Question 1

A basketball is inflated to a pressure of 1.90 atm in a 24.0°C garage.What is the pressure of the basketball outside where the temperature is - 1.00°C?

Accepted Answer

We can use the combined gas law to solve this problem:

(P1/T1) = (P2/T2)

Where P1 is the initial pressure (1.90 atm), T1 is the initial temperature in Kelvin (24.0°C + 273 = 297 K), P2 is the pressure outside the garage, and T2 is the temperature outside in Kelvin (-1.00°C + 273 = 272 K).

Solving for P2, we get:

P2 = (P1/T1) x T2 = (1.90 atm/297 K) x 272 K = 1.738 atm or 1.74 atm (rounded to two significant figures).

Therefore, the best choice is A) 1.74 atm.

Question 2

A steel bottle contains argon gas at STP.What is the final pressure if the temperature is changed to 145°C?

Accepted Answer

STP stands for Standard Temperature and Pressure, which is 0°C and 1 atm. Therefore, the initial pressure of the argon gas is 1 atm. 
Using the formula P1/T1 = P2/T2, where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature, we can calculate the final pressure: 
P2 = (P1 x T2)/T1
P2 = (1 atm x 418 K) / 273 K
P2 = 1.53 atm
Therefore, the correct choice is (D) 1.53 atm.

Question 3

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 742 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?

Accepted Answer

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

Question 4

A 2.75-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?

Accepted Answer

Using the ideal gas law, we can calculate the number of moles of gas lost:PV = nRTn = PV/RTn1 = (225 kPa * 2.75 L) / (0.0821 L kPa/mol K * 298 K) = 0.0335 moln2 = (185 kPa * 2.75 L) / (0.0821 L kPa/mol K * 283 K) = 0.0272 molThe number of moles of gas lost is:nlost = n1 - n2 = 0.0335 mol - 0.0272 mol = 0.0063 molTherefore, the correct answer is A) 0.0063 mol.

Question 5

What is the Celsius temperature of 100.0 g of chlorine gas in a 55.0-L container at 800 mm Hg?

Accepted Answer

Using the ideal gas law, PV = nRT, and converting units appropriately, the temperature is calculated to be approximately 228°C.

Question 6

Effusion of a 1:1 mixture of two gases,represented by unshaded and shaded spheres in the diagram below,through a small pinhole produces the result shown below.The shaded spheres have a molecular mass of 20 amu.Which gas molecules have the higher average speed and what is the molecular mass of the unshaded molecules?

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. Since both gases are in a 1:1 mixture, they will have the same rate of effusion. Therefore, the unshaded molecules must have a lower molar mass than the shaded ones in order to compensate for their higher average speed. The only choice that satisfies this is A, where the unshaded molecules have a molecular mass of 14 amu.

Question 7

How many molecules of N₂ are in a 400.0 mL container at 780 mm Hg and 135°C?

Accepted Answer

To solve this problem, we need to use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n: n = (PV)/(RT) Plugging in the given values: P = 780 mm Hg = 780/760 atm (converting to atm) V = 400.0 mL = 0.4 L (converting to L) T = 135°C = 408 K (converting to Kelvin) R = 0.0821 L atm/(mol K) (gas constant) We also need to know the molecular weight of N₂, which is (14 x 2) + (32 x 4) + (12 x 4) + (16 x 5) = 408 g/mol. Now we can calculate n: n = ((780/760) x 0.4) / (0.0821 x 408) n = 0.015 mol Finally, we can convert moles to molecules by multiplying by Avogadro's number (6.022 x 10^23 molecules/mol): 0.015 mol x 6.022 x 10^23 molecules/mol = 7.1385 x 10^21 molecules This is closest to answer choice B: 7.38 x 10^21 molecules.

Question 8

A 45.0-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?

Accepted Answer

Using the ideal gas law, we can solve for the number of moles of acetylene gas in the tank:

PV = nRT

n = PV/RT

where P = 1.39 atm, V = 45.0 L, R = 0.0821 L atm/mol K (universal gas constant), and T = 20°C + 273.15 = 293.15 K.

The molar mass of acetylene (C2H2) is:

2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol

So the mass of acetylene in grams is:

mass = n x molar mass

Substituting the given values and simplifying:

n = (1.39 atm x 45.0 L)/(0.0821 L atm/mol K x 293.15 K) = 1.92 mol

mass = 1.92 mol x 26.04 g/mol = 50.3 g

Therefore, the answer is choice C) 67.8 g.

Question 9

A gas occupies 22.4 L at STP and 19.0 L at 100°C and 1.50 atm pressure.How many moles of gas did the system gain or lose?

Accepted Answer

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

Question 10

A 0.286-g sample of gas occupies 125 mL at 60.cm of Hg and 25°C.What is the molar mass of the gas?

Accepted Answer

The answer of A 0.286-g sample of gas occupies 125...

Question 11

What is the volume of 30.0 g of argon gas at 157°C and 2.50 kPa pressure?

Accepted Answer

The answer of What is the volume of 30.0 g...

Question 12

Three identical flasks contain three different gases at standard temperature and pressure.Flask A contains C₂H₄,flask B contains O₃,and flask C contains F₂.Which flask contains the largest number of molecules?

Accepted Answer

At standard temperature and pressure (STP), one mole of any ideal gas occupies the same volume (22.4 L) and contains the same number of molecules (Avogadro's number). Since the flasks are identical and at STP, they all contain the same number of molecules.

Question 13

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

Accepted Answer

The answer of How many grams of NOgas are...

Question 14

A gas bottle contains 0.250 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?

Accepted Answer

To find the moles of gas added, use the ideal gas law relation: P1/n1 = P2/n2. Convert the initial pressure to atm: 730 mm Hg = 0.9605 atm. Then, solve for n2: n2 = (P2 * n1) / P1 = (1.15 atm * 0.250 mol) / 0.9605 atm = 0.2993 mol. The moles added are 0.2993 mol - 0.250 mol = 0.0493 mol.

Question 15

A balloon filled with helium gas at 20 °C occupies 4.91 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?

Accepted Answer

Using the combined gas law, (P1V1/T1) = (P2V2/T2), where T is in Kelvin, we can solve for V2. Converting temperatures to Kelvin, T1 = 293 K and T2 = 77 K. Substituting the values, (1.00 atm * 4.91 L / 293 K) = (5.20 atm * V2 / 77 K), solving for V2 gives approximately 0.25 L.

Question 16

A 4.00-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?

Accepted Answer

The answer of A 4.00-L flask contains nitrogen gas at...

Question 17

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 753 mm Hg and atmospheric pressure is 724 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?

Accepted Answer

The answer of A container filled with gas is connected...

Question 18

Effusion of a 1:1 mixture of two gases,represented by unshaded and shaded spheres in the diagram below,through a small pinhole produces the result shown below.The shaded spheres have a molecular mass of 32 amu.Which gas molecules have the higher average speed and what is the molecular mass of the unshaded molecules?

Accepted Answer

The answer of Effusion of a 1:1 mixture of two...

Question 19

If the pressure in a gas container that is connected to an open-end U-tube manometer is 116 kPa and the pressure of the atmosphere at the open end of the tube is 752 mm Hg,the level of mercury in the tube will

Accepted Answer

The answer of If the pressure in a gas container...

Question 20

The volume of 350.mL of gas at 25°C is decreased to 135 mL at constant pressure.What is the final temperature of the gas?

Accepted Answer

We can use the formula for Charles's Law, which states that the volume of a gas is directly proportional to its temperature at constant pressure. Mathematically, V1/T1 = V2/T2.

Plugging in the given values, we get:

350 mL / (25°C + 273.15)K = 135 mL / T2

Simplifying, we get:

T2 = 135 mL * (25°C + 273.15)K / 350 mL

T2 = 298.15K

Converting this temperature back to Celsius, we get:

T2 = 298.15K - 273.15 = 25°C

Therefore, the final temperature of the gas is 25°C, which corresponds to answer choice A.

Quiz 9: Gases: Their Properties and Behavior