Solids and liquids have volumes that tend to remain constant with changes in their environment. In contrast, the volumes of gases vary with changes in the temperature and/or pressure. This chapter addresses the factors that influence the volume of gases.
The general definition of pressure is the force acting on a unit area of surface. Mathematically,
More specifically,
and, by the above, the pascal is defined as
The pressure exerted by a column of fluid is
More specifically,
Because air has weight, it exerts a pressure. The composition of the air tends to vary, leading to changes in the weight of the air (changes in atmospheric pressure). Standard pressure, also called a standard atmosphere, is set at 1 atm pressure. The value of 1 atm is defined as 101,325 Pa. Another definition of the standard atmosphere is the pressure exerted by a column of mercury 760 mm high when at 0°C and at sea level. The mm Hg is also referred to as the torr (1 mm Hg = 1 torr) and 1 atm pressure then becomes 760 torr. The bar is often used as a measure of pressure (1 bar = 105 Pa = 1 atm).
Note that there is a small built-in error when using 1 atm = 1 bar (1325 out of 100,000 or 1.3%).
The pressure of a gas can be measured by attaching a manometer to the vessel containing the gas. A manometer is a tube (U-shaped in our examples) containing a liquid, usually mercury. The height of the liquid is read in mm Hg (i.e torr) pressure units.
The closed-tube manometer, Fig. 5-1(a), is originally filled all the way to the closed end so that the difference in mercury levels is the absolute value of the pressure of the gas. An open-tube manometer shows the difference between the pressure of the gas and the barometric pressure—Fig. 5-1(b) shows less than barometric pressure and Fig. 5-1(c) shows a pressure higher than barometric pressure.
Fig. 5-1
There are mechanical and electronic measuring devices. A familiar one is the tire gauge, which provides a relative reading because the reading is in excess of the existing barometric pressure (a flat tire contains air at the ambient pressure). Gauges that read absolute pressure (starting with 0 pressure, rather than above ambient) are identified as such on the face of the gauge.
The temperature of 273.15 K (0°C) and 1 atm pressure are taken as standard conditions (S.T.P.). Unless extreme accuracy is called for, 273.15 K is usually rounded to 273 K. Further, many problems are written using torr rather that atm (1 atm = 760 torr), as many of the measuring devices read in either mm Hg or torr (1 mm Hg = 1 torr).
There are three laws to describe the behavior of an unchanging mass of a gas: Boyle’s law, Charles’ law, and Gay-Lussac’s law. These laws detail the effects of volume, temperature, and pressure and the interrelationships of these factors. Each of the three laws holds a single variable (V or P or T) constant with the relationship between the other two addressed. Gases that behave exactly as predicted are referred to as ideal gases (perfect gases). Most gases do not respond exactly as predicted by the laws because the laws do not account for forces between the molecules of the gases. However, the gas laws are a good starting place for predictions of the behavior of a fixed amount of a gas beginning with the initial conditions and looking to the final conditions after a change in a variable.
Boyle’s law describes the effects of volume and pressure—temperature is constant. The law states that the volume of a gas varies inversely with the pressure at a constant temperature.
Charles’ law describes the effects of volume and temperature—pressure is constant. The law states that the volume of the gas varies directly with the temperature in the Kelvin scale (absolute temperature).
Note: Mathematical relationships involving temperatures most often require the temperatures to be stated in the Kelvin scale. This is because the temperature-dependent phenomena generally conform to neither Celsius nor Fahrenheit relationships. This is true of gas behavior with reference to temperature.
Gay-Lussac’s law describes the effects of pressure and temperature—volume is constant. The law states that the pressure of the gas varies directly with the absolute temperature.
The individual gas laws can be used to derive a law involving all three influencing factors—volume, pressure, and temperature—into one law. Since the three laws deal with a fixed amount of gas (no change in mass), so does the combined gas law.
The density (D) of a gas varies inversely to the volume, as shown in equation (a). We can substitute a rearrangement of the combined gas law from equation (b) to provide equation (c), which describes changes in density with temperature and pressure.
(a) 
(b) 
(c) 
Dalton’s law states that the sum of the pressures of the gases in a mixture is the total pressure exerted—the volume and temperature are constant (Ptotal = P1 + P2 + P3 + …). The pressure of each gas is the partial pressure of that gas. As with the other gas laws, this law only holds true when working with ideal gases; however, this law may be applied to the problems in this book for predicting results.
A gas can be collected by bubbling through a liquid. Since liquids tend to evaporate, the liquid’s vapor mixes with the gas being collected contributing to the total pressure (Dalton’s law). Gas collection is often over water and the vapor pressure of the water, even though relatively low, must be subtracted from the total to provide the pressure of the gas collected; the result is the pressure of the “dry” gas.
If the gas is collected over another volatile liquid, the vapor pressure of that liquid is subtracted from the total pressure. The partial pressures of volatile liquids are constant and dependent on temperature—the partial pressures can be obtained from reference sources.
The laws discussed above are strictly valid only for ideal gases. The fact that a gas can be liquefied if it is compressed and cooled sufficiently is an indication that the gas is not ideal at high pressures and low temperatures. The predictions made by the application of the laws are most accurate at low pressure and high temperatures, conditions far from those of the change from a gas to a liquid.
5.1. Calculate the difference in pressure between the top and bottom of a vessel exactly 76 cm deep when filled at 25°C with (a) water, then emptied and refilled with (b) mercury. The density of water at 25°C is 0.997 g/cm3 and that of mercury is 13.53 g/cm3.
Using the equivalent values in SI units for height and density,
(a)
(b)
5.2. How high a column of air would be necessary to cause the barometer to read 76 cm of mercury, if the atmosphere were of uniform density 1.2kg/m3? The density of mercury is 13.53 × 103 kg/m3.
Actually, the density of the air decreases with increasing height so that the atmosphere must extend much beyond 8.6 km. Modern aircraft routinely fly higher than 8.6 km (28,000 ft) and there are modified aircraft that fly in excess of 60,000 feet.
5.3. A mass of oxygen occupies 5.00 L under 740 torr pressure. What is the volume of the same mass of gas under the conditions of standard pressure with a constant temperature?
Figure 5-2 provides us with one way the experiment could be performed. We need to know that standard pressure is 760 torr. Boyle’s law is applied as follows:
Fig. 5-2
An interesting point about this problem is that any units of volume could have been used (qt, gal, etc.) since the conversions are performed by factors (multiplied by fractions) and would cancel. The same applies to units of pressure.
5.4. A mass of neon occupies 200 cm3 at 100°C. Find its volume at 0°C under constant pressure.
Figure 5-3 illustrates the change. Applying Charles’ law is appropriate for these data and the question.
Fig. 5-3
Absolute temperatures, the Kelvin scale, must be used when performing gas law calculations.
5.5. A steel tank contains carbon dioxide at 27°C and 12.0 atm. Determine the internal gas pressure when the tank and contents are heated to 100°C. The volume of the tank does not change.
See Fig. 5-4. Gay-Lussac’s law gives us
Fig. 5-4
5.6. Diesel engines run without the use of a spark plug because the fuel/air mixture is heated during compression to the flash point. Suppose the 6-cylinder, 6.0-L engine takes in the fuel/air mixture at 1 atm and 25°C, but is capable of 13.5 atm compression at the 220°C required to ignite the particular mixture. As the design engineer, calculate the volume per cylinder of the gas/air mixture that is necessary.
Since we have all three factors involved in gas action/reaction, we can apply the combined gas law. The 6-cylinder engine, 6.0 liters total volume, has 1.0-L cylinders, which is the value of V1. The remainder of the factors are stated. Remembering that all gas law problems require temperature expressed in the Kelvin scale, you can apply the combined gas law to determine the answer.
5.7. Given 20.0 L of ammonia at 5°C and 760 torr, determine its volume at 30°C and 800 torr.
See Fig. 5-5 for some visual help. This problem is set up in the same manner as Problem 5.6.
Fig. 5-5
5.8. One liter of gas, originally at 1.00 atm and –20°C, was warmed to 40°C. How many atmospheres pressure must it then be subjected to in order to reduce its volume to one-half liter?
The solution for this problem is derived from the use of the combined gas law.
5.9. Suppose just before taking a trip you filled your tires to 30 pounds pressure (psi; lb/in2) and the temperature that day was 27°F. After the trip, you checked the pressure and found it to be 34.2 psi. Estimate the temperature in °F of the air in the tires. Assume that the tire gauge reads relative pressure, that the volume of the tires remains constant, and that the ambient pressure was 1.00 atm that day.
Problem 1.9 contains the calculation showing that 1 atm is equal to 14.7 psi. Making the conversion, we find that the beginning pressure was (30.0 + 14.7) psi and that the tire pressure at the end of the trip was (34.2 + 14.7)psi. Since absolute zero is –460°F (Fig. 1-1), the absolute temperature is (27 + 460) Fahrenheit degree intervals. Applying Gay-Lussac’s law, which determines temperature and pressure changes with no volume changes,
T2 = T1 × P2/P1 = (27 + 460)(34.2 + 14.7)/(30.0 + 14.7) = 533 Fahrenheit degree intervals
Final temperature = Fahrenheit degree intervals – absolute zero = 533 – 460 = 73°F
Note that the absolute temperatures were used in this problem, as is required by gas law problems. The difference between this problem and others is that the scale is Fahrenheit rather than Kelvin.
5.10. A container holds 6.00 g of CO2 at 150°C and 100 kPa pressure. How many grams of CO2 will it hold at 30°C and the same pressure?
Charles’ law will organize the information and provide us with the solution. However, we must assume that CO2 behaves consistently with the change in the number of molecules (change in mass). Now, we can let the volume of the container be V1 and the volume of the 6.00 g of CO2 be V2.
Since 6.00 g of CO2 occupies 0.716 V1, to fill the container (V1) will require
6.00g/0.716 = 8.38 g CO2
5.11. The density of helium is 0.1786 kg/m3 at S.T.P. If a given mass of helium at S.T.P. is allowed to expand to 1.500 times the initial volume by changing the temperature and pressure, what will be the resultant density?
The density of a gas varies inversely with the volume.
Note that the problem could have been posed in terms of liters because 1 kg/m3 = 1 g/L.
5.12. The density of oxygen is 1.43 g/L at S.T.P. What is the density of oxygen at 17°C and 700 torr?
The combined gas law shows that the density of an ideal gas varies inversely with the absolute temperature and directly with the pressure.
5.13. A mixture of gases at 760 torr contains 65.0% nitrogen, 15.0% oxygen, and 20.0% carbon dioxide by volume. What is the pressure of each gas in torr?
Dalton’s law of partial pressure states that the total pressure is the result of the sum of the pressures of the component gases in a mixture. Therefore, if the total pressure is 760 torr, then the pressure of the nitrogen would be 65.0% of 760 torr. Using this approach results in the pressures of each of the gases.
Notice that the sum of the pressures (494 + 114 + 152) is 760, which is our check on the work.
5.14. In a gaseous mixture at 20°C the partial pressures of the components are: hydrogen, 200 torr; carbon dioxide, 150 torr; methane, 320 torr; and ethylene, 105 torr. What is (a) the total pressure of the mixture and (b) the volume percent of hydrogen?
Total pressure = the sum of the partial pressure = 200 + 150 + 320 + 105 = 755 torr
(b) By applying Boyle’s law, it can be shown that the fraction of the total pressure (all the gases in the mixture occupy the same volume—constant volume) is the same as the fraction of the total volume (when each gas is at the same pressure).
5.15. A 200-mL flask contained oxygen at 200 torr, and a 300-mL flask contained nitrogen at 100 torr (Fig. 5-6). The two flasks were then interconnected so that gases flowed together in the combined volume. Assuming no change in temperature, (a) what was the partial pressure of each gas in the final mixture and (b) what was the total pressure?
Fig. 5-6
Since the final volume is the sum of the two volumes, it is 500 mL. And, applying Dalton’s law of partial pressure:
(a)
(b)
Total pressure = sum of the partial pressures = 80 torr + 60 torr = 140 torr
5.16. Exactly 100 cm3 of oxygen are collected over water at 23°C and 800 torr. Compute the standard volume (volume at S.T.P.) of the dry oxygen. Vapor pressure of water at 23°C is 21.1 torr.
The gas collected in this experiment is not just oxygen, but also contains some water vapor which must be mathematically removed to determine the pressure of just the oxygen.
Then, for dry oxygen, V1 = 100 cm3, T1 = 23 + 273 = 296 K, and P = 779 torr. In order to solve for the S.T.P. conditions, we can use the second set of variables (V2, T2, and P2) in the combined gas law and solve it for V2.
5.17. In a basal metabolism measurement timed at exactly 6 minutes, a patient exhaled 52.5 L of air, measured over water at 20°C. The vapor pressure of water at 20°C is 17.5 torr. The barometric pressure was 750 torr. The exhaled air analyzed at 16.75 volume % oxygen, and the inhaled air at 20.32 volume % oxygen (both on a dry basis). Neglecting any solubility of the gases in water and any difference in total volumes of inhaled and exhaled air, calculate the rate of oxygen consumption by the patient in cm3 (S.T.P.) per minute.
5.18. A quantity of gas is collected in a graduated tube over mercury. The volume of gas at 20°C is 50.0 cm3, and the level of the mercury in the tube is 200 mm above the outside mercury level (see Fig. 5-7). The barometer reads 750 torr. Find the volume at S.T.P. The vapor pressure of mercury is not significant at this temperature.
Fig. 5-7
After a gas is collected over a liquid, the receiver is often adjusted so that the level of the liquid inside and outside is the same. When this cannot be done conveniently, it is necessary to calculate the effect of the difference in levels.
Since the level of mercury inside the tube is 200 mm higher than outside, the pressure of the gas is 200 mm Hg = 200 torr less than the atmospheric pressure of 750 torr.
5.19. A portable air tank has a note on its gauge stating that 125 psi cannot be exceeded. What is the total pressure in (a) atm, (b) torr, (c) mm Hg, and (d) millibars?
Ans. (a) 9.5 atm; (b) 722.6 torr; (c) 7222.6 mm Hg; (d) 9500 millibars
5.20. Provide Boyles’ and Charles’ gas laws starting with the combined gas law equation.
Ans.
5.21. The vapor pressure of water at 25°C is 23.8 torr. Express this in (a) atm and (b) kPa.
Ans. (a) 0.0313 atm; (b) 3.17kPa
5.22. Camphor has been found to undergo a crystalline modification at a temperature of 148°C and a pressure of 3.09 × 109 N/m2. What is the transition pressure in atmospheres?
Ans. 3.05 × 104 atm
5.23. When solving gas law problems using the combined gas law, the pressure and volume units do not have to be as indicated by the authors of the laws and by the ideal gas law; they don’t even have to be in the metric system. However, temperature must be in the Kelvin scale. Explain.
Ans. The conversion factors used to change units to liters (V) and atm (P) will cancel since they are used in the same manner on both sides of the equation (multiplication/division in the same location on both sides with no addition/subtraction). However, the conversions from either °F or °C to K include addition/subtraction, which does not cancel.
5.24. A hydrogen-filled weather balloon occupies 48.0 ft3 at ground level where the pressure is 753 torr. Compute its volume at the top of the mountain where it is about to be launched when the temperature is the same as at ground level, but the pressure is only 652 torr. (We don’t need to account for the elasticity of the skin.)
Ans. 55.4 ft3
5.25. During a fire, the gases in a room will expand and can eventually break out of tight-fitting windows. Suppose a room 10 feet by 12 feet with a 7-foot ceiling originally at 1 atm and 25°C (room/lab conditions) begins to burn. The temperature in the room rises to 400°C (approximately 750°F). What will be the amount of pressure exerted by the heated air (no change in volume, yet)?
Ans. 2.26 atm
5.26. Ten liters of hydrogen under 1 atm pressure are contained in a cylinder which has a movable piston. The piston is moved in until the same mass of gas occupies 2 L with no temperature change. Find the pressure in the cylinder.
Ans. 5 atm
5.27. Chlorine gas is evolved at the anode of a commercial electrolysis cell at the rate of 3.65 L/min at a temperature of 647°C. On its way to the intake pump, it is cooled to 63°C. Calculate its rate of intake to the pump assuming the pressure has remained constant.
Ans. 1.33 L/min
5.28. A quantity of hydrogen is confined in a platinum chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is 1000 torr. (a) What is the Celsius temperature when the pressure manometer indicates an absolute pressure of 100 torr? (b) What pressure will be indicated when the chamber is brought to 100°C?
Ans. (a) –246°C; (b) 1366 torr
5.29. Since helium gas is a precious resource, the contents of a small passenger balloon were pumped to storage during an overhaul. Normally the balloon contains 18,700 ft3 of helium at 31 °C and 1.00 atm (14.7 lb/in2 or 14.7 psi). How many steel cylinders will be required to store it if each cylinder holds 2.50 ft3, the storage temperature is a constant 11°C, and the cylinders are rated safe up to 2000 psi?
Ans. 52 cylinders (calculated answer: 51.4 cylinders)
5.30. A gas at 50°C and 785 torr occupies 350 mL. What volume will the gas occupy at S.T.P.?
Ans. 306 mL
5.31. A reaction for the production of hydrogen gas suitable for use in a hydrogen-powered vehicle is given below. What volume of H2 can be prepared from 1.00 kg CaH2 (the final conditions are 25°C and 1 atm)?
Ans. 1160L
5.32. The gas evolved in a reaction was collected in a vessel like that shown in Fig. 5-7. Its volume was 47.3 cm3 when first observed at 752 torr and 26°C. The Hg level in the bulb was 279 mm below that in the collection tube. The next day, with the temperature at 17°C and the pressure at 729 torr, the volume was measured again, but first the bulb was adjusted to equalize the levels. What volume reading is expected?
Ans. 29.8 cm3
5.33. Exactly 500 cm3 of nitrogen are collected over water at 25°C and 755 torr. The gas is saturated with water vapor. Compute the volume of the nitrogen in the dry condition at S.T.P. The vapor pressure of water at 25°C is 23.8 torr.
Ans. 441 cm3
5.34. A dry gas occupied 127 cm3 at S.T.P. If this same mass of gas were collected over water at 23°C (total gas pressure 745 torr), what volume would it occupy? Water’s vapor pressure at 23°C is 21 torr.
Ans. 145 cm3
5.35. A mass of gas occupies 0.825 L at –30°C and 556 Pa. What is the pressure if the volume becomes 1 L and the temperature 20°C?
Ans. 553 Pa
5.36. A 57.3-L pressure vessel has a safety valve set to open at 875 kPa. A chemical reaction is expected to yield 472 L of gaseous product at S.T.P. Would it be a good idea to store this in the vessel described if the ambient temperature can rise to 105°F?
Ans. No; the final pressure could be 959 kPa.
5.37. An attempt to identify a colorless liquid is made by heating it to 100°C at 754 torr in a partially sealed 250 flask to replace the air in the flask with only the vapor as the liquid is noted to boil at about 65°C. The volume of the flask is 271 mL, determined by filling it with water and measuring the volume. The mass of the flask has increased by 0.284 g. What is the molecular mass of the unknown liquid?
Ans. 32 g/mol; the BP and the molecular mass are clues to the identity, which might be methyl alcohol, CH3OH.
5.38. The density of a gas at S.T.P. provides an easy measurement of molar mass (refer to Problem 6.1). For a normally liquid substance, however, it is necessary to measure the gas density at an elevated temperature and reduced pressure. Calculate the density in g/L (same as kg/m3) at S.T.P. of a gas which has a density of 3.45 g/L at 90°C and 638 torr.
Ans. 5.59 g/L
5.39. Airplanes are known to have less lift in warm weather than in cold. Compare the density of air at 30°C to air at S.T.P.
Ans. The density of the air will be 0.9 of the density at S.T.P., resulting in a loss of lift.
5.40. A container holds 2.55 g of neon at S.T.P. What mass of neon will it hold at 100°C and 10.0 atm?
Ans. 18.7g
5.41. Glass tubing is being chosen to produce a neon sign, but the glass must support 2.5 atm without breaking. The design for the sign requires the use of 10.5 g Ne gas in a total of 6.77 L volume for the entire sign. Operating temperature is expected to reach the highest at 78°C. Will this glass take the strain or should another tubing be located?
Ans. This tubing has sufficient strength to hold the pressure (2.2 atm is the expected), but not with much reserve. Common practice in design should allow for the chance that the pressure might increase; therefore, considering the use of a stronger tubing would be a prudent practice.
5.42. At the top of a mountain, the thermometer reads 10°C and the barometer reads 700 mm Hg. At the bottom of the same mountain, the temperature is 30°C and the pressure is 760 mm Hg. Compare the density of the air at the top with that at the bottom of the mountain.
Ans. 0.986 at the top and 1.000 at the bottom
5.43. A volume of 95 cm3 of nitrous oxide at 27°C is collected over mercury in a graduated tube. The level of mercury inside the tube is 60 mm above the outside level when the barometer reads 750 torr. (a) Calculate the volume of the gas at S.T.P. (b) What volume would the gas occupy at 40°C with a barometric pressure of 745 torr and the mercury level inside the tube 25 mm below the outside?
Ans. (a) 78 cm3; (b) 89 cm3
5.44. At a certain altitude in the upper atmosphere, the temperature is estimated to be –100°C and the density just 10–9 that of the earth’s atmosphere at S.T.P. Assuming a uniform atmospheric composition, what is the pressure in torr at this altitude?
Ans. 4.82 × 10–7 torr
5.45. At 0°C the density of nitrogen at 1 atm is 1.25 kg/m3. The nitrogen which occupied 1500 cm3 at S.T.P. was compressed at 0°C to 575 atm and the gas volume was observed to be 3.92cm3, in violation of Boyle’s law. What is the final density of this nonideal gas?
Ans. 478 kg/m3
5.46. One cylinder of a particular 8-cylinder engine has the largest volume of 625 mL. The fuel-air mixture in that cylinder (original pressure 1 atm) is compressed to 85 mL, then is ignited. (a) If the gases are present at the largest volume at 1 atm, what is the pressure on compression before ignition? (b) Express the compression ratio (volume ratio).
Ans. (a) 7.67 atm; (b) 7.35:1
5.47. Let us reconsider Problem 5.46 a little more realistically. The fuel-air mixture enters the engine at 18°C and, after compression, it is at 121°C just before being burned. If the original pressure before compression is 1 atm, (a) what is the pressure just before ignition? (b) Would the engine designed using the parameters in Problem 5.46 be suitable for the conditions of this problem?
Ans. 9.96 atm; (b) Considering that the pressure here is nearly 30% greater, some strengthening of the cylinder would be a wise move.
5.48. The respiration of a suspension of yeast cells was measured by observing the decrease in pressure of gas above the cell suspension. The apparatus shown in Fig. 5-8 was arranged so that the gas was confined to a constant volume of 16.0 cm3 and the entire pressure change was caused by the uptake of oxygen by the cells. The pressure was measured with the use of a manometer, the fluid of which had a density of 1.034 g/cm3. The supply of liquid was adjustable so that the level on the closed side was kept constant. The entire apparatus was maintained at 37°C. In a 30-minute observation period, the fluid in the open side of the manometer dropped 37 mm. Neglecting the solubility of oxygen in the yeast suspension, compute the rate of oxygen consumption by the cells in cubic millimeters of O2 (S.T.P.) per hour.
Fig. 5-8
Ans. 105mm3/hr
5.49. A mixture of N2, NO, and NO2 was analyzed by selective absorption of the oxides of nitrogen. The initial volume of the mixture was 2.74 cm3. After treatment with water, which absorbed the NO2, the volume was 2.02 cm3. A solution of FeSO4 in water was then shaken with the residual gas to absorb the NO, after which the volume was 0.25 cm3. All the volumes were measured at the same pressure. Neglecting water vapor, what was the volume percentage of each gas in the original mixture?
Ans. 9.1% N2; 64.6% NO; 26.3% NO2
5.50. HCN can be toxic at a concentration of 150 ppm. (a) What is the percent composition of the HCN in the air at this concentration? (b) What is the partial pressure of HCN at this level if the total pressure is 1 atm?
Ans. (a) 0.015% HCN; (b) 0.00015 atm
5.51. A handball with the internal volume of 60 cm3 was filled with air to a pressure of 1.35 atm. A dishonest player filled a syringe to the 25-cm3 mark with air at 1.00 atm and injected it into the handball. Calculate the pressure inside the tampered handball, assuming no volume change.
Ans. 1.77 atm
5.52. A 250-mL flask contained krypton at 500 torr. A 450-mL flask contained helium at 950 torr. The contents of the two flasks were mixed by opening a stopcock connecting them. Assuming that all operations were carried out at a constant temperature, calculate the final pressure and the volume percent of each gas in the mixture. Neglect the volume of the stopcock.
Ans. 789 torr; 22.6% Kr and 77.4% He
5.53. A glass vacuum tube was sealed at the factory at 750°C with a residual pressure of air of 4.5 × 10–7 torr. Then a metal “getter” was used to remove all the oxygen (which is 21% by volume of air). What was the final pressure in the tube at 22°C?
Ans. 1.03 × 10–7torr
5.54. The vapor pressure of water at 80°C is 355 torr. A 100-mL vessel contained water-saturated oxygen at 80°C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50-mL vessel at the same temperature. Assuming no condensation, (a) What were the partial pressures of oxygen and of the water vapor? (b) What was the total pressure in the final equilibrium state?
Ans. (a) 810 torr, 355 torr; (b) 1165 torr
5.55. A gas sample was collected in an apparatus similar to that in Fig. 5-7 except that the confining liquid was water. At 17°C, the volume of gas was 67.3 cm3, the barometric pressure was 723 torr, and the water level in the bulb was 210 mm below that in the collection tube. Later that day the room warmed to 34°C and the barometer went up to 741 torr. The experimenter slowly adjusted the bulb to equalize the levels. What was the new volume reading? The vapor pressure of water is 14.5 torr at 17°C and 39.9 torr at 34°C. The density of mercury is 13.6 times that of water.
Ans. 70.4 cm3