. Then solve for P_two:

 [latex]\begin{assortment}{l}\\ \\ \frac{0.981\text{atm}\times 100.21\text{50}}{294\text{K}}=\frac{{P}_{2}\times 144.53\text{L}}{278.24\text{atm}}\\ {P}_{two}=0.644\text{atm}\terminate{array}[/latex]

30. The pressure decreases by a factor of iii.

Glossary

absolute zero
temperature at which the volume of a gas would exist zero according to Charles's police.

Amontons's law
(as well, Gay-Lussac's law) force per unit area of a given number of moles of gas is directly proportional to its kelvin temperature when the book is held constant

Avogadro's law
volume of a gas at constant temperature and force per unit area is proportional to the number of gas molecules

Boyle'south police
volume of a given number of moles of gas held at constant temperature is inversely proportional to the force per unit area under which it is measured

Charles's law
volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant

ideal gas
hypothetical gas whose physical backdrop are perfectly described by the gas laws

ideal gas constant (R)
constant derived from the ideal gas equation R = 0.08226 L atm mol^–i K^–1 or 8.314 L kPa mol^–1 K^–i

ideal gas law
relation between the pressure level, book, amount, and temperature of a gas under conditions derived past combination of the uncomplicated gas laws

standard conditions of temperature and pressure (STP)
273.15 K (0 °C) and 1 atm (101.325 kPa)

standard molar book
volume of 1 mole of gas at STP, approximately 22.iv L for gases behaving ideally

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Source: https://courses.lumenlearning.com/wsu-sandbox2/chapter/relating-pressure-volume-amount-and-temperature-the-ideal-gas-law-needs-formulas/

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