Question 1

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^$\circ$C will be

Accepted Answer

The volume of the dry ice block is 25.0 cm * 25.0 cm * 25.0 cm = 15625 cm³ = 15.625 L. The mass of the dry ice is 15.625 L * 1.56 g/mL = 24,375 g. The number of moles of CO₂ is 24,375 g / 44.01 g/mol = 554.1 mol. The volume of the chamber is 4.00 m * 5.00 m * 3.00 m = 60,000 L. Using the ideal gas law, PV = nRT, with R = 0.0821 L·atm/mol·K and T = 298 K, the pressure P = (554.1 mol * 0.0821 L·atm/mol·K * 298 K) / 60,000 L = 0.225 atm = 171 mmHg.

Question 2

1.000 atm of dry nitrogen, placed in a container having a pinhole opening in its side, leaks from the container 3.55 times faster than does 1.000 atm of an unknown gas placed in this same apparatus. Which of these species could be the unknown gas

Accepted Answer

The rate of effusion is inversely proportional to the square root of the molar mass (Graham's law). Since nitrogen effuses 3.55 times faster, the unknown gas must have a molar mass approximately (3.55)^2 times that of nitrogen (28 g/mol), which is about 353 g/mol. UF6 has a molar mass close to this value.

Question 3

Charles Law states that the volume of a fixed amount of gas maintained at constant pressure is inversely proportional to the absolute temperature of the gas.

Accepted Answer

Charles Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, not inversely proportional.

Question 4

A 2.50-L flask contains a mixture of methane (CH₄) and propane (C₃H₈) at a pressure of 1.45 atm and 20^$\circ$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

Accepted Answer

The answer of A 2.50-L flask contains a mixture of...

Question 5

1.000 atm of oxygen gas, placed in a container having a pinhole opening in its side, leaks from the container 2.14 times faster than does 1.000 atm of an unknown gas placed in this same apparatus. Which of these species could be the unknown gas

Accepted Answer

The rate of effusion (leakage) of a gas is inversely proportional to the square root of its molar mass. This means that the gas with the smaller molar mass will effuse faster. Among the given options, only SF6 has a smaller molar mass than oxygen, and therefore it could be the unknown gas. The other options, Cl2, Kr, UF6, and Xe, all have larger molar masses than oxygen and would therefore effuse more slowly, contradicting the given information.

Question 6

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

Accepted Answer

The answer of What is the pressure (in atmospheres) of...

Question 7

What is the pressure (mmHg) 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

The pressure of the gas can be calculated using the equation:

Pgas = Patm + ρgh

where Pgas is the pressure of the gas, Patm is the atmospheric pressure, ρ is the density of the mercury, g is the acceleration due to gravity, and h is the height difference between the two arms of the manometer.

Substituting the given values:

Pgas = 742 + (13.6 x 9.81 x 0.167)
Pgas = 574.88 ≈ 575 mmHg

Therefore, the pressure of the gas trapped in the open-tube mercury manometer is approximately 575 mmHg. The correct answer is C.

Question 8

What is the pressure (mmHg) 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

The pressure of the gas can be calculated using the equation: 
Pgas = Patm + h

where Pgas is the pressure of the gas, Patm is the atmospheric pressure, and h is the height difference between the two arms of the manometer.

In this case, the height difference is given as h = 17.3 cm. And the atmospheric pressure is given as Patm = 751 mmHg.

To convert the height from cm to mmHg, we need to multiply it by the conversion factor 1 cm = 10 mmHg. So, h = 17.3 cm x 10 mmHg/cm = 173 mmHg.

Therefore, the pressure of the gas is:
Pgas = Patm + h = 751 mmHg + 173 mmHg = 924 mmHg.

The correct answer is A) 173 mmHg.

Question 9

A "gas" is a substance in which the molecules are separated on the average by distances that are large compared with the sizes of the molecules.

Accepted Answer

This is the definition of a gas according to the kinetic theory of gases.

Question 10

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,

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 11

Boyle's Law states that the volume of a fixed amount of gas maintained at constant temperature is directly proportional to the gas pressure

Accepted Answer

Boyle's Law states that the volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure.

Question 12

The magnitude of the van der Waals "a" constant reflects the strength of the attractions between molecules of a given type of gas.

Accepted Answer

This is a true statement. The "a" constant in the van der Waals equation represents the strength of intermolecular attractions between molecules of a gas. A higher value of "a" indicates stronger attractions and thus greater potential for the gas to condense or liquefy at a given temperature and pressure.

Question 13

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.)

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 14

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

Accepted Answer

The answer of The mole fraction of oxygen molecules in...

Question 15

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

Accepted Answer

The answer of The mole fraction of oxygen molecules in...

Question 16

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 its side. Under these circumstances,

Accepted Answer

The answer of A spacecraft is filled with 0.500 atm...

Question 17

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^$\circ$C and 749 mmHg can be removed per kilogram of sodium hydroxide that reacts

Accepted Answer

The first step is to write the balanced chemical equation for the reaction between CO₂ and sodium hydroxide: CO₂ + 2 NaOH → Na2CO3 + H2O From this equation, we can see that one mole of CO₂ reacts with two moles of NaOH to produce one mole of CO2. Therefore, we need to know how many moles of NaOH are present in one kilogram of NaOH. The molar mass of NaOH is 40.00 g/mol, so one kilogram contains 25 mol: 1 kg NaOH × (1000 g/kg) / (40.00 g/mol) = 25 mol NaOH Since two moles of NaOH react with one mole of CO₂ to produce one mole of CO2, we can remove 12.5 mol of CO₂ with one kilogram of NaOH. Using the ideal gas law, we can calculate the volume of CO2 produced: PV = nRT V = nRT/P V = (12.5 mol)(0.0821 L·atm/mol·K)(298 K)/(749 mmHg × 101.3 kPa/1 atm) V = 310 L Therefore, the answer is D, 310 L of CO2 can be removed per kilogram of NaOH that reacts.

Question 18

The standard temperature and standard pressure in units of ^$\circ$C and atm are 25^$\circ$C and 1 atm pressure.

Accepted Answer

The standard temperature is 0°C (32°F), not 25°C, and the standard pressure is indeed 1 atm.

Question 19

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

Accepted Answer

We can use the following formula to calculate the pressure of the gas:

P_gas = P_atm + (h/13.6)

Where P_atm is the atmospheric pressure (which we assume to be 1 atm), h is the height difference between the two arms of the manometer, and 13.6 is the density of mercury in g/cm^3.

Plugging in the values, we get:

P_gas = 1 atm + (23.6 cm /13.6 cm/g)

P_gas = 1.73 atm

However, we need to remember that the pressure of the gas is balanced by the pressure of the mercury column in the other arm. Since the mercury level is higher on the left side of the manometer, we need to subtract the difference in height between the two arms:

P_gas = 1.73 atm - (23.6 cm / 13.6 cm/g)

P_gas = 0.311 atm

Therefore, the pressure of the gas trapped in the manometer is 0.311 atm, which corresponds to choice B.

Question 20

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^$\circ$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

Accepted Answer

The answer of A sample of mercury(II) oxide is placed...

Quiz 5: Gases