Question 1

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr)when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.

Accepted Answer

Boyle's Law states that P1V1 = P2V2 for a constant temperature. Using the given values: (667 torr)(10.0 L) = (P2)(7.88 L), solving for P2 gives approximately 846 torr.

Question 2

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 11.8 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was __________ mm Hg.

Accepted Answer

In a closed-end manometer, the pressure of the gas is equal to the difference in mercury levels. Therefore, the pressure of the argon is 11.8 cm, which is equivalent to 118 mm Hg (since 1 cm Hg = 10 mm Hg).

Question 3

A sample of He gas (2.35 mol)occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is __________ L at 423 K and 1.00 atm.

Accepted Answer

Using Charles's Law (V1/T1 = V2/T2), where V1 = 57.9 L, T1 = 300.0 K, and T2 = 423 K, we solve for V2: V2 = V1 * (T2/T1) = 57.9 L * (423 K / 300.0 K) = 81.6 L.

Question 4

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00°C is __________ atm. (a = 6.49 L²-atm/mol², b = 0.0562 L/mol)

Accepted Answer

The answer of Using the van der Waals equation, the...

Question 5

If 3.21 mol of a gas occupies 56.2 L at 44°C and 793 torr, 5.21 mol of this gas occupies __________ L under these conditions.

Accepted Answer

Using the ideal gas law, the volume is directly proportional to the number of moles when temperature and pressure are constant. Therefore, (5.21 mol / 3.21 mol) * 56.2 L = 91.2 L.

Question 6

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0°C. The total pressure in the container was __________ atm.

Accepted Answer

The answer of CO (5.00 g)and CO₂ (5.00 g)were placed...

Question 7

A sample of He gas (2.0 mmol)effused through a pinhole in 53 s. The same amount of an unknown gas, under the same conditions, effused through the pinhole in 248 s. The molecular mass of the unknown gas is __________ g/mol.

Accepted Answer

According to Graham's Law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Using this law, we can set up the following equation:

(rate of He gas effusion) / (rate of unknown gas effusion) = sqrt(Molar Mass of unknown gas) / sqrt(Molar Mass of He gas)

Plugging in the given values, we get:

(2.0 mmol / 53 s) / (2.0 mmol / 248 s) = sqrt(Molar Mass of unknown gas) / sqrt(4.0 g/mol)

Simplifying this equation gives:

sqrt(Molar Mass of unknown gas) = sqrt(4.0 g/mol) * (2.0 mmol / 53 s) / (2.0 mmol / 248 s)

Solving for the unknown molar mass gives:

Molar Mass of unknown gas = (4.0 g/mol) * (248 s / 53 s)² = 88 g/mol

Therefore, the correct answer is (C) 88.

Question 8

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100°C is __________ atm. (a = 0.211 L²-atm/mol², b = 0.0171 L/mol)

Accepted Answer

The answer of Using the van der Waals equation, the...

Question 9

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

Accepted Answer

The answer of CO (5.00 g)and CO₂ (5.00 g)were placed...

Question 10

A sample of an ideal gas (3.00 L)in a closed container at 25.0°C and 76.0 torr is heated to 340°C. The pressure of the gas at this temperature is __________ torr.

Accepted Answer

The answer of A sample of an ideal gas (3.00...

Question 11

A sample of gas (24.2 g)initially at 6.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was __________ atm.

Accepted Answer

The answer of A sample of gas (24.2 g)initially at...

Question 12

The root-mean-square speed of CO at 113°C is __________ m/s.

Accepted Answer

Using the root-mean-square speed equation, 
u(rms) = sqrt[(3RT)/M] 
where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

M(CO) = 12.01 g/mol + 15.99 g/mol = 28.00 g/mol

Converting the temperature to Kelvin: 
113°C + 273.15 = 386.15 K

Substituting the values and solving for u(rms):
u(rms) = sqrt[(3 x 8.31 J/mol.K x 386.15 K)/28.00 g/mol] 
u(rms) = 586 m/s

Therefore, the answer is C.

Question 13

A sample of N₂ gas (2.0 mmol)effused through a pinhole in 5.5 s. It will take __________ s for the same amount of CH₄ to effuse under the same conditions.

Accepted Answer

The effusion rate of a gas is inversely proportional to the square root of its molar mass. Since the gases in question have the same volume and pressure, the ratio of their effusion rates is given by the ratio of the square roots of their molar masses. The molar mass of N₂ is 60 g/mol (28 + 32). The molar mass of CH₄ is 76 g/mol (12 + 1 + 32 + 1 + 16 + 1 + 14). So, the effusion rate of CH₄ relative to N₂ is: sqrt(60/76) ≈ 0.876 Therefore, it will take about 5.5 x 0.876 = 4.81 s for the same amount of CH₄ to effuse under the same conditions. The closest choice is D, 4.2 s.

Question 14

A balloon originally had a volume of 4.39 L at 44°C and a pressure of 729 torr. The balloon must be cooled to __________°C to reduce its volume to 3.78 L (at constant pressure).

Accepted Answer

The answer of A balloon originally had a volume of...

Question 15

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L)the gas will occupy when the temperature is increased to 121°C while maintaining the pressure at 629 torr.

Accepted Answer

Charles's law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature. We can use the formula:

V1/T1 = V2/T2

Where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Substituting the values given in the problem, we get:

V1 = 10.0 L
T1 = 25.0 + 273 = 298 K
P = 629 torr
T2 = 121.0 + 273 = 394 K

V2 = (V1/T1) x T2 = (10.0/298) x 394 = 13.2 L

Therefore, the volume of the gas will be 13.2 L when the temperature is increased to 121°C while maintaining the pressure at 629 torr.

The correct answer is B) 13.2.

Question 16

A sample of O₂ gas (2.0 mmol)effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO₂ to effuse under the same conditions.

Accepted Answer

The answer of A sample of O₂ gas (2.0 mmol)effused...

Question 17

A gas originally at 27°C and 1.00 atm pressure in a 3.9 L flask is cooled at constant pressure until the temperature is 11°C. The new volume of the gas is __________ L.

Accepted Answer

According to the Charles's law, at constant pressure, the volume of a gas is directly proportional to its absolute temperature. Therefore, we can use the following equation to solve for the new volume:

(V1/T1) = (V2/T2)

where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature, respectively.

Plugging in the given values, we get:

(3.9 L / 300 K) = (V2 / 284 K)

Solving for V2, we get:

V2 = (3.9 L / 300 K) x 284 K = 3.7 L

Therefore, the new volume of the gas is 3.7 L.

The answer is B.

Question 18

SO₂ (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

Accepted Answer

The answer of SO₂ (5.00 g)and CO₂ (5.00 g)were placed...

Question 19

CO (5.00 g)and CO₂ (5.00 g)were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO in the container was __________ atm.

Accepted Answer

The molar mass of CO is 28.01 g/mol. Therefore, 5.00 g of CO is equivalent to $ \frac{5.00 \, \text{g}}{28.01 \, \text{g/mol}} = 0.1785 \, \text{mol} $. Using the ideal gas law $ PV = nRT $, where $ P $ is the pressure in atm, $ V $ is the volume in liters (0.750 L), $ n $ is the number of moles of gas (0.1785 mol), $ R $ is the ideal gas constant (0.0821 L·atm/(mol·K)), and $ T $ is the temperature in Kelvin (50.0 °C = 323.15 K), we can solve for $ P $. $$ P = \frac{nRT}{V} = \frac{(0.1785 \, \text{mol})(0.0821 \, \text{L·atm/(mol·K)})(323.15 \, \text{K})}{0.750 \, \text{L}} = 6.29 \, \text{atm} $$

Question 20

A pressure of 1.25 atm is the same as a pressure of __________ of mm Hg.

Accepted Answer

To convert from atm to mm Hg, we use the conversion factor 1 atm = 760 mm Hg. Therefore, we can set up a proportion:
1.25 atm / 1 = x mm Hg / 760 mm Hg
Cross-multiplying and solving for x, we get:
x = 950 mm Hg
Therefore, the pressure of 1.25 atm is equivalent to 950 mm Hg.

Quiz 10: Gases

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 11.8 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was __________ mm Hg.

A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is __________ L at 423 K and 1.00 atm.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00°C is __________ atm. (a = 6.49 L²-atm/mol², b = 0.0562 L/mol)

If 3.21 mol of a gas occupies 56.2 L at 44°C and 793 torr, 5.21 mol of this gas occupies __________ L under these conditions.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The total pressure in the container was __________ atm.

A sample of He gas (2.0 mmol) effused through a pinhole in 53 s. The same amount of an unknown gas, under the same conditions, effused through the pinhole in 248 s. The molecular mass of the unknown gas is __________ g/mol.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100°C is __________ atm. (a = 0.211 L²-atm/mol², b = 0.0171 L/mol)

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

A sample of an ideal gas (3.00 L) in a closed container at 25.0°C and 76.0 torr is heated to 340°C. The pressure of the gas at this temperature is __________ torr.

A sample of gas (24.2 g) initially at 6.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was __________ atm.

The root-mean-square speed of CO at 113°C is __________ m/s.

A sample of N₂ gas (2.0 mmol) effused through a pinhole in 5.5 s. It will take __________ s for the same amount of CH₄ to effuse under the same conditions.

A balloon originally had a volume of 4.39 L at 44°C and a pressure of 729 torr. The balloon must be cooled to __________°C to reduce its volume to 3.78 L (at constant pressure) .

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L) the gas will occupy when the temperature is increased to 121°C while maintaining the pressure at 629 torr.

A sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO₂ to effuse under the same conditions.

A gas originally at 27°C and 1.00 atm pressure in a 3.9 L flask is cooled at constant pressure until the temperature is 11°C. The new volume of the gas is __________ L.

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO in the container was __________ atm.

A pressure of 1.25 atm is the same as a pressure of __________ of mm Hg.

Quiz 10: Gases

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 11.8 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was __________ mm Hg.

A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is __________ L at 423 K and 1.00 atm.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00°C is __________ atm. (a = 6.49 L2-atm/mol2, b = 0.0562 L/mol)

If 3.21 mol of a gas occupies 56.2 L at 44°C and 793 torr, 5.21 mol of this gas occupies __________ L under these conditions.

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0°C. The total pressure in the container was __________ atm.

A sample of He gas (2.0 mmol) effused through a pinhole in 53 s. The same amount of an unknown gas, under the same conditions, effused through the pinhole in 248 s. The molecular mass of the unknown gas is __________ g/mol.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100°C is __________ atm. (a = 0.211 L2-atm/mol2, b = 0.0171 L/mol)

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO2 in the container was __________ atm.

A sample of an ideal gas (3.00 L) in a closed container at 25.0°C and 76.0 torr is heated to 340°C. The pressure of the gas at this temperature is __________ torr.

A sample of gas (24.2 g) initially at 6.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was __________ atm.

The root-mean-square speed of CO at 113°C is __________ m/s.

A sample of N2 gas (2.0 mmol) effused through a pinhole in 5.5 s. It will take __________ s for the same amount of CH4 to effuse under the same conditions.

A balloon originally had a volume of 4.39 L at 44°C and a pressure of 729 torr. The balloon must be cooled to __________°C to reduce its volume to 3.78 L (at constant pressure) .

A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L) the gas will occupy when the temperature is increased to 121°C while maintaining the pressure at 629 torr.

A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse under the same conditions.

A gas originally at 27°C and 1.00 atm pressure in a 3.9 L flask is cooled at constant pressure until the temperature is 11°C. The new volume of the gas is __________ L.

SO2 (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO2 in the container was __________ atm.

CO (5.00 g) and CO2 (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO in the container was __________ atm.

A pressure of 1.25 atm is the same as a pressure of __________ of mm Hg.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00°C is __________ atm. (a = 6.49 L²-atm/mol², b = 0.0562 L/mol)

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The total pressure in the container was __________ atm.

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100°C is __________ atm. (a = 0.211 L²-atm/mol², b = 0.0171 L/mol)

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

A sample of N₂ gas (2.0 mmol) effused through a pinhole in 5.5 s. It will take __________ s for the same amount of CH₄ to effuse under the same conditions.

A sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO₂ to effuse under the same conditions.

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO₂ in the container was __________ atm.

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 mL container at 50.0°C. The partial pressure of CO in the container was __________ atm.