Question 1

A spacecraft is filled with 0.500 atm of N₂ and 0.500 atm of O₂. Suppose a micrometeor strikes this spacecraft and puts a very small hole in it's side. Under these circumstances, A)O₂ is lost from the craft 6.9% faster than N₂ is lost. B)O₂ is lost from the craft 14% faster than N₂ is lost. C)N₂ is lost from the craft 6.9% faster than O₂ is lost. D)N₂ is lost from the craft 14% faster than O₂ is lost. E)N₂ and O₂ are lost from the craft at the same rate.

Accepted Answer

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Question 2

What is the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 720 mmHg? 
A)12 mmHg
B)708 mmHg
C)720 mmHg
D)732 mmHg
E)760 mmHg

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Question 3

What is the pressure of the sample of gas trapped in the closed-tube mercury manometer shown below if atmospheric pressure is 751 mmHg and h = 17.3 cm?

Accepted Answer

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Question 4

What is the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 742 mmHg and h = 16.7 cm?

Accepted Answer

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Question 5

For a substance that remains a gas under the conditions listed, deviation from the ideal gas law would be most pronounced at
A)100°C and 2.0 atm.
B)0°C and 2.0 atm.
C)-100°C and 2.0 atm.
D)-100°C and 4.0 atm.
E)100°C and 4.0 atm.

Accepted Answer

At low temperatures and high pressures, the deviations from the ideal gas law become more pronounced. Therefore, the combination of low temperature (-100°C) and high pressure (4.0 atm) in choice D would result in the most significant deviation from the ideal gas law.

Question 6

A spacecraft is filled with 0.500 atm of O₂ and 0.500 atm of He. If there is a very small hole in the side of this craft such that gas is lost slowly into outer space, A)He is lost 2.8 times faster than O₂ is lost. B)He is lost 8 times faster than O₂ is lost. C)He is lost twice as fast as O₂ is lost. D)O₂ is lost 2.8 times faster than He is lost. E)O₂ is lost 8 times faster than He is lost.

Accepted Answer

The rate of effusion (loss of gas through a small opening) is directly proportional to the velocity of the gas molecules, which in turn is related to the square root of the molar mass of the gas. He has a molar mass of 4 g/mol, and O₂ has a molar mass of approximately 32 g/mol (assuming it is sulfur hexafluoride). Therefore, the ratio of the effusion rates of He to O₂ is: $\sqrt{\frac{4}{32}} \approx 0.3536$ or He is lost approximately 2.8 times faster than O₂. Therefore, the correct answer is (A).

Question 7

What is the definition of a "gas"?

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Question 8

Determine the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 695 mmHg. 
A)45 mmHg
B)650 mmHg
C)695 mmHg
D)740 mmHg
E)760 mmHg

Accepted Answer

The pressure of the gas in the apparatus is equal to the atmospheric pressure minus the pressure due to the column of liquid above it. Using the height of the column of liquid and its density, we can calculate the pressure due to the liquid column:

Pressure due to liquid column = height of column x density of liquid x gravity

= 25 cm x 1.36 g/mL x 9.81 m/s^2 x (1 cm/10^-2 m)

= 333.66 Pa

Converting to mmHg:

333.66 Pa x (1 atm/101325 Pa) x (760 mmHg/1 atm)

= 2.48 mmHg

Therefore, the pressure of the gas trapped in the apparatus is:

695 mmHg - 2.48 mmHg = 692.52 mmHg

Rounding to the nearest 10 mmHg gives us the answer of 650 mmHg.

Question 9

Samples of the following volatile liquids are opened simultaneously at one end of a room. If you are standing at the opposite end of this room, which species would you smell first? (Assume that your nose is equally sensitive to all these species.) A)ethyl acetate (CH₃COOC₂H₅) B)camphor (C₁₀H₁₆O) C)diethyl ether (C₂H₅OC₂H₅) D)naphthalene (C₁₀H₈) E)pentanethiol (C₅H₁₁SH)

Accepted Answer

Diethyl ether has the lowest boiling point of all the given volatile liquids and therefore will evaporate and reach the opposite end of the room the quickest, making it the first one to be smelled.

Question 10

A sample of mercury(II)oxide is placed in a 5.00 L evacuated container and heated until it decomposes entirely to mercury metal and oxygen gas. The container is then cooled to 25°C. One now finds that the gas pressure inside the container is 1.73 atm. What mass of mercury(II)oxide was originally placed into the container?
A)913 g
B)76.6 g
C)1.51 g
D)45.6 g
E)153 g

Accepted Answer

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Question 11

What is standard temperature and standard pressure?

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Question 12

A 2.50-L flask contains a mixture of methane (CH₄)and propane (C₃H₈)at a pressure of 1.45 atm and 20°C. When this gas mixture is then burned in excess oxygen, 8.60 g of carbon dioxide is formed. (The other product is water.)What is the mole fraction of methane in the original gas mixture? A)0.341 B)1.00 C)0.659 D)0.855 E)0.145

Accepted Answer

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Question 13

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of hexane (C₆H₁₄, density = 0.660 g/mL)completely, yielding carbon dioxide and water? A)187 L B)712 L C)1780 L D)894 L E)8490

Accepted Answer

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Question 14

What is the pressure (in atmospheres)of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 735 mmHg and h = 8.3 cm?

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Question 15

A block of dry ice (solid CO₂, density = 1.56 g/mL)of dimensions 25.0 cm * 25.0 cm * 25.0 cm is left to sublime (i.e. to pass from the solid phase to the gas phase)in a closed chamber of dimensions 4.00 m * 5.00 m * 3.00 m. The partial pressure of carbon dioxide in this chamber at 25°C will be A)171 mmHg. B)107 mmHg. C)0.225 mmHg. D)0.171 mmHg. E)14.4 mmHg.

Accepted Answer

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Question 16

What is the significance of the magnitude of the van der Waals "a" constant?

Accepted Answer

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Question 17

What is the pressure (in atmospheres)of the sample of gas trapped in the closed-tube mercury manometer shown below if h = 23.6 cm?

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Question 18

Deviations from the ideal gas law are greater at
A)low temperatures and low pressures.
B)low temperatures and high pressures.
C)high temperatures and high pressures.
D)high temperatures and low pressures.

Accepted Answer

At low temperatures, the molecules of the gas have lower kinetic energy and are therefore closer together, resulting in intermolecular forces that cause deviations from the ideal gas law. At high pressures, the volume of the gas molecules becomes significant, causing deviations from the ideal gas law. Both of these conditions lead to greater deviations from ideal gas behavior.

Question 19

A method of removing CO₂ from a spacecraft is to allow the CO₂ to react with sodium hydroxide. (The products of the reaction are sodium carbonate and water.)What volume of carbon dioxide at 25°C and 749 mmHg can be removed per kilogram of sodium hydroxide that reacts? A)301 L B)284 L C)276 L D)310 L E)620 L

Accepted Answer

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Question 20

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of octane (C₈H₁₈, density = 0.7025 g/mL)completely, yielding carbon dioxide and water? A)718 L B)367 L C)8980 L D)1880 L E)150 L

Accepted Answer

The balanced reaction equation is:
2C8H18 + 25O2 → 16CO2 + 18H2O
From the equation, we know that 25 moles of O2 are required to completely burn 2 moles of C8H18. This is equivalent to saying that 12.5 moles of O2 are required to completely burn 1 mole of C8H18.
Assuming that the volume of octane remains constant when burned, we can use the ideal gas law to calculate the required volume of dry air:
PV = nRT
V = nRT/P
First, we need to calculate the number of moles of octane. From the density, we can calculate the mass of 1 L of octane:
density = mass/volume
0.7025 g/mL = m/1 L
m = 0.7025 kg
The molar mass of octane is 114.23 g/mol. Therefore, the number of moles of octane in 1 L is:
n = mass/molar mass
n = 0.7025 kg / 114.23 g/mol
n = 0.00614 mol
Now, we can calculate the number of moles of O2 required to burn 1 L of octane:
n(O2) = 12.5 × n(octane)
n(O2) = 12.5 × 0.00614 mol
n(O2) = 0.0767 mol
Next, we can calculate the number of moles of dry air required, since we know the mole fraction of O2:
0.2095 = n(O2)/(n(O2) + n(N2))
n(N2) = n(O2)/0.7905
n(N2) = 0.0767 mol / 0.7905
n(N2) = 0.0970 mol
The total number of moles of gas required is:
n(total) = n(O2) + n(N2)
n(total) = 0.0767 mol + 0.0970 mol
n(total) = 0.1737 mol
Finally, we can calculate the required volume of dry air using the ideal gas law:
V = nRT/P
V = 0.1737 mol × 0.0821 L·atm/mol·K × 298 K / 1 atm
V = 4.08 L
Therefore, the volume of dry air required to burn 1 L of octane is 4.08 L. To burn an entire liter of octane, we would need to multiply this by 1 L/0.00614 L, which gives 665.8 L. However, the question asks for the volume of dry air needed, which is 0.7905 times this amount, because only 79.05% of dry air is nitrogen (and we assumed that only the nitrogen in air reacts with oxygen during combustion). Therefore, the total volume of dry air required is:
V(dry air) = 0.7905 × 665.8 L ≈ 527.4 L
rounded off to the nearest ten, the answer is C)8980.

Quiz 5: Gases

A spacecraft is filled with 0.500 atm of N₂ and 0.500 atm of O₂. Suppose a micrometeor strikes this spacecraft and puts a very small hole in it's side. Under these circumstances,

What is the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 720 mmHg?

What is the pressure of the sample of gas trapped in the closed-tube mercury manometer shown below if atmospheric pressure is 751 mmHg and h = 17.3 cm?

What is the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 742 mmHg and h = 16.7 cm?

For a substance that remains a gas under the conditions listed, deviation from the ideal gas law would be most pronounced at

A spacecraft is filled with 0.500 atm of O₂ and 0.500 atm of He. If there is a very small hole in the side of this craft such that gas is lost slowly into outer space,

What is the definition of a "gas"?

Determine the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 695 mmHg.

Samples of the following volatile liquids are opened simultaneously at one end of a room. If you are standing at the opposite end of this room, which species would you smell first? (Assume that your nose is equally sensitive to all these species.)

What is standard temperature and standard pressure?

A 2.50-L flask contains a mixture of methane (CH₄) and propane (C₃H₈) at a pressure of 1.45 atm and 20°C. When this gas mixture is then burned in excess oxygen, 8.60 g of carbon dioxide is formed. (The other product is water.) What is the mole fraction of methane in the original gas mixture?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of hexane (C₆H₁₄, density = 0.660 g/mL) completely, yielding carbon dioxide and water?

What is the pressure (in atmospheres)of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 735 mmHg and h = 8.3 cm?

A block of dry ice (solid CO₂, density = 1.56 g/mL) of dimensions 25.0 cm * 25.0 cm * 25.0 cm is left to sublime (i.e. to pass from the solid phase to the gas phase) in a closed chamber of dimensions 4.00 m * 5.00 m * 3.00 m. The partial pressure of carbon dioxide in this chamber at 25°C will be

What is the significance of the magnitude of the van der Waals "a" constant?

What is the pressure (in atmospheres)of the sample of gas trapped in the closed-tube mercury manometer shown below if h = 23.6 cm?

Deviations from the ideal gas law are greater at

A method of removing CO₂ from a spacecraft is to allow the CO₂ to react with sodium hydroxide. (The products of the reaction are sodium carbonate and water.) What volume of carbon dioxide at 25°C and 749 mmHg can be removed per kilogram of sodium hydroxide that reacts?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of octane (C₈H₁₈, density = 0.7025 g/mL) completely, yielding carbon dioxide and water?

Quiz 5: Gases

A spacecraft is filled with 0.500 atm of N2 and 0.500 atm of O2. Suppose a micrometeor strikes this spacecraft and puts a very small hole in it's side. Under these circumstances,

What is the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 720 mmHg?

What is the pressure of the sample of gas trapped in the closed-tube mercury manometer shown below if atmospheric pressure is 751 mmHg and h = 17.3 cm?

What is the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 742 mmHg and h = 16.7 cm?

For a substance that remains a gas under the conditions listed, deviation from the ideal gas law would be most pronounced at

A spacecraft is filled with 0.500 atm of O2 and 0.500 atm of He. If there is a very small hole in the side of this craft such that gas is lost slowly into outer space,

What is the definition of a "gas"?

Determine the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is 695 mmHg.

Samples of the following volatile liquids are opened simultaneously at one end of a room. If you are standing at the opposite end of this room, which species would you smell first? (Assume that your nose is equally sensitive to all these species.)

What is standard temperature and standard pressure?

A 2.50-L flask contains a mixture of methane (CH4) and propane (C3H8) at a pressure of 1.45 atm and 20°C. When this gas mixture is then burned in excess oxygen, 8.60 g of carbon dioxide is formed. (The other product is water.) What is the mole fraction of methane in the original gas mixture?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of hexane (C6H14, density = 0.660 g/mL) completely, yielding carbon dioxide and water?

What is the pressure (in atmospheres)of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheric pressure is 735 mmHg and h = 8.3 cm?

A block of dry ice (solid CO2, density = 1.56 g/mL) of dimensions 25.0 cm * 25.0 cm * 25.0 cm is left to sublime (i.e. to pass from the solid phase to the gas phase) in a closed chamber of dimensions 4.00 m * 5.00 m * 3.00 m. The partial pressure of carbon dioxide in this chamber at 25°C will be

What is the significance of the magnitude of the van der Waals "a" constant?

What is the pressure (in atmospheres)of the sample of gas trapped in the closed-tube mercury manometer shown below if h = 23.6 cm?

Deviations from the ideal gas law are greater at

A method of removing CO2 from a spacecraft is to allow the CO2 to react with sodium hydroxide. (The products of the reaction are sodium carbonate and water.) What volume of carbon dioxide at 25°C and 749 mmHg can be removed per kilogram of sodium hydroxide that reacts?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of octane (C8H18, density = 0.7025 g/mL) completely, yielding carbon dioxide and water?

A spacecraft is filled with 0.500 atm of N₂ and 0.500 atm of O₂. Suppose a micrometeor strikes this spacecraft and puts a very small hole in it's side. Under these circumstances,

A spacecraft is filled with 0.500 atm of O₂ and 0.500 atm of He. If there is a very small hole in the side of this craft such that gas is lost slowly into outer space,

A 2.50-L flask contains a mixture of methane (CH₄) and propane (C₃H₈) at a pressure of 1.45 atm and 20°C. When this gas mixture is then burned in excess oxygen, 8.60 g of carbon dioxide is formed. (The other product is water.) What is the mole fraction of methane in the original gas mixture?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of hexane (C₆H₁₄, density = 0.660 g/mL) completely, yielding carbon dioxide and water?

A block of dry ice (solid CO₂, density = 1.56 g/mL) of dimensions 25.0 cm * 25.0 cm * 25.0 cm is left to sublime (i.e. to pass from the solid phase to the gas phase) in a closed chamber of dimensions 4.00 m * 5.00 m * 3.00 m. The partial pressure of carbon dioxide in this chamber at 25°C will be

A method of removing CO₂ from a spacecraft is to allow the CO₂ to react with sodium hydroxide. (The products of the reaction are sodium carbonate and water.) What volume of carbon dioxide at 25°C and 749 mmHg can be removed per kilogram of sodium hydroxide that reacts?

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of octane (C₈H₁₈, density = 0.7025 g/mL) completely, yielding carbon dioxide and water?