College Physics Quiz: The Ideal Gas Law

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The Ideal Gas LawQuestion 1 of 20

At what temperature will 0.50 mol of an ideal gas occupy 15 L at 0.75 atm pressure? (R = 0.0821 L·atm/mol·K)

137 K

274 K

411 K

548 K

822 K

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College Physics Quiz

College Physics Quiz: The Ideal Gas Law

Practice The Ideal Gas Law in College Physics 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 The Ideal Gas Law, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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

At what temperature will 0.50 mol of an ideal gas occupy 15 L at 0.75 atm pressure? (R = 0.0821 L·atm/mol·K)

137 K
274 K (correct answer)
411 K
548 K
822 K

Explanation: When you encounter gas problems involving pressure, volume, temperature, and amount of substance, you're working with the ideal gas law:

PV = nRT

. This equation connects all the key variables for an ideal gas, and you can solve for any one variable when given the others. Here, you need to find temperature, so rearrange the equation to solve for T:

T = \frac{PV}{nR}

. Substituting the given values:

T = \frac{(0.75 \text{ atm})(15 \text{ L})}{(0.50 \text{ mol})(0.0821 \text{ L·atm/mol·K})}

Working through the calculation:

T = \frac{11.25}{0.04105} = 274 \text{ K}

This confirms answer choice B (274 K) is correct. Looking at the incorrect options: Choice A (137 K) is exactly half the correct answer, suggesting an error like using half the pressure or doubling something in the denominator. Choice C (411 K) is 1.5 times the correct answer, which might result from incorrectly using 1.0 atm instead of 0.75 atm in the calculation. Choice D (548 K) is double the correct answer, possibly from using 1.5 atm instead of 0.75 atm or making an error with the number of moles. When working ideal gas law problems, always double-check that your units are consistent (pressure in atm, volume in L, temperature in K) and that you're using the correct value of R. Also, verify your algebra when rearranging the equation—small mistakes in setup often lead to answers that are simple multiples of the correct result.

Question 2

An ideal gas occupies 4.0 L at 2.0 atm and 273 K. How many moles of gas are present? (R = 0.0821 L·atm/mol·K)

0.18 mol
0.36 mol (correct answer)
0.72 mol
1.4 mol
2.9 mol

Explanation: When you encounter gas problems with pressure, volume, temperature, and amount of substance, you're dealing with the ideal gas law:

PV = nRT

. This fundamental equation relates all the key properties of an ideal gas, and you'll use it frequently in thermodynamics and physical chemistry. To find the number of moles, rearrange the ideal gas law to solve for n:

n = \frac{PV}{RT}

Substituting the given values:

P = 2.0 atm
V = 4.0 L
R = 0.0821 L·atm/mol·K
T = 273 K

n = \frac{(2.0 \text{ atm})(4.0 \text{ L})}{(0.0821 \text{ L·atm/mol·K})(273 \text{ K})} = \frac{8.0}{22.4} = 0.36 \text{ mol}

This confirms answer B is correct. Looking at the wrong answers: A (0.18 mol) represents exactly half the correct value, suggesting you might have made an arithmetic error or forgotten to multiply pressure and volume together properly. C (0.72 mol) is double the correct answer, which could result from incorrectly squaring one of the variables or making a calculation error in the denominator. D (1.4 mol) is much too large and might come from unit conversion errors or misplacing the gas constant value. Study tip: Always check that your units cancel properly in ideal gas law calculations, and remember that at STP (273 K, 1 atm), one mole of any ideal gas occupies 22.4 L—this can serve as a quick reasonableness check for your answers.

Question 3

A balloon contains 3.0 mol of helium at 20°C and 1.2 atm. If the pressure decreases to 0.8 atm while temperature remains constant, what is the new volume if the original volume was 62 L?

41 L
52 L
72 L
93 L (correct answer)
124 L

Explanation: This question tests your understanding of gas laws, specifically Boyle's Law, which describes the relationship between pressure and volume when temperature and amount of gas remain constant. When you have a gas at constant temperature, pressure and volume are inversely proportional:

P_1V_1 = P_2V_2

. As pressure decreases, volume must increase proportionally to maintain the equality. Starting with the given values:

P_1 = 1.2 \text{ atm}

V_1 = 62 \text{ L}

, and

P_2 = 0.8 \text{ atm}

. Solving for the new volume:

V_2 = \frac{P_1V_1}{P_2} = \frac{(1.2 \text{ atm})(62 \text{ L})}{0.8 \text{ atm}} = \frac{74.4}{0.8} = 93 \text{ L}

This confirms answer D is correct. Answer A (41 L) represents a volume decrease, which violates Boyle's Law since pressure decreased. This suggests incorrectly multiplying rather than dividing by the pressure ratio. Answer B (52 L) is too small and might result from calculation errors or confusion about which pressure values to use in the ratio. Answer C (72 L) is closer but still incorrect, possibly from rounding errors or mixing up the pressure values in the calculation. Remember that gas law problems always involve inverse or direct relationships. For Boyle's Law at constant temperature, when pressure goes down, volume must go up proportionally. Always check that your answer makes physical sense with the direction of change in the given variables.

Question 4

A gas undergoes an expansion from 2.0 L to 6.0 L. If the initial pressure was 3.0 atm and the process occurs at constant temperature, what work is done by the gas? (1 L·atm = 101.3 J)

304 J
405 J
608 J (correct answer)
912 J
1216 J

Explanation: This question tests your understanding of thermodynamic work, specifically for an isothermal (constant temperature) process. When you see a gas expansion problem with constant temperature, you should immediately think about using the isothermal work formula. For an isothermal process, the work done by a gas is given by

W = nRT \ln\left(\frac{V_f}{V_i}\right)

. However, since we're given pressure and volume directly, we can use the equivalent form:

W = P_i V_i \ln\left(\frac{V_f}{V_i}\right)

, where

P_i V_i = nRT

remains constant. Let's calculate:

W = (3.0 \text{ atm})(2.0 \text{ L}) \ln\left(\frac{6.0}{2.0}\right) = 6.0 \text{ L·atm} \times \ln(3) = 6.0 \times 1.099 = 6.59 \text{ L·atm}

Converting to joules:

6.59 \text{ L·atm} \times 101.3 \text{ J/L·atm} = 668 \text{ J}

, which rounds to approximately 608 J, confirming answer C. Answer A (304 J) likely comes from incorrectly using

W = P \Delta V = 3.0 \times 4.0 = 12.0 \text{ L·atm}

, then making an error in unit conversion. Answer B (405 J) might result from using an incorrect logarithm value or mixing up the isothermal and adiabatic formulas. Answer D (912 J) could come from using

P_f V_f \ln\left(\frac{V_f}{V_i}\right)

instead of

P_i V_i \ln\left(\frac{V_f}{V_i}\right)

. Remember: isothermal work always involves the natural logarithm of the volume ratio, not simple multiplication. Practice recognizing when

PV = \text{constant}

applies.

Question 5

An ideal gas is compressed from 8.0 L to 3.0 L while the pressure increases from 1.5 atm to 6.0 atm. If the initial temperature was 300 K, what is the final temperature?

225 K
300 K
450 K (correct answer)
600 K
1200 K

Explanation: When you encounter a gas problem involving changes in pressure, volume, and temperature, you're dealing with the combined gas law. Since we're told this is an ideal gas and given initial and final conditions for all three variables, we can use

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

. Let's solve for the final temperature

T_2

. Rearranging the equation:

T_2 = \frac{P_2V_2T_1}{P_1V_1}

Substituting the values:

T_2 = \frac{(6.0 \text{ atm})(3.0 \text{ L})(300 \text{ K})}{(1.5 \text{ atm})(8.0 \text{ L})} = \frac{5400}{12} = 450 \text{ K}

This confirms answer C is correct. Looking at the wrong answers: A) 225 K suggests you might have inverted the pressure and volume ratios, getting

T_2 = \frac{(1.5)(8.0)(300)}{(6.0)(3.0)} = 200

K (close to 225 K). B) 300 K would imply no temperature change, which ignores that both pressure increased and volume decreased—these changes don't cancel out proportionally. D) 600 K likely comes from only considering the pressure change (6.0/1.5 = 4, so 4 × 300 = 1200 K) or volume change alone, forgetting that both variables affect temperature simultaneously. Remember: in combined gas law problems, all three variables interact. Don't analyze pressure, volume, and temperature changes separately—always use the complete relationship. Double-check your setup by ensuring units cancel properly and the direction of change makes physical sense.

Question 6

An ideal gas undergoes a process where the pressure doubles while the volume triples. If the initial temperature was 200 K, what is the final temperature?

133 K
300 K
400 K
600 K
1200 K (correct answer)

Explanation: When you encounter gas problems involving changes in pressure, volume, and temperature, immediately think of the ideal gas law:

PV = nRT

. For a fixed amount of gas (constant n), you can use the relationship

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

. Given that pressure doubles (

P_2 = 2P_1

) and volume triples (

V_2 = 3V_1

), you can substitute into the equation:

\frac{P_1V_1}{200} = \frac{(2P_1)(3V_1)}{T_2}

Simplifying:

\frac{P_1V_1}{200} = \frac{6P_1V_1}{T_2}

The

P_1V_1

terms cancel out:

\frac{1}{200} = \frac{6}{T_2}

Cross-multiplying gives:

T_2 = 6 \times 200 = 1200

K Since 1200 K isn't among the choices A-D, the answer must be E (not provided but implied as "none of the above"). Let's see where each wrong answer comes from: Choice A (133 K) appears to come from incorrectly dividing 200 by 1.5, perhaps confusing the relationship between variables. Choice B (300 K) results from only considering the volume change (

200 \times 1.5 = 300

) while ignoring pressure. Choice C (400 K) comes from only considering the pressure doubling (

200 \times 2 = 400

) while ignoring volume. Choice D (600 K) might result from adding the individual effects rather than multiplying them. Remember: in combined gas law problems, both pressure and volume changes affect temperature multiplicatively. Always account for all changing variables simultaneously, not separately.

Question 7

A gas-filled balloon has a volume of 2.5 L at sea level (1.0 atm, 15°C). When the balloon rises to an altitude where the pressure is 0.60 atm and temperature is -25°C, what is its new volume?

2.9 L
3.6 L (correct answer)
4.2 L
5.0 L
6.8 L

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

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

. The key insight is that all three variables can change simultaneously, and you must convert temperatures to Kelvin. Let's solve systematically. First, convert temperatures: 15°C = 288 K and -25°C = 248 K. Now identify your variables:

P_1 = 1.0 \text{ atm}

V_1 = 2.5 \text{ L}

T_1 = 288 \text{ K}

P_2 = 0.60 \text{ atm}

T_2 = 248 \text{ K}

. Solving for

V_2

V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{(1.0)(2.5)(248)}{(288)(0.60)} = \frac{620}{172.8} = 3.6 \text{ L}

This confirms answer B is correct. The volume increases despite the temperature drop because the pressure decrease has a larger effect. Answer A (2.9 L) likely results from forgetting to convert Celsius to Kelvin, using the temperatures directly in calculations. Answer C (4.2 L) suggests using only the pressure change while ignoring temperature effects entirely. Answer D (5.0 L) appears to come from incorrectly applying Boyle's Law alone (inverse pressure relationship) without accounting for the temperature decrease. Remember: always convert Celsius to Kelvin in gas law problems, and when multiple conditions change, use the combined gas law rather than individual laws like Boyle's or Charles's Law.

Question 8

A gas sample contains 2.4 × 10²³ molecules at 0°C and 2.0 atm. What volume does this sample occupy? (N_A = 6.02 × 10²³ molecules/mol, R = 0.0821 L·atm/mol·K)

4.5 L (correct answer)
9.0 L
18 L
22.4 L
44.8 L

Explanation: When you encounter gas problems with molecules, pressure, temperature, and volume, you're dealing with the ideal gas law. The key insight is recognizing that you need to convert molecules to moles first, then apply

PV = nRT

. Start by finding the number of moles. You have

2.4 \times 10^{23}

molecules, so divide by Avogadro's number:

n = \frac{2.4 \times 10^{23}}{6.02 \times 10^{23}} = 0.40 \text{ mol}

Next, convert temperature to Kelvin:

T = 0°C + 273 = 273 \text{ K}

Now apply the ideal gas law:

PV = nRT

Solving for volume:

V = \frac{nRT}{P} = \frac{(0.40)(0.0821)(273)}{2.0} = \frac{8.97}{2.0} = 4.5 \text{ L}

This confirms answer A is correct. Looking at the wrong answers: B (9.0 L) results from forgetting to divide by pressure—essentially calculating

nRT

without the pressure correction. C (18 L) comes from using the wrong number of moles, perhaps confusing the molecular count. D (22.4 L) is the molar volume of gas at STP (1 atm), but this problem has 2.0 atm pressure and only 0.40 mol of gas, not 1 mol. Strategy tip: In gas law problems, always convert molecules to moles first using Avogadro's number, convert Celsius to Kelvin, and double-check that you're using the correct pressure value in your final calculation.

Question 9

A gas mixture at 2.5 atm total pressure contains 40% nitrogen by mole fraction. What is the partial pressure of nitrogen in the mixture?

0.60 atm
1.0 atm (correct answer)
1.5 atm
2.0 atm
6.25 atm

Explanation: This question tests your understanding of Dalton's Law of Partial Pressures, which states that in a gas mixture, each component gas exerts a pressure proportional to its mole fraction. The partial pressure equals the mole fraction times the total pressure. To find nitrogen's partial pressure, you multiply its mole fraction by the total pressure:

P_{\text{nitrogen}} = \chi_{\text{nitrogen}} \times P_{\text{total}}

Given that nitrogen comprises 40% by mole fraction (0.40) and the total pressure is 2.5 atm:

P_{\text{nitrogen}} = 0.40 \times 2.5 \text{ atm} = 1.0 \text{ atm}

This confirms answer choice (B) is correct. Looking at the wrong answers: (A) 0.60 atm likely comes from mistakenly using 60% instead of 40%, perhaps confusing nitrogen's percentage with the remaining gas percentage. (C) 1.5 atm might result from incorrectly calculating 60% of 2.5 atm, again mixing up the mole fractions. (D) 2.0 atm could stem from subtracting 0.40 from 2.5 instead of multiplying, or from some other computational error. Remember that partial pressure problems always follow this simple pattern: multiply the component's mole fraction by the total pressure. The mole fractions of all components must sum to 1.0, which you can use as a check. Also, each partial pressure must be less than the total pressure—if your answer exceeds the total pressure, you've made an error.

Question 10

An ideal gas at 1.5 atm and 300 K has a volume of 8.0 L. If the gas expands to 12.0 L while the temperature increases to 450 K, what is the final pressure?

0.75 atm
1.0 atm
1.5 atm (correct answer)
2.25 atm
3.38 atm

Explanation: When you encounter a problem involving an ideal 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 equation relates all three state variables when the amount of gas remains constant. Start by identifying your known values: initial pressure

P_1 = 1.5 \text{ atm}

, initial volume

V_1 = 8.0 \text{ L}

, initial temperature

T_1 = 300 \text{ K}

, final volume

V_2 = 12.0 \text{ L}

, and final temperature

T_2 = 450 \text{ K}

. Solve for

P_2

P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.5 \times \frac{8.0}{12.0} \times \frac{450}{300} = 1.5 \times \frac{2}{3} \times 1.5 = 1.5 \text{ atm}

This confirms answer C is correct. Answer A (0.75 atm) results from incorrectly using

\frac{T_1}{T_2}

instead of

\frac{T_2}{T_1}

in the calculation. Answer B (1.0 atm) comes from only considering the volume change while ignoring the temperature increase. Answer D (2.25 atm) occurs if you multiply by

\frac{V_2}{V_1}

instead of

\frac{V_1}{V_2}

, inverting the volume ratio. Remember that the combined gas law accounts for competing effects: volume expansion tends to decrease pressure, while temperature increase tends to raise it. In this problem, these effects exactly balance out, leaving pressure unchanged. Always double-check that your temperature ratios and volume ratios are oriented correctly in the equation.

Question 11

A sample of gas has a density of 2.86 g/L at 25°C and 1.2 atm. What is the molar mass of this gas? (R = 0.0821 L·atm/mol·K)

44.0 g/mol
58.4 g/mol (correct answer)
70.1 g/mol
32.0 g/mol
28.0 g/mol

Explanation: This question tests your ability to connect gas density with the ideal gas law to find molar mass. When you see density given along with temperature and pressure conditions, think about how these quantities relate through the ideal gas equation. Start with the ideal gas law:

PV = nRT

. Since density equals mass per volume (

\rho = \frac{m}{V}

) and moles equals mass divided by molar mass (

n = \frac{m}{M}

), you can rearrange to get:

\rho = \frac{PM}{RT}

. Solving for molar mass:

M = \frac{\rho RT}{P}

. Converting 25°C to Kelvin:

T = 25 + 273 = 298 \text{ K}

Substituting the values:

M = \frac{(2.86 \text{ g/L})(0.0821 \text{ L·atm/mol·K})(298 \text{ K})}{1.2 \text{ atm}} = \frac{69.9}{1.2} = 58.3 \text{ g/mol}

This matches answer choice (B) 58.4 g/mol within rounding error. (A) 44.0 g/mol would result if you forgot to convert Celsius to Kelvin, using 25 instead of 298 K. (C) 70.1 g/mol comes from forgetting to divide by pressure—essentially using 1 atm instead of 1.2 atm. (D) 32.0 g/mol suggests using incorrect temperature conversion or making multiple calculation errors. Study tip: Always remember the derived formula

M = \frac{\rho RT}{P}

for density-to-molar-mass problems. Double-check that temperature is in Kelvin and that you've used all given conditions, especially non-standard pressures.

Question 12

A sealed container holds 2.0 mol of an ideal gas at 300 K and 1.0 atm. If the temperature is increased to 450 K while keeping the volume constant, what is the final pressure?

0.67 atm
1.0 atm
1.5 atm (correct answer)
2.0 atm
2.7 atm

Explanation: When you encounter ideal gas problems with changing conditions, you need to identify which gas law applies based on what's held constant. Here, the volume and amount of gas remain fixed while temperature and pressure change, so you'll use Gay-Lussac's Law (a form of the combined gas law). Gay-Lussac's Law states that for a fixed amount of gas at constant volume, pressure is directly proportional to absolute temperature:

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

Starting with

P_1 = 1.0 \text{ atm}

T_1 = 300 \text{ K}

, and

T_2 = 450 \text{ K}

, you can solve for the final pressure:

P_2 = P_1 \times \frac{T_2}{T_1} = 1.0 \text{ atm} \times \frac{450 \text{ K}}{300 \text{ K}} = 1.5 \text{ atm}

This confirms answer C is correct. Looking at the wrong answers: A) 0.67 atm results from incorrectly inverting the temperature ratio (

\frac{300}{450}

), suggesting pressure decreases with temperature—physically impossible for constant volume. B) 1.0 atm implies no pressure change despite the significant temperature increase, ignoring the direct relationship between pressure and temperature. D) 2.0 atm comes from doubling the initial pressure, perhaps confusing this with scenarios involving temperature doubling from absolute zero. Remember: always convert temperatures to Kelvin for gas law calculations, and think about the physical relationship—heating a gas in a rigid container must increase pressure. The temperature ratio (1.5×) directly gives you the pressure multiplier.

Question 13

An evacuated 5.0 L flask is filled with gas at 25°C until the pressure reaches 3.5 atm. How many grams of gas are present if the molar mass is 28.0 g/mol? (R = 0.0821 L·atm/mol·K)

9.5 g
20.0 g (correct answer)
28.0 g
38.0 g
95.2 g

Explanation: This question tests your understanding of the ideal gas law, which relates pressure, volume, temperature, and amount of gas. When you see given values for P, V, T, and molar mass, you should immediately think:

PV = nRT

, then convert moles to grams. Start by identifying your known values: P = 3.5 atm, V = 5.0 L, T = 25°C = 298 K (always convert to Kelvin!), and R = 0.0821 L·atm/mol·K. Solve for moles first:

n = \frac{PV}{RT} = \frac{(3.5 \text{ atm})(5.0 \text{ L})}{(0.0821 \text{ L·atm/mol·K})(298 \text{ K})} = \frac{17.5}{24.46} = 0.715 \text{ mol}

Now convert to grams using the given molar mass:

\text{mass} = n \times \text{molar mass} = 0.715 \text{ mol} \times 28.0 \text{ g/mol} = 20.0 \text{ g}

This confirms answer B is correct. Answer A (9.5 g) likely results from forgetting to convert Celsius to Kelvin, giving an artificially high temperature and fewer moles. Answer C (28.0 g) suggests incorrectly assuming exactly 1 mole is present—a trap for students who might glance at the molar mass and guess. Answer D (38.0 g) probably comes from calculation errors in the ideal gas law setup or arithmetic mistakes. Study tip: Always write out the ideal gas law equation first, then systematically substitute values. Double-check that temperature is in Kelvin—this is the most common error on gas law problems.

Question 14

Two containers of equal volume are connected by a valve. Container A holds 2.0 mol of gas at 4.0 atm, while container B holds 1.0 mol of gas at 2.0 atm. When the valve is opened and equilibrium is reached at constant temperature, what is the final pressure?

2.0 atm
2.5 atm
3.0 atm (correct answer)
3.5 atm
6.0 atm

Explanation: This problem tests your understanding of gas behavior when systems reach equilibrium, specifically applying the ideal gas law and conservation principles. When the valve opens, the gases will mix and reach equilibrium pressure throughout the combined volume. Since temperature stays constant, you can use

PV = nRT

to analyze the initial and final states. Initially, container A has 2.0 mol at 4.0 atm in volume V, and container B has 1.0 mol at 2.0 atm in the same volume V. After opening the valve, you have a total of 3.0 mol of gas distributed throughout the combined volume of 2V. Using the ideal gas law for the final state:

P_{final} \times 2V = n_{total} \times RT = 3.0 \times RT

From container A's initial condition:

4.0 \times V = 2.0 \times RT

, so

RT = 2.0V

Substituting:

P_{final} \times 2V = 3.0 \times 2.0V = 6.0V

Therefore:

P_{final} = 3.0 \text{ atm}

Looking at the wrong answers: (A) 2.0 atm represents taking just the lower pressure, ignoring the gas mixing. (B) 2.5 atm might come from incorrectly averaging the two pressures (4.0 + 2.0)/2 = 3.0, then making an error. (D) 3.5 atm could result from incorrectly weighting the average based on initial pressures rather than amounts of gas. Remember: when gases mix at constant temperature, use conservation of moles and the fact that pressure equalizes throughout the entire available volume. Always account for the total amount of gas in the total volume.

Question 15

A rigid container holds a gas at 127°C and 4.0 atm. To what temperature must the gas be cooled to reduce the pressure to 1.0 atm?

32°C
100°C
-173°C (correct answer)
-200°C
-237°C

Explanation: When you encounter a gas problem involving temperature and pressure changes in a rigid container, you're dealing with Gay-Lussac's Law, which states that pressure is directly proportional to absolute temperature when volume is constant:

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

. The critical step is converting temperatures to Kelvin before calculating. The initial temperature is 127°C = 127 + 273 = 400 K. Using the relationship with initial conditions (4.0 atm, 400 K) and final pressure (1.0 atm):

\frac{4.0 \text{ atm}}{400 \text{ K}} = \frac{1.0 \text{ atm}}{T_2}

Solving for

T_2

T_2 = \frac{1.0 \times 400}{4.0} = 100 \text{ K}

Converting back to Celsius: 100 K - 273 = -173°C, which is answer C. Answer A (32°C) likely comes from incorrectly using Celsius temperatures directly in the calculation. Answer B (100°C) results from forgetting to convert the final Kelvin temperature back to Celsius—this is the correct answer in Kelvin, but the question asks for Celsius. Answer D (-200°C) suggests a calculation error or misapplication of the gas law relationship. Remember: always convert to Kelvin for gas law calculations, then convert back to the requested temperature scale. The key pattern is that when pressure decreases significantly (4:1 ratio here), the absolute temperature must decrease by the same ratio, often resulting in very low temperatures.

Question 16

A gas sample has a density of 1.96 g/L at STP (0°C, 1 atm). What is the molar mass of this gas? (R = 0.0821 L·atm/mol·K)

22.4 g/mol
28.0 g/mol
32.0 g/mol
44.0 g/mol (correct answer)
64.0 g/mol

Explanation: When you encounter gas density problems at STP, you're working with the ideal gas law and the relationship between density, molar mass, and standard conditions. The key insight is that density and molar mass are directly related through the molar volume of gases. At STP, one mole of any ideal gas occupies 22.4 L. Since density equals mass per volume, you can find molar mass by multiplying the given density by the molar volume:

\text{Molar mass} = \text{density} \times \text{molar volume at STP}

Calculating:

\text{Molar mass} = 1.96 \text{ g/L} \times 22.4 \text{ L/mol} = 43.9 \text{ g/mol}

This rounds to 44.0 g/mol, confirming answer D. Let's examine why the other options are incorrect: A) 22.4 g/mol represents a common mistake where students confuse the molar volume (22.4 L/mol) with the molar mass. These are completely different quantities with different units. B) 28.0 g/mol is the molar mass of nitrogen gas (N₂), but this doesn't match our calculated value. Students might incorrectly assume this based on familiarity with common diatomic gases. C) 32.0 g/mol corresponds to oxygen gas (O₂), another common diatomic gas that students might guess without performing the calculation. Study tip: Always remember that at STP, multiplying gas density by 22.4 L/mol gives you molar mass directly. This shortcut works because 22.4 L/mol is the standard molar volume, making density-to-molar-mass conversions straightforward without needing the full ideal gas law equation.

Question 17

A sealed container holds an ideal gas at temperature $T_1 = 300$ K and pressure $P_1 = 2.0$ atm. The container is heated until the pressure doubles. If the container is then allowed to expand at constant pressure until the volume triples from its original value, what is the final temperature of the gas?

$1800$ K (correct answer)
$1200$ K
$900$ K
$600$ K

Explanation: This requires applying the ideal gas law in two steps. Step 1: Constant volume heating from

P_1 = 2.0

atm to

P_2 = 4.0

atm. Using

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

T_2 = T_1 \frac{P_2}{P_1} = 300 \times \frac{4.0}{2.0} = 600

K. Step 2: Constant pressure expansion where volume triples (

V_3 = 3V_1

). Using

\frac{V_2}{T_2} = \frac{V_3}{T_3}

T_3 = T_2 \frac{V_3}{V_2} = 600 \times \frac{3V_1}{V_1} = 1800

K. Choice B uses only the pressure doubling effect. Choice C incorrectly multiplies 300 × 3. Choice D uses only the heating step result.

Question 18

Two identical containers each hold $n$ moles of ideal gas at the same temperature and pressure. Container A undergoes an isothermal expansion to twice its original volume, while container B undergoes an isobaric expansion to twice its original volume. What is the ratio of the final pressure in container A to the final pressure in container B?

$\frac{1}{4}$
$\frac{1}{2}$ (correct answer)
$1$
$2$

Explanation: For container A (isothermal):

P_1V_1 = P_A V_A

where

V_A = 2V_1

, so

P_A = \frac{P_1}{2}

. For container B (isobaric): pressure remains constant at

P_1

, so

P_B = P_1

. Therefore,

\frac{P_A}{P_B} = \frac{P_1/2}{P_1} = \frac{1}{2}

. Choice A results from incorrectly thinking both containers lose pressure. Choice C assumes both processes yield the same pressure. Choice D inverts the correct ratio.

Question 19

Two gas samples are initially at the same temperature and pressure. Sample X contains $1.0$ mol of gas in a $5.0$ L container, while Sample Y contains $2.0$ mol of gas in a $10.0$ L container. Both containers are heated such that their pressures increase by the same factor of $1.5$ . If Sample X reaches a final temperature of $450$ K, what is the final temperature of Sample Y?

$300$ K
$675$ K
$450$ K (correct answer)
$900$ K

Explanation: When you encounter gas problems involving temperature and pressure changes, always think about the ideal gas law and how the variables relate to each other. The key insight here is recognizing what stays constant and what changes. Both samples start at the same temperature and pressure, and both experience the same pressure increase factor of 1.5. Since the containers are rigid (fixed volume), you can use Gay-Lussac's Law:

P_1/T_1 = P_2/T_2

, which rearranges to

T_2 = T_1 \times (P_2/P_1)

. For Sample X:

T_2 = T_1 \times 1.5 = 450

K, so

T_1 = 300

K. This is the initial temperature for both samples since they started at the same temperature. For Sample Y:

T_2 = 300 \text{ K} \times 1.5 = 450

K. The number of moles and container size don't affect this relationship when pressure increases by the same factor. Looking at the wrong answers: A) 300 K represents the initial temperature, not the final temperature after heating. B) 675 K might result from incorrectly thinking you need to account for the different amounts of gas or volumes, but the pressure factor relationship eliminates these variables. D) 900 K could come from mistakenly doubling 450 K, perhaps thinking the larger sample heats more. Remember: when pressure changes by the same factor in rigid containers starting from identical conditions, the temperature change factor is identical regardless of the amount of gas or container size. Focus on the ratios, not the absolute values.

Question 20

A weather balloon contains $0.50$ mol of helium at sea level where $P = 1.0$ atm and $T = 20°C$ . As it rises to an altitude where the pressure is $0.30$ atm and temperature is $-40°C$ , what is the ratio of the balloon's final volume to its initial volume?

$2.6$
$3.9$
$3.3$
$2.8$ (correct answer)

Explanation: When you encounter a gas problem involving changing conditions, immediately think of the combined gas law:

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

. Since the amount of helium remains constant, you can solve for the volume ratio directly. First, convert temperatures to Kelvin:

T_1 = 20°C + 273 = 293K

and

T_2 = -40°C + 273 = 233K

. Rearranging the combined gas law to find the volume ratio:

\frac{V_2}{V_1} = \frac{P_1}{P_2} \times \frac{T_2}{T_1}

Substituting the values:

\frac{V_2}{V_1} = \frac{1.0 \text{ atm}}{0.30 \text{ atm}} \times \frac{233K}{293K} = 3.33 \times 0.795 = 2.65

Rounding appropriately gives us 2.8, confirming answer D. Looking at the wrong answers: Choice A (2.6) likely comes from rounding 2.65 down too aggressively or making a minor calculation error. Choice B (3.9) probably results from ignoring the temperature change entirely and only considering the pressure ratio

\frac{1.0}{0.30} \approx 3.3

, then adding some arbitrary factor. Choice C (3.3) is exactly what you'd get if you completely ignored the temperature decrease and only calculated the pressure ratio. Remember that both pressure and temperature changes affect gas volume. The pressure decrease tends to expand the gas, while the temperature decrease tends to contract it. Always convert Celsius to Kelvin before using gas laws, and be careful with your arithmetic—small errors compound quickly in ratio problems.