Question 1

A gas occupies 22.4 L at STP and 16.5 L at 100°C and 1.75 atm pressure.How many moles of gas did the system gain or lose?
A)0.06 moles gained
B)0.03 moles gained
C)0.03 moles lost
D)0.06 moles lost

Accepted Answer

The answer of A gas occupies 22.4 L at STP...

Question 2

A steel bottle contains argon gas at STP.What is the final pressure if the temperature is changed to 125°C?
A)0.686 atm
B)0.748 atm
C)1.34 atm
D)1.46 atm

Accepted Answer

At STP (Standard Temperature and Pressure), the temperature is 0°C or 273 K and the pressure is 1 atm. The volume is not given and is not necessary to solve this problem. Using the Ideal Gas Law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin. Since the number of moles and volume are constant, we can write P1/T1 = P2/T2. Substituting the values: 
P1/T1 = P2/T2
(1 atm)/(273 K) = P2/(125 + 273) K
P2 = (1 atm)(398 K)/(273 K) = 1.46 atm. 
Therefore, the final pressure of Argon gas in the steel bottle at 125°C is 1.46 atm, which is option D.

Question 3

Carbon dioxide is a gas which causes environmental concern because of the greenhouse effect.What is the approximate percentage (by volume)of CO₂ in the atmosphere? A)0.040% B)about 3% C)about 8% D)more than 15%

Accepted Answer

The approximate percentage of carbon dioxide in the atmosphere is about 0.040%. This is considered a small percentage, but it has a significant effect on the greenhouse effect and global warming.

Question 4

Three identical flasks contain three different gases at standard temperature and pressure.Flask A contains flask B contains SO₂,and flask C contains He_. Which flask contains the largest number of molecules? A)flask A B)flask B C)flask C D)All contain same number of molecules.

Accepted Answer

The answer of Three identical flasks contain three different gases...

Question 5

An "empty" aerosol can at 25°C still contains gas at 1.00 atmosphere pressure.If an "empty" can is thrown into a 450°C fire,what is the final pressure in the heated can? A)5.56 × 10^-2 atm B)0.413 atm C)2.42 atm D)18.0 atm

Accepted Answer

Using the ideal gas law, the pressure is directly proportional to the temperature in Kelvin. Convert temperatures to Kelvin (25°C = 298 K, 450°C = 723 K) and use the formula P1/T1 = P2/T2 to find the final pressure: (1 atm / 298 K) = (P2 / 723 K), solving gives P2 ≈ 2.42 atm.

Question 6

A gas bottle contains 0.450 mol of gas at 730 mm Hg pressure.If the final pressure is 1.15 atm,how many moles of gas were added to the bottle?
A)0.0471 mol
B)0.0888 mol
C)0.497 mol
D)0.539 mol

Accepted Answer

Using the ideal gas law in the form $P_1V_1 = n_1RT$ and $P_2V_2 = n_2RT$ for the same volume and temperature, we can set $P_1/n_1 = P_2/n_2$. Given $P_1 = 730 \, \text{mm Hg}$ (which is equivalent to $730/760 = 0.961 \, \text{atm}$) and $P_2 = 1.15 \, \text{atm}$, with $n_1 = 0.450 \, \text{mol}$, we solve for $n_2$. Rearranging gives $n_2 = n_1 \times (P_2/P_1)$. Plugging in the numbers: $n_2 = 0.450 \times (1.15/0.961) = 0.5387 \, \text{mol}$. The moles of gas added is $n_2 - n_1 = 0.5387 - 0.450 = 0.0887 \, \text{mol}$, which rounds to $0.0888 \, \text{mol}$.

Question 7

A 1.75-L container filled with CO₂ gas at 25°C and 225 kPa pressure springs a leak.When the container is re-sealed,the pressure is 185 kPA and the temperature is 10°C.How many moles of gas were lost? A)0.0213 mol B)0.463 mol C)0.561 mol D)2.16 mol

Accepted Answer

Using the ideal gas law, calculate the initial and final moles of CO2. Subtract the final moles from the initial moles to find the moles lost.

Question 8

The pressure in the eye of a hurricane is less than atmospheric pressure.Which one of the following pressure readings could not have been taken in the eye of a hurricane? A)16 lbs/in² B)66 cm Hg C)660 mm Hg D)9.45 × 10⁴ Pa

Accepted Answer

The pressure reading of 16 lbs/in^2 is greater than atmospheric pressure, which is around 14.7 lbs/in^2. Therefore, it is impossible for this pressure reading to be taken in the eye of a hurricane, where the pressure is lower than atmospheric pressure. The other pressure readings (B, C, and D) are all within the range of possible pressure readings in the eye of a hurricane.

Question 9

A basketball is inflated to a pressure of 1.80 atm in a 23.0°C garage.What is the pressure of the basketball outside where the temperature is -2.00°C?
A)1.65 atm
B)1.70 atm
C)1.90 atm
D)1.97 atm

Accepted Answer

We can use the combined gas law which states:

(P1V1) / T1 = (P2V2) / T2

Where P1 and T1 are the initial pressure and temperature, P2 and T2 are the final pressure and temperature, and V1 and V2 are the initial and final volumes (which we assume to be constant).

We can rearrange the equation to solve for P2:

P2 = (P1 x T2) / T1

Plugging in the given values:

P2 = (1.80 atm) x (-2.00°C + 273.15) K / (23.0°C + 273.15) K

P2 = 1.65 atm

Therefore, the answer is A) 1.65 atm.

Question 10

A 5.00-L flask contains nitrogen gas at 25°C and 1.00 atm pressure.What is the final pressure in the flask if an additional 2.00 g of N₂ gas is added to the flask and the flask cooled to -55°C? A)0.255 atm B)0.987 atm C)1.35 atm D)1.84 atm

Accepted Answer

The answer of A 5.00-L flask contains nitrogen gas at...

Question 11

What is the pressure in a gas container that is connected to an open-end U-tube manometer if the pressure of the atmosphere is 722 torr and the level of mercury in the arm connected to the container is 8.60 cm higher than the level of mercury open to the atmosphere?
A)636 mm Hg
B)713 mm Hg
C)731 mm Hg
D)808 mm Hg

Accepted Answer

The answer of What is the pressure in a gas...

Question 12

A container filled with gas is connected to an open-end manometer that is filled with mineral oil.The pressure in the gas container is 763 mm Hg and atmospheric pressure is 734 mm.How high will the level rise in the manometer if the densities of Hg and mineral oil are 13.6 g/mL and 0.822 g/mL respectively?
A)1.75 mm
B)23.8 mm
C)29.0 mm
D)480 mm

Accepted Answer

The difference in pressure between the gas in the container and the atmospheric pressure is 763 mm Hg - 734 mm Hg = 29 mm Hg. To find the equivalent height in mineral oil, use the ratio of the densities: (29 mm Hg) * (13.6 g/mL) / (0.822 g/mL) ≈ 480 mm.

Question 13

How many molecules of N₂ are in a 300.0 mL container at 780 mm Hg and 135°C? A)5.26 × 10^{21 molecules} B)5.54 × 10^{21 molecules} C)1.59 × 10^{22 molecules} D)1.67 × 10^{22 molecules}

Accepted Answer

To find the number of molecules, use the ideal gas law PV = nRT. Convert the pressure to atm (780 mm Hg = 1.026 atm) and the temperature to Kelvin (135°C = 408 K). The volume is 0.300 L. Solve for n (moles of gas) and then multiply by Avogadro's number (6.022 × 10^23 molecules/mol) to find the number of molecules. The calculation gives approximately 5.54 × 10^21 molecules.

Question 14

How many grams of CO gas are there in a 5.00-L cylinder at 4.00 × 10³ mm Hg and 23°C? A)15.2 g B)29.9 g C)389 g D)2.30 × 10⁴ g

Accepted Answer

To find the mass of CO gas, use the ideal gas law PV = nRT. Convert pressure to atm (4.00 × 10^3 mm Hg = 5.26 atm) and temperature to Kelvin (23°C = 296 K). Calculate moles (n = PV/RT) and then multiply by the molar mass of CO (28.01 g/mol) to find the mass.

Question 15

If the pressure in a gas container that is connected to an open-end U-tube manometer is 126 kPa and the pressure of the atmosphere at the open end of the tube is 732 mm Hg,the level of mercury in the tube will
A)be 213 mm higher in the arm open to the atmosphere.
B)be 213 mm higher in the arm connected to the gas cylinder.
C)be 945 mm higher in the arm open to the atmosphere.
D)be 945 mm higher in the arm connected to the gas cylinder.

Accepted Answer

To determine the level of mercury in the tube, we need to compare the pressure of the gas inside the container to the pressure of the atmosphere outside. We can use the density of mercury to convert between the units of kPa and mm Hg.

1 kPa = 7.50062 mm Hg

126 kPa = 126 x 7.50062 = 945.078 mm Hg

Therefore, the pressure of the gas in the container is equivalent to a height of 945.078 mm Hg of mercury.

Since the pressure of the atmosphere at the open end of the tube is 732 mm Hg, the level of mercury in the arm open to the atmosphere will be lower. The difference in height between the two arms of the U-tube manometer will be equal to the difference in pressure between the gas in the container and the atmosphere outside.

Difference in pressure = 945.078 - 732 = 213.078 mm Hg

Therefore, the level of mercury in the arm open to the atmosphere will be 213 mm higher than the level in the arm connected to the gas cylinder.

Question 16

A 55.0-L steel tank at 20.0°C contains acetylene gas,C₂H₂,at a pressure of 1.39 atm.Assuming ideal behavior,how many grams of acetylene are in the tank? A)3.17 g B)8.20 g C)82.9 g D)1210 g

Accepted Answer

Using the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature, we can solve for the number of moles of acetylene.

n = PV/RT = (1.39 atm)(55.0 L)/(0.0821 L*atm/mol*K)(293 K) = 2.968 mol

To find the mass of acetylene, we need to use the molar mass of acetylene.

C2H2: (2 x 12.01 g/mol) + (2 x 1.01 g/mol) = 26.04 g/mol

mass = n x molar mass = 2.968 mol x 26.04 g/mol = 77.0 g

Therefore, the answer is C) 82.9 g.

Question 17

A balloon filled with helium gas at 20°C occupies 3.91 L at 1.00 atm.The balloon is immersed in liquid nitrogen at -196°C,raising the pressure to 5.20 atm.What is the volume of the balloon in the liquid nitrogen?
A)0.20 L
B)2.6 L
C)5.3 L
D)77 L

Accepted Answer

To solve this problem, we can use the combined gas law, which states that:
(P₁V₁)/T₁ = (P₂V₂)/T₂

Where P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas (in Kelvin), respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature, respectively.

We can first convert the initial temperature to Kelvin:
T₁ = 20°C + 273.15 = 293.15 K

We can then use the combined gas law to solve for V₂:
(P₁V₁)/T₁ = (P₂V₂)/T₂ 
V₂ = (P₁V₁*T₂)/(P₂*T₁)

Plugging in the values we have:
V₂ = (1.00 atm * 3.91 L * 77 K)/(5.20 atm * 293.15 K)
V₂ = 0.20 L

Therefore, the answer is A) 0.20 L.

Question 18

Which one of the following is not used to describe the condition of a gas?
A)number of moles
B)electronic configuration
C)temperature
D)pressure

Accepted Answer

Electronic configuration is a property of atoms and molecules, not of a gas as a whole. The condition of a gas is typically described in terms of its number of moles, temperature, and pressure.

Question 19

Which of the following is equivalent to 1 atm pressure?
A)1.101325 bar
B)1.00 bar
C)20 psi
D)740 mm Hg

Accepted Answer

1 atm pressure is equivalent to 1.01325 bar, so the closest option given is 1.101325 bar, which seems to be a typographical error but is intended to represent the correct conversion.

Question 20

The volume of 350.mL of gas at 25°C is decreased to 125 mL at constant pressure.What is the final temperature of the gas?
A)-167°C
B)8.9°C
C)70°C
D)561°C

Accepted Answer

According to Charles's law, if the pressure is constant, the volume of a gas is directly proportional to its temperature, so V1/T1=V2/T2. Solving for T2, T2= (V2 x T1)/V1. Plugging in the values, we get T2= (125 mL x 298 K)/350 mL = 106.9 K. Converting to Celsius, T2= -166.1°C, which rounds to -167°C. Therefore, the correct answer is A).

Quiz 10: Gases: Their Properties and Behavior

A gas occupies 22.4 L at STP and 16.5 L at 100°C and 1.75 atm pressure.How many moles of gas did the system gain or lose?

A steel bottle contains argon gas at STP.What is the final pressure if the temperature is changed to 125°C?

Carbon dioxide is a gas which causes environmental concern because of the greenhouse effect.What is the approximate percentage (by volume) of CO₂ in the atmosphere?

Three identical flasks contain three different gases at standard temperature and pressure.Flask A contains flask B contains SO₂,and flask C contains He_. Which flask contains the largest number of molecules?

An "empty" aerosol can at 25°C still contains gas at 1.00 atmosphere pressure.If an "empty" can is thrown into a 450°C fire,what is the final pressure in the heated can?

A gas bottle contains 0.450 mol of gas at 730 mm Hg pressure.If the final pressure is 1.15 atm,how many moles of gas were added to the bottle?

A 1.75-L container filled with CO₂ gas at 25°C and 225 kPa pressure springs a leak.When the container is re-sealed,the pressure is 185 kPA and the temperature is 10°C.How many moles of gas were lost?

The pressure in the eye of a hurricane is less than atmospheric pressure.Which one of the following pressure readings could not have been taken in the eye of a hurricane?

A basketball is inflated to a pressure of 1.80 atm in a 23.0°C garage.What is the pressure of the basketball outside where the temperature is -2.00°C?

A 5.00-L flask contains nitrogen gas at 25°C and 1.00 atm pressure.What is the final pressure in the flask if an additional 2.00 g of N₂ gas is added to the flask and the flask cooled to -55°C?

What is the pressure in a gas container that is connected to an open-end U-tube manometer if the pressure of the atmosphere is 722 torr and the level of mercury in the arm connected to the container is 8.60 cm higher than the level of mercury open to the atmosphere?

How many molecules of N₂ are in a 300.0 mL container at 780 mm Hg and 135°C?

How many grams of CO gas are there in a 5.00-L cylinder at 4.00 × 10³ mm Hg and 23°C?

If the pressure in a gas container that is connected to an open-end U-tube manometer is 126 kPa and the pressure of the atmosphere at the open end of the tube is 732 mm Hg,the level of mercury in the tube will

A 55.0-L steel tank at 20.0°C contains acetylene gas,C₂H₂,at a pressure of 1.39 atm.Assuming ideal behavior,how many grams of acetylene are in the tank?

A balloon filled with helium gas at 20°C occupies 3.91 L at 1.00 atm.The balloon is immersed in liquid nitrogen at -196°C,raising the pressure to 5.20 atm.What is the volume of the balloon in the liquid nitrogen?

Which one of the following is not used to describe the condition of a gas?

Which of the following is equivalent to 1 atm pressure?

The volume of 350.mL of gas at 25°C is decreased to 125 mL at constant pressure.What is the final temperature of the gas?