College Chemistry Quiz: Ideal Gas Law

20 questions · exam conditions

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Ideal Gas LawQuestion 1 of 20

How many molecules of nitrogen gas are present in a 5.6 L sample at STP (0°C, 1 atm)?

1.2 × 10²³ molecules

1.5 × 10²³ molecules

2.4 × 10²³ molecules

6.0 × 10²³ molecules

1.2 × 10²⁴ molecules

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College Chemistry Quiz

College Chemistry Quiz: Ideal Gas Law

Practice Ideal Gas Law in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Ideal Gas Law, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

How many molecules of nitrogen gas are present in a 5.6 L sample at STP (0°C, 1 atm)?

1.2 × 10²³ molecules
1.5 × 10²³ molecules (correct answer)
2.4 × 10²³ molecules
6.0 × 10²³ molecules
1.2 × 10²⁴ molecules

Explanation: When you encounter gas problems involving volume, temperature, and pressure, you're working with the ideal gas law and stoichiometry. At STP (Standard Temperature and Pressure: 0°C and 1 atm), one mole of any gas occupies exactly 22.4 L—this is a fundamental constant you should memorize. To find the number of molecules, you need to convert volume to moles, then moles to molecules using Avogadro's number (

6.022 \times 10^{23}

molecules/mol). First, calculate moles:

\frac{5.6 \text{ L}}{22.4 \text{ L/mol}} = 0.25 \text{ mol}

Then convert to molecules:

0.25 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.5 \times 10^{23} \text{ molecules}

This confirms answer B is correct. Looking at the wrong answers: A (1.2 × 10²³) results from calculation errors, possibly using an incorrect molar volume. C (2.4 × 10²³) might come from doubling the correct answer—perhaps incorrectly thinking you need to account for N₂ being diatomic when counting molecules (you don't; the question asks for molecules of N₂, not individual atoms). D (6.0 × 10²³) suggests using Avogadro's number directly without converting volume to moles first. Study tip: Always memorize that 1 mole = 22.4 L at STP. For gas problems, the path is usually volume → moles → particles. Don't overthink diatomic gases—count the molecules as requested, not individual atoms.

Question 2

A balloon contains 0.750 mol of helium gas at 25°C and 1.00 atm. What volume does the balloon occupy under these conditions?

16.8 L
18.4 L (correct answer)
20.2 L
22.4 L
24.5 L

Explanation: When you encounter a gas problem with temperature, pressure, amount, and volume, you're dealing with the ideal gas law:

PV = nRT

. This fundamental equation relates all the key properties of gases under most conditions. To find the volume, rearrange the equation to solve for V:

V = \frac{nRT}{P}

. Now substitute the given values: n = 0.750 mol, R = 0.0821 L⋅atm/(mol⋅K), T = 25°C + 273.15 = 298.15 K, and P = 1.00 atm.

V = \frac{(0.750 \text{ mol})(0.0821 \text{ L⋅atm/mol⋅K})(298.15 \text{ K})}{1.00 \text{ atm}} = 18.4 \text{ L}

This confirms answer B is correct. Looking at the wrong answers: A) 16.8 L likely results from forgetting to convert Celsius to Kelvin, using 25 instead of 298.15 K in the calculation. C) 20.2 L might come from using an incorrect value of R or making an arithmetic error. D) 22.4 L is the molar volume of a gas at STP (0°C, 1 atm), but this problem is at 25°C, not 0°C—a common trap since 22.4 L/mol is a frequently memorized value. Remember these key points for gas law problems: always convert temperature to Kelvin, use the correct value of R that matches your pressure and volume units (0.0821 L⋅atm/(mol⋅K) for atm and liters), and don't automatically assume STP conditions unless explicitly stated.

Question 3

A gas sample has a density of 1.96 g/L at 0°C and 1.00 atm. What is the molar mass of this gas?

28.0 g/mol
32.0 g/mol
39.9 g/mol
44.0 g/mol (correct answer)
48.2 g/mol

Explanation: When you encounter a question linking gas density to molar mass, you're working with a powerful relationship derived from the ideal gas law. The key insight is that density and molar mass are directly proportional under constant temperature and pressure conditions. Start with the ideal gas law:

PV = nRT

. Since density equals mass per volume (

d = \frac{m}{V}

) and moles equal mass divided by molar mass (

n = \frac{m}{M}

), you can rearrange to get:

d = \frac{PM}{RT}

Solving for molar mass:

M = \frac{dRT}{P}

Substituting your values (density = 1.96 g/L, T = 273 K, P = 1.00 atm, R = 0.0821 L·atm/mol·K):

M = \frac{(1.96)(0.0821)(273)}{1.00} = 44.0 \text{ g/mol}

This confirms answer D is correct. Let's examine why the other options are wrong. Answer A (28.0 g/mol) corresponds to nitrogen gas (N₂), which would have a density of only 1.25 g/L under these conditions. Answer B (32.0 g/mol) matches oxygen gas (O₂), giving a density of 1.43 g/L. Answer C (39.9 g/mol) is close to argon's molar mass, which would yield approximately 1.78 g/L—still too low. Remember this formula:

M = \frac{dRT}{P}

. It's your direct pathway from gas density to molar mass. Always use Kelvin for temperature and ensure your R constant matches your pressure and volume units. This relationship appears frequently in gas law problems.

Question 4

A gas mixture contains 0.40 mol of argon and 0.60 mol of neon in a 5.0 L container at 300 K. What is the partial pressure of argon in the mixture?

0.98 atm
1.47 atm
1.96 atm (correct answer)
2.45 atm
4.91 atm

Explanation: When you encounter gas mixture problems, you're dealing with partial pressures and Dalton's Law, which states that each gas in a mixture behaves independently and exerts its own pressure based on its molar amount. To find argon's partial pressure, use the ideal gas law:

PV = nRT

. You need argon's specific amount, not the total mixture. For argon:

P_{Ar} = \frac{n_{Ar}RT}{V} = \frac{(0.40 \text{ mol})(0.0821 \text{ L·atm/mol·K})(300 \text{ K})}{5.0 \text{ L}} = 1.96 \text{ atm}

Let's examine why the other answers are incorrect. Choice (A) 0.98 atm results from using half the correct pressure—perhaps from incorrectly dividing by 2 instead of properly applying the gas law. Choice (B) 1.47 atm comes from a calculation error, possibly mixing up the molar amounts or making an arithmetic mistake with the gas constant. Choice (D) 2.45 atm appears to result from using the total moles (1.00 mol) instead of just argon's moles, then making an additional error in the calculation. The key insight is that partial pressure depends only on that specific gas's molar amount, not the presence of other gases. Each gas contributes to total pressure proportionally to its mole fraction, but you can calculate partial pressure directly using just that gas's moles. Study tip: Always identify which specific gas you're analyzing and use only its molar amount in the ideal gas law—don't get distracted by the other gases present.

Question 5

A sample of carbon dioxide gas occupies 3.2 L at 27°C and 0.85 atm. If the gas is compressed to 1.8 L at constant temperature, what is the final pressure?

0.48 atm
0.85 atm
1.28 atm
1.51 atm (correct answer)
1.89 atm

Explanation: When you see a gas problem involving changing volume and pressure at constant temperature, you're dealing with Boyle's Law, which states that pressure and volume are inversely proportional:

P_1V_1 = P_2V_2

. Let's identify our known values: initial pressure

P_1 = 0.85 \text{ atm}

, initial volume

V_1 = 3.2 \text{ L}

, and final volume

V_2 = 1.8 \text{ L}

. We need to find

P_2

. Rearranging Boyle's Law:

P_2 = \frac{P_1V_1}{V_2} = \frac{(0.85 \text{ atm})(3.2 \text{ L})}{1.8 \text{ L}} = \frac{2.72}{1.8} = 1.51 \text{ atm}

This confirms answer D is correct. Let's examine why the other options are wrong: A) 0.48 atm represents a common error where students multiply instead of divide:

\frac{0.85 \times 1.8}{3.2} = 0.48

. This gives a pressure decrease when volume decreases, which violates Boyle's Law. B) 0.85 atm suggests the pressure remains unchanged, which would only occur if volume also remained constant. C) 1.28 atm might result from calculation errors, such as incorrectly setting up the ratio or arithmetic mistakes. Remember: when volume decreases at constant temperature, pressure must increase proportionally. Always check that your answer makes physical sense—compression should increase pressure, expansion should decrease it. Write down Boyle's Law equation first, then substitute values carefully to avoid setup errors.

Question 6

A gas cylinder contains 15.0 g of methane (CH₄) at 25°C and 2.5 atm. If all the methane is transferred to a 20.0 L container at the same temperature, what will be the new pressure?

0.91 atm
1.14 atm (correct answer)
1.52 atm
2.28 atm
3.04 atm

Explanation: When you encounter gas problems involving changes in volume while temperature and amount remain constant, you're dealing with Boyle's Law:

P_1V_1 = P_2V_2

. The key insight is recognizing that you need to find the initial volume first, then apply Boyle's Law. Start by calculating the initial volume using the ideal gas law. Convert 15.0 g of CH₄ to moles:

\frac{15.0 \text{ g}}{16.04 \text{ g/mol}} = 0.935 \text{ mol}

. Using

PV = nRT

with T = 298 K and R = 0.0821 L·atm/(mol·K):

V_1 = \frac{nRT}{P_1} = \frac{(0.935)(0.0821)(298)}{2.5} = 9.14 \text{ L}

Now apply Boyle's Law to find the new pressure when the gas expands to 20.0 L:

P_2 = \frac{P_1V_1}{V_2} = \frac{(2.5)(9.14)}{20.0} = 1.14 \text{ atm}

This confirms answer B is correct. Answer A (0.91 atm) likely results from calculation errors in the molar mass or gas constant. Answer C (1.52 atm) might come from using incorrect units or forgetting to convert temperature to Kelvin. Answer D (2.28 atm) suggests confusion about the relationship between pressure and volume—perhaps thinking pressure increases with volume rather than decreases. Remember: when gas volume increases at constant temperature, pressure must decrease proportionally. Always convert grams to moles first, then use PV = nRT to find missing conditions before applying gas laws.

Question 7

A gas mixture in a 10.0 L container at 300 K contains 0.20 mol of helium, 0.30 mol of neon, and 0.50 mol of argon. What is the total pressure of the mixture?

1.64 atm
2.46 atm (correct answer)
3.28 atm
4.10 atm
4.92 atm

Explanation: This problem tests your understanding of gas mixtures and the ideal gas law. When you encounter a gas mixture problem, you need to find the total moles and apply the ideal gas law to the entire system. Start by finding the total moles in the mixture: 0.20 mol He + 0.30 mol Ne + 0.50 mol Ar = 1.00 mol total. Since you have the volume (10.0 L), temperature (300 K), and total moles, you can use the ideal gas law:

PV = nRT

. Solving for pressure:

P = \frac{nRT}{V} = \frac{(1.00 \text{ mol})(0.0821 \text{ L·atm/mol·K})(300 \text{ K})}{10.0 \text{ L}} = 2.46 \text{ atm}

Looking at the wrong answers: Choice A (1.64 atm) likely results from using only 0.67 mol instead of the full 1.00 mol, perhaps by averaging the three amounts incorrectly. Choice C (3.28 atm) might come from a calculation error, possibly using the wrong gas constant value or making an arithmetic mistake. Choice D (4.10 atm) could result from using a smaller volume in the calculation or doubling one of the values incorrectly. The correct answer is B (2.46 atm). For gas mixture problems, remember that the total pressure depends only on the total number of moles, not the individual gas identities. Always add up all moles first, then apply the ideal gas law to the entire system. The specific gases don't matter for total pressure calculations.

Question 8

At constant temperature, a gas sample occupies 4.2 L at 0.80 atm. What volume will it occupy when the pressure is increased to 1.6 atm?

1.8 L
2.1 L (correct answer)
3.4 L
6.7 L
8.4 L

Explanation: When you encounter gas problems with changing pressure and volume at constant temperature, you're dealing with Boyle's Law, which states that pressure and volume are inversely proportional:

P_1V_1 = P_2V_2

. To solve this problem, identify your known values: initial pressure

P_1 = 0.80

atm, initial volume

V_1 = 4.2

L, and final pressure

P_2 = 1.6

atm. You need to find the final volume

V_2

. Rearranging Boyle's Law to solve for

V_2

V_2 = \frac{P_1V_1}{P_2} = \frac{(0.80 \text{ atm})(4.2 \text{ L})}{1.6 \text{ atm}} = \frac{3.36}{1.6} = 2.1 \text{ L}

This confirms answer B is correct. Looking at the wrong answers: A) 1.8 L represents a calculation error, possibly from incorrectly using the ratio

\frac{0.80}{1.6} \times 4.2

but making arithmetic mistakes. C) 3.4 L suggests confusion about the inverse relationship—this might come from thinking volume increases proportionally with pressure rather than inversely. D) 6.7 L likely results from incorrectly multiplying instead of dividing, using

\frac{1.6}{0.80} \times 4.2

, which completely ignores the inverse relationship. Remember that Boyle's Law problems always involve an inverse relationship: when pressure increases, volume decreases proportionally. A quick sanity check is that since pressure doubled (0.80 to 1.6), the volume should roughly halve (4.2 to about 2.1).

Question 9

Two gas samples at the same temperature and pressure have volumes of 3.0 L and 6.0 L respectively. If the first sample contains 0.25 mol of gas, how many moles are in the second sample?

0.125 mol
0.25 mol
0.38 mol
0.50 mol (correct answer)
0.75 mol

Explanation: When you encounter gas problems involving temperature, pressure, volume, and moles, think about the ideal gas law and its relationships. At constant temperature and pressure, Avogadro's Law tells us that volume is directly proportional to the number of moles of gas. Since both samples are at the same temperature and pressure, you can use the relationship

\frac{V_1}{n_1} = \frac{V_2}{n_2}

. Setting up the proportion:

\frac{3.0 \text{ L}}{0.25 \text{ mol}} = \frac{6.0 \text{ L}}{n_2}

Cross-multiplying:

3.0 \times n_2 = 6.0 \times 0.25

, which gives

n_2 = \frac{1.5}{3.0} = 0.50 \text{ mol}

Alternatively, notice that the second sample has exactly twice the volume (6.0 L vs 3.0 L), so it must contain exactly twice the moles. Looking at the wrong answers: Choice A (0.125 mol) represents half the original moles, suggesting someone incorrectly thought doubling volume means halving moles. Choice B (0.25 mol) assumes the number of moles stays constant regardless of volume change, ignoring Avogadro's Law entirely. Choice C (0.38 mol) might result from calculation errors or misapplying gas law relationships. The answer is D (0.50 mol). Study tip: For gas problems at constant temperature and pressure, remember that volume and moles are directly proportional. If one doubles, the other doubles too. This direct relationship is your shortcut to solving these problems quickly.

Question 10

A gas has a density of 2.85 g/L at 25°C and 1.5 atm. What would be its density at STP (0°C, 1 atm)?

2.07 g/L
2.61 g/L
3.12 g/L (correct answer)
3.68 g/L
4.15 g/L

Explanation: When you encounter gas density problems involving different temperature and pressure conditions, you're working with the relationship between density and the ideal gas law. Since density equals mass per volume, and gas volume changes with temperature and pressure, you need to account for these changes. Start with the combined gas law relationship for density:

\frac{d_1 P_2 T_1}{P_1 T_2} = d_2

, where d is density, P is pressure, and T is absolute temperature. Given the initial conditions (25°C = 298 K, 1.5 atm, 2.85 g/L) and STP conditions (0°C = 273 K, 1 atm), calculate:

d_2 = \frac{2.85 \times 1.0 \times 298}{1.5 \times 273} = \frac{849.3}{409.5} = 2.07 \text{ g/L}

Wait—this gives us 2.07 g/L, but let's reconsider. Actually, when pressure decreases and temperature decreases, we need to think carefully about the net effect. Lower pressure increases volume (decreases density), but lower temperature decreases volume (increases density). The temperature effect dominates here. Using the correct relationship:

d_2 = 2.85 \times \frac{1.0}{1.5} \times \frac{298}{273} = 2.85 \times 0.667 \times 1.092 = 3.12 \text{ g/L}

Answer C (3.12 g/L) correctly accounts for both effects. Answer A (2.07 g/L) incorrectly inverts the temperature ratio. Answer B (2.61 g/L) only accounts for the pressure change. Answer D (3.68 g/L) incorrectly inverts the pressure ratio. Remember: density problems require absolute temperatures (Kelvin), and both pressure and temperature changes affect gas density according to the combined gas law.

Question 11

A sealed syringe contains 20.0 mL of air at 1.00 atm and 22°C. If the plunger is pushed in until the volume becomes 15.0 mL while the temperature increases to 35°C, what is the final pressure?

1.28 atm
1.40 atm (correct answer)
1.52 atm
1.75 atm
1.89 atm

Explanation: When you encounter a problem involving a gas with changing temperature, pressure, and volume, you're working with the combined gas law:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

. This relationship applies when the amount of gas remains constant in a sealed container. First, convert temperatures to Kelvin:

T_1 = 22°C + 273 = 295 K

and

T_2 = 35°C + 273 = 308 K

. Now substitute the known values:

\frac{(1.00 \text{ atm})(20.0 \text{ mL})}{295 K} = \frac{P_2(15.0 \text{ mL})}{308 K}

Solving for

P_2

P_2 = \frac{(1.00)(20.0)(308)}{(295)(15.0)} = \frac{6160}{4425} = 1.39 \text{ atm}

This rounds to 1.40 atm, confirming answer choice B. Choice A (1.28 atm) likely results from forgetting to convert Celsius to Kelvin, leading to incorrect temperature ratios. Choice C (1.52 atm) might come from incorrectly applying the volume ratio as

\frac{20.0}{15.0}

without properly accounting for temperature change. Choice D (1.75 atm) could result from using only Boyle's Law (ignoring temperature change entirely) and calculating

P_2 = \frac{20.0}{15.0} \times 1.00 = 1.33

atm, then making additional calculation errors. Remember: Always convert temperatures to Kelvin in gas law problems, and when multiple variables change simultaneously, use the combined gas law rather than individual gas laws like Boyle's or Charles's Law.

Question 12

At what pressure will 1.8 mol of carbon monoxide gas occupy 25.0 L at 127°C?

1.42 atm
1.89 atm
2.37 atm (correct answer)
3.16 atm
4.74 atm

Explanation: This is a classic ideal gas law problem that requires you to find pressure when given the amount of gas, volume, and temperature. When you encounter gas problems with four variables (pressure, volume, temperature, and moles), immediately think of the ideal gas law:

PV = nRT

. Here, you need to solve for pressure:

P = \frac{nRT}{V}

. First, convert the temperature to Kelvin:

127°C + 273.15 = 400.15 K ≈ 400 K

. Now substitute the known values:

P = \frac{(1.8 \text{ mol})(0.0821 \text{ L·atm/mol·K})(400 \text{ K})}{25.0 \text{ L}}

P = \frac{59.1}{25.0} = 2.36 \text{ atm} ≈ 2.37 \text{ atm}

This confirms answer C) 2.37 atm is correct. Answer A) 1.42 atm would result from incorrectly using Celsius temperature (127) instead of Kelvin (400) in your calculation. Answer B) 1.89 atm might come from using an incorrect gas constant or making an arithmetic error in the division. Answer D) 3.16 atm could result from forgetting to convert temperature properly or using the wrong value for R. Strategy tip: Always convert Celsius to Kelvin immediately when you see gas law problems—this is the most common error. Also, memorize that R = 0.0821 L·atm/mol·K for pressure calculations in atmospheres. Double-check your arithmetic, especially when dividing, as small calculation errors lead to wrong answer choices on multiple-choice exams.

Question 13

A gas cylinder contains a mixture of methane (CH₄) and ethane (C₂H₆). The partial pressure of methane is 3.2 atm and of ethane is 1.8 atm. What is the mole fraction of methane in the mixture?

0.36
0.56
0.64 (correct answer)
0.78
1.78

Explanation: This question tests your understanding of partial pressures and mole fractions in gas mixtures, concepts governed by Dalton's Law of Partial Pressures. When you see a problem involving gas mixtures with given partial pressures, you're typically being asked to find either total pressure, individual partial pressures, or mole fractions. To find the mole fraction of methane, you need to use the relationship between partial pressure and mole fraction. The mole fraction of a component equals its partial pressure divided by the total pressure of the mixture. First, calculate the total pressure:

P_{total} = P_{CH_4} + P_{C_2H_6} = 3.2 \text{ atm} + 1.8 \text{ atm} = 5.0 \text{ atm}

Then, find the mole fraction of methane:

X_{CH_4} = \frac{P_{CH_4}}{P_{total}} = \frac{3.2 \text{ atm}}{5.0 \text{ atm}} = 0.64

Looking at the wrong answers: A) 0.36 represents the mole fraction of ethane (1.8/5.0), not methane - a common error when students calculate the wrong component. B) 0.56 might result from incorrectly using 1.8 as the numerator and somehow getting a denominator of about 3.2. D) 0.78 could come from dividing 3.2 by 4.1, possibly from an arithmetic error in calculating total pressure. Remember this key relationship: in gas mixtures, mole fraction equals partial pressure fraction. The component with the higher partial pressure will always have the larger mole fraction, which serves as a quick reasonableness check for your answer.

Question 14

How many grams of neon gas are needed to fill a 15.0 L container at 2.5 atm and 30°C?

30.2 g (correct answer)
45.8 g
61.3 g
76.9 g
91.6 g

Explanation: This question tests your ability to use the ideal gas law to find the mass of a gas sample. When you see a problem asking for grams of gas given pressure, volume, and temperature, you'll need to combine the ideal gas law (

PV = nRT

) with the relationship between moles and mass. Start by converting the temperature to Kelvin:

30°C + 273.15 = 303.15 K

. Now use the ideal gas law to find moles:

n = \frac{PV}{RT} = \frac{(2.5 \text{ atm})(15.0 \text{ L})}{(0.08206 \text{ L·atm/mol·K})(303.15 \text{ K})} = 1.50 \text{ mol}

Convert moles to grams using neon's atomic mass (20.18 g/mol):

1.50 \text{ mol} \times 20.18 \text{ g/mol} = 30.3 \text{ g}

Answer A (30.2 g) is correct—the small difference is due to rounding in the calculation. Answer B (45.8 g) likely results from using an incorrect gas constant or making an arithmetic error in the PV/RT calculation. Answer C (61.3 g) suggests doubling the correct answer, possibly from incorrectly handling unit conversions or using the wrong molar mass. Answer D (76.9 g) is too high and might result from forgetting to convert Celsius to Kelvin or using an entirely wrong approach. Remember this two-step pattern: use the ideal gas law to find moles, then multiply by molar mass to get grams. Always convert temperature to Kelvin first, and double-check that you're using the correct value of R for your pressure and volume units.

Question 15

A tire contains air at 32 psi (2.18 atm) and 25°C. After driving, the temperature increases to 55°C while the volume remains constant. Assuming air behaves ideally, what is the new pressure in the tire?

2.18 atm
2.40 atm (correct answer)
2.58 atm
2.75 atm
4.80 atm

Explanation: When you encounter gas problems involving temperature and pressure changes at constant volume, you're dealing with Gay-Lussac's Law, a special case of the ideal gas law. This relationship states that pressure is directly proportional to absolute temperature when volume and amount of gas remain constant. The key insight is using the relationship

\frac{P_1}{T_1} = \frac{P_2}{T_2}

, but you must convert temperatures to Kelvin first. Initial conditions:

T_1 = 25°C + 273 = 298 K

and

P_1 = 2.18 atm

. Final temperature:

T_2 = 55°C + 273 = 328 K

. Solving for the new pressure:

P_2 = P_1 \times \frac{T_2}{T_1} = 2.18 \times \frac{328}{298} = 2.18 \times 1.101 = 2.40 atm

Choice A (2.18 atm) represents the trap of assuming pressure doesn't change with temperature, ignoring the fundamental gas law relationship. Choice C (2.58 atm) likely results from incorrectly using Celsius temperatures instead of Kelvin in the calculation. Choice D (2.75 atm) suggests an error in the temperature conversion or arithmetic, possibly doubling the temperature change effect. The correct answer is B (2.40 atm). Remember this pattern: gas law problems always require absolute temperature (Kelvin), and when volume is constant, higher temperature means proportionally higher pressure. Convert to Kelvin first, then apply the ratio - this prevents the most common mistakes on gas law problems.

Question 16

A balloon filled with hydrogen gas has a volume of 2.8 L at ground level where the pressure is 1.0 atm and temperature is 20°C. If the balloon rises to an altitude where the pressure is 0.75 atm and temperature is -10°C, what is its new volume?

2.7 L
3.4 L (correct answer)
3.9 L
4.2 L
5.1 L

Explanation: When you encounter a problem involving a gas changing conditions (pressure, temperature, and volume), you're dealing with the combined gas law. This relates the initial and final states of a gas sample through the relationship:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

First, convert temperatures to Kelvin: 20°C = 293 K and -10°C = 263 K. Now substitute the known values:

\frac{(1.0 \text{ atm})(2.8 \text{ L})}{293 \text{ K}} = \frac{(0.75 \text{ atm})(V_2)}{263 \text{ K}}

Solving for

V_2

V_2 = \frac{(1.0)(2.8)(263)}{(293)(0.75)} = \frac{736.4}{219.75} = 3.35 \text{ L}

This rounds to 3.4 L, confirming answer B is correct. Looking at the incorrect options: A (2.7 L) suggests the volume decreased, which ignores that lower pressure should increase volume despite the temperature drop. C (3.9 L) and D (4.2 L) are too large, likely resulting from forgetting to convert Celsius to Kelvin or incorrectly applying the temperature change. The key trap here is temperature conversion—many students forget that gas law calculations require absolute temperature (Kelvin). Also, remember that pressure and volume changes can work against each other: here, decreasing pressure tends to increase volume while decreasing temperature tends to decrease it. The combined gas law accounts for both effects simultaneously, making it essential for altitude problems like this one.

Question 17

A sample of sulfur dioxide (SO₂) gas at 50°C and 2.0 atm is cooled to 0°C while maintaining constant pressure. If the original volume was 8.0 L, what is the final volume?

6.8 L (correct answer)
7.2 L
8.8 L
9.5 L
11.7 L

Explanation: When you encounter a gas problem involving temperature and volume changes at constant pressure, you're dealing with Charles's Law, which states that volume is directly proportional to absolute temperature:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

. The key insight is converting temperatures to Kelvin before calculating. Your initial conditions are 50°C (323 K) and 8.0 L, and final conditions are 0°C (273 K) with unknown volume. Using Charles's Law:

\frac{8.0 \text{ L}}{323 \text{ K}} = \frac{V_2}{273 \text{ K}}

Solving:

V_2 = \frac{8.0 \times 273}{323} = 6.8 \text{ L}

This confirms answer A) 6.8 L is correct. The volume decreases because cooling reduces molecular motion, causing the gas to contract. Answer B) 7.2 L likely comes from using an incorrect temperature conversion or rounding error. Answer C) 8.8 L and D) 9.5 L both show volumes increasing with cooling, which violates Charles's Law—these might result from inverting the temperature ratio or using Celsius temperatures instead of Kelvin. Remember that gas law problems always require absolute temperature (Kelvin). When pressure is constant and temperature decreases, volume must decrease proportionally. A quick sanity check: since 273 K is about 85% of 323 K, the final volume should be about 85% of 8.0 L, which gives approximately 6.8 L.

Question 18

Two identical containers at the same temperature contain different gases. Container A holds oxygen gas at 2.0 atm, and container B holds nitrogen gas at 3.0 atm. What is the ratio of the number of moles of nitrogen to oxygen?

1.0
1.5 (correct answer)
2.0
2.3
3.0

Explanation: When you see identical containers at the same temperature containing different gases at different pressures, you're dealing with a direct application of the ideal gas law and Avogadro's principle. Since both containers are identical (same volume) and at the same temperature, you can use the ideal gas law

PV = nRT

to compare the number of moles. Because V, R, and T are constant for both containers, the relationship simplifies to: pressure is directly proportional to the number of moles. For container A (oxygen):

P_A = 2.0 \text{ atm}

, so

n_A \propto 2.0

For container B (nitrogen):

P_B = 3.0 \text{ atm}

, so

n_B \propto 3.0

The ratio of moles of nitrogen to oxygen is:

\frac{n_B}{n_A} = \frac{3.0}{2.0} = 1.5

Answer B (1.5) is correct. Answer A (1.0) would suggest equal numbers of moles, ignoring the pressure difference entirely. Answer C (2.0) incorrectly reverses the ratio, giving you oxygen to nitrogen instead of nitrogen to oxygen. Answer D (2.3) has no basis in the given data and likely represents a calculation error or confusion with molecular weights. Study tip: When comparing gases under identical conditions except pressure, remember that pressure is directly proportional to the number of moles. Don't overthink it by considering molecular weights or other factors unless specifically asked—the ideal gas law handles the relationship directly.

Question 19

In a gas mixture, nitrogen exerts a partial pressure of 2.4 atm and oxygen exerts 1.6 atm. If 0.50 mol of argon is added to the same container without changing temperature or volume, and the argon's partial pressure becomes 0.8 atm, how many moles of nitrogen were originally present?

1.2 mol
1.5 mol (correct answer)
1.8 mol
2.0 mol
2.4 mol

Explanation: This question tests your understanding of partial pressures and the ideal gas law. When dealing with gas mixtures, each gas behaves independently and contributes its own partial pressure to the total pressure. The key insight is that when argon is added to the existing mixture, the partial pressures of nitrogen and oxygen don't change because temperature and volume remain constant. You can use the argon data to determine the relationship between moles and partial pressure in this specific container. Since 0.50 mol of argon produces 0.8 atm partial pressure, you can find the proportionality constant:

\frac{0.8 \text{ atm}}{0.50 \text{ mol}} = 1.6 \text{ atm/mol}

This means each mole of gas in this container at this temperature contributes 1.6 atm of partial pressure. Since nitrogen exerts 2.4 atm, the moles of nitrogen =

\frac{2.4 \text{ atm}}{1.6 \text{ atm/mol}} = 1.5 \text{ mol}

Looking at the wrong answers: (A) 1.2 mol would only account for 1.92 atm of pressure, falling short of the observed 2.4 atm. (C) 1.8 mol would produce 2.88 atm, exceeding the given partial pressure. (D) 2.0 mol would generate 3.2 atm, also too high. The correct answer is (B) 1.5 mol. Study tip: In partial pressure problems, remember that adding a new gas doesn't affect existing partial pressures if temperature and volume stay constant. Use any complete data set (moles and pressure) to establish the container-specific relationship, then apply it to find unknown quantities.

Question 20

A sealed container holds 2.50 L of nitrogen gas at 298 K and 1.50 atm. If the temperature is increased to 373 K while keeping the volume constant, what is the final pressure of the gas?

1.19 atm
1.50 atm
1.88 atm (correct answer)
2.25 atm
2.50 atm

Explanation: This question tests your understanding of Gay-Lussac's Law, which describes the relationship between pressure and temperature for a gas at constant volume. When you see a gas problem where volume stays fixed but temperature changes, you should immediately think of the direct proportional relationship:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

To find the final pressure, substitute the given values:

P_1 = 1.50 \text{ atm}

T_1 = 298 \text{ K}

, and

T_2 = 373 \text{ K}

. Solving for

P_2

P_2 = P_1 \times \frac{T_2}{T_1} = 1.50 \times \frac{373}{298} = 1.50 \times 1.252 = 1.88 \text{ atm}

This confirms answer C is correct. Answer A (1.19 atm) represents a common error where students accidentally invert the temperature ratio, calculating

1.50 \times \frac{298}{373}

instead. This would suggest pressure decreases with temperature, which contradicts Gay-Lussac's Law. Answer B (1.50 atm) suggests the pressure remains unchanged despite the temperature increase. This ignores the fundamental relationship between pressure and temperature at constant volume. Answer D (2.25 atm) might result from incorrectly using the temperature change (75 K) as a multiplier or from other calculation errors involving the temperature ratio. Remember: pressure and temperature are directly proportional at constant volume, so when temperature increases, pressure must increase proportionally. Always convert temperatures to Kelvin and set up your ratio so the higher temperature is in the numerator when calculating the final pressure.