Gas Laws & Kinetic Theory - DAT Survey of the Natural Sciences

A sample of an ideal gas is heated from 100 °C to 200 °C in a container that expands to keep the pressure constant. How does the density of the gas change?
Answer: The density decreases because the volume increases. When you encounter gas problems involving temperature and pressure changes, think about how the ideal gas law connects these variables with volume and density. Since pressure remains constant while temperature increases, you need to apply Gay-Lussac's Law (a form of the ideal gas law): $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ at constant pressure. First, convert to absolute temperature: 100°C = 373 K and 200°C = 473 K. The volume ratio becomes $\frac{V_2}{V_1} = \frac{473 K}{373 K} = 1.27$ , meaning volume increases by 27%. Since density equals mass divided by volume ( $\rho = \frac{m}{V}$ ), and the mass of gas remains constant, density is inversely proportional to volume. When volume increases, density decreases proportionally. The density becomes $\frac{373}{473} = 0.79$ times the original density. Choice A incorrectly links molecular speed to density—while molecules do move faster at higher temperatures, this doesn't directly determine density. Choice B makes the classic error of using Celsius temperatures directly in gas law calculations; doubling from 100°C to 200°C doesn't double the absolute temperature (373 K to 473 K). Choice C misunderstands the relationship between variables—constant pressure doesn't mean constant density when temperature changes. The correct answer is D: density decreases because volume increases. Study tip: Always convert Celsius to Kelvin in gas law problems, and remember that density and volume are inversely related when mass stays constant. Questions often try to trick you with Celsius temperature relationships.
A sample of methane (CH₄, molar mass = 16.0 g/mol) effuses through a porous barrier in 5.0 minutes. How long would it take for the same number of moles of sulfur dioxide (SO₂, molar mass = 64.0 g/mol) to effuse under identical conditions?
Answer: 10.0 minutes When you encounter gas effusion problems, you're dealing with Graham's Law, which relates the rates of effusion to the molar masses of gases. Graham's Law states that the rate of effusion is inversely proportional to the square root of molar mass: $\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}$ Since rate is inversely related to time (faster rate = less time), we can write: $\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$ For this problem, methane (CH₄) has a molar mass of 16.0 g/mol and takes 5.0 minutes to effuse. Sulfur dioxide (SO₂) has a molar mass of 64.0 g/mol. Substituting into Graham's Law: $\frac{t_{SO_2}}{t_{CH_4}} = \sqrt{\frac{64.0}{16.0}} = \sqrt{4} = 2$ Therefore: $t_{SO_2} = 2 \times 5.0 \text{ min} = 10.0 \text{ minutes}$ Answer A (2.5 minutes) incorrectly suggests SO₂ effuses faster than CH₄, which violates the principle that heavier molecules effuse more slowly. Answer B (5.0 minutes) assumes equal effusion rates despite different molar masses—this would only be true if both gases had identical molar masses. Answer D (20.0 minutes) represents using the ratio of molar masses directly (64/16 = 4) instead of taking the square root. Remember: heavier molecules always effuse more slowly, and you must take the square root of the molar mass ratio. Graham's Law problems frequently appear on standardized tests, so practice identifying the square root relationship rather than using direct proportions.
According to the kinetic molecular theory of ideal gases, which of the following statements is a fundamental postulate?
Answer: Collisions between gas particles and with container walls are perfectly elastic. The kinetic molecular theory provides a fundamental framework for understanding ideal gas behavior by making several key assumptions about gas particles and their interactions. When you encounter questions about this theory, focus on the core postulates that define how ideal gases behave at the molecular level. The theory assumes that gas particles undergo perfectly elastic collisions, meaning no kinetic energy is lost when particles collide with each other or the container walls. This postulate is essential because it ensures that the total kinetic energy of the gas system remains constant at a given temperature, which is necessary for the gas laws to work. Answer D correctly identifies this fundamental assumption. Let's examine why the other options contradict the theory's postulates. Choice A is incorrect because ideal gas theory assumes gas particles have negligible volume compared to the container—they're treated as point masses. Choice B contains a critical error: average kinetic energy is proportional to absolute temperature (Kelvin), not Celsius temperature. The relationship is $KE_{avg} = \frac{3}{2}kT$ where T must be in Kelvin. Choice C directly contradicts another key postulate—the theory assumes no intermolecular forces exist between gas particles except during brief collisions. For DAT success, memorize the five main postulates of kinetic molecular theory: negligible particle volume, no intermolecular forces, random motion, elastic collisions, and average kinetic energy proportional to absolute temperature. Questions often test whether you can distinguish between ideal gas assumptions and real gas behaviors.
At a constant temperature and pressure, 5.0 L of N₂ gas contains 'n' moles. What volume would 2n moles of Ar gas occupy under the same conditions?
Answer: 10.0 L When you encounter gas problems involving volume and moles at constant temperature and pressure, you're working with Avogadro's Law, which states that equal volumes of gases contain equal numbers of molecules under identical conditions. This means volume is directly proportional to the number of moles. Since temperature and pressure remain constant, you can set up a direct proportion: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$ . Given that 5.0 L of N₂ contains n moles, you have $\frac{5.0 \text{ L}}{n} = \frac{V_2}{2n}$ . Cross-multiplying gives you $V_2 = \frac{5.0 \text{ L} \times 2n}{n} = 10.0 \text{ L}$ . The identity of the gas (N₂ vs. Ar) doesn't matter under these conditions—only the number of moles matters. Choice A (2.5 L) incorrectly suggests that doubling the moles halves the volume, which represents an inverse relationship. Choice B (5.0 L) fails to account for the doubled amount of gas—this would be correct only if you had n moles of Ar instead of 2n moles. Choice D (20.0 L) results from incorrectly multiplying both the original volume and the mole ratio (5.0 × 2 × 2), treating the problem as if both factors change the volume independently. Remember that Avogadro's Law creates a direct relationship: double the moles, double the volume. The type of gas is irrelevant as long as temperature and pressure stay constant. Watch for this proportional thinking in gas law problems.
A balloon contains 2.50 L of helium gas at 27°C. If the balloon is cooled to -73°C at constant pressure, what is its new volume?
Answer: 1.67 L This question tests Charles's Law, which describes the relationship between gas volume and temperature at constant pressure. When you see a gas problem involving temperature and volume changes, think about how gas particles move more slowly when cooled, allowing the gas to compress. To solve this, you need Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ . The critical step is converting temperatures to Kelvin, since gas laws require absolute temperature. Convert 27°C to 300 K and -73°C to 200 K by adding 273. Setting up the equation: $\frac{2.50 \text{ L}}{300 \text{ K}} = \frac{V_2}{200 \text{ K}}$ Solving for $V_2$ : $V_2 = \frac{2.50 \times 200}{300} = 1.67 \text{ L}$ This confirms answer B is correct. Answer A (0.93 L) represents too dramatic a volume decrease, likely from a calculation error or incorrect temperature conversion. Answer C (3.75 L) suggests the volume increased with cooling, which violates Charles's Law—gases contract when cooled, not expand. Answer D (6.76 L) shows an even larger increase, possibly from using Celsius temperatures directly in the calculation instead of converting to Kelvin. Remember that gas law problems always require absolute temperature (Kelvin). A quick check: since temperature decreased from 300 K to 200 K (a 2/3 ratio), the volume should also decrease proportionally to about 2/3 of the original, which matches our answer of 1.67 L.
A 1.0 L container at 300 K holds 1.0 atm of N₂ and 0.5 atm of O₂. A spark causes the reaction 2 N₂(g) + O₂(g) → 2 N₂O(g) to go to completion. What is the final total pressure in the container, assuming the temperature returns to 300 K?
Answer: 1.0 atm When you see gas reactions with pressure changes, you need to track both the stoichiometry and the limiting reactant to determine how many moles of gas remain after reaction. Start by identifying your limiting reactant. The balanced equation $2 \text{N}_2(g) + \text{O}_2(g) → 2 \text{N}_2\text{O}(g)$ shows you need 2 moles of N₂ for every 1 mole of O₂. Since partial pressures are proportional to moles at constant temperature and volume, you have 1.0 atm N₂ and 0.5 atm O₂. The stoichiometry requires 1.0 atm O₂ to consume all the N₂, but you only have 0.5 atm O₂. Therefore, O₂ is limiting. With 0.5 atm O₂ as your limiting reactant, the reaction will consume 1.0 atm N₂ (following the 2:1 ratio) and produce 1.0 atm N₂O. After reaction: 0 atm O₂ remains, 0 atm N₂ remains (1.0 - 1.0 = 0), and 1.0 atm N₂O is produced. Total final pressure = 1.0 atm. Answer A (0.5 atm) incorrectly assumes only the limiting reactant amount of product forms. Answer C (1.5 atm) wrongly adds all initial pressures without accounting for consumption. Answer D (2.0 atm) treats this as if no reaction occurred, ignoring the chemical change entirely. Strategy tip: In gas reaction problems, always identify the limiting reactant first, then track each species through the stoichiometry. The total moles of gas often change in reactions, so don't assume initial and final pressures are simply related.
Which of the following gases will have the highest root-mean-square speed at 25 °C?
Answer: H₂ (Molar Mass ≈ 2 g/mol) When you encounter questions about molecular speeds, you're dealing with kinetic molecular theory. The key insight is that at any given temperature, lighter molecules move faster than heavier ones because they all have the same average kinetic energy. The root-mean-square (RMS) speed is given by: $v_{rms} = \sqrt{\frac{3RT}{M}}$ Where R is the gas constant, T is temperature, and M is molar mass. Since temperature is constant at 25°C for all gases, the RMS speed is inversely proportional to the square root of molar mass. Simply put: lighter molecules = higher speeds. Looking at the molar masses: H₂ has by far the smallest molar mass at 2 g/mol, so it will have the highest RMS speed. This makes D the correct answer. Choice A (CO₂) is wrong because at 44 g/mol, it's the heaviest molecule listed, giving it the slowest RMS speed. Choice B (O₂) at 32 g/mol is lighter than CO₂ but still much heavier than H₂, so it moves significantly slower. Choice C (N₂) at 28 g/mol is the second-lightest after H₂, making it faster than CO₂ and O₂, but still much slower than H₂. For DAT questions about molecular speeds or effusion rates, always look for the lightest molecule first. The relationship is so strong that you rarely need to calculate - just identify the smallest molar mass. Remember: light and fast, heavy and slow.
What volume of CO₂ gas, measured at STP, is produced from the complete thermal decomposition of 50.0 g of calcium carbonate (CaCO₃, Molar Mass ≈ 100 g/mol) according to the reaction: CaCO₃(s) → CaO(s) + CO₂(g)?
Answer: 11.2 L This question tests your ability to connect stoichiometry with gas law calculations—a common combination on the DAT. When you see thermal decomposition with gas production and "at STP," you need to use molar relationships and the standard molar volume of gases. Start with the balanced equation: CaCO₃(s) → CaO(s) + CO₂(g). This shows a 1:1 mole ratio between calcium carbonate and carbon dioxide produced. First, convert grams to moles: $\frac{50.0 \text{ g CaCO₃}}{100 \text{ g/mol}} = 0.50 \text{ mol CaCO₃}$ Since the mole ratio is 1:1, you'll produce 0.50 mol CO₂. At STP (standard temperature and pressure), one mole of any gas occupies 22.4 L. Therefore: $0.50 \text{ mol CO₂} \times \frac{22.4 \text{ L}}{1 \text{ mol}} = 11.2 \text{ L}$ Answer choice A (5.6 L) represents exactly half the correct volume—you might get this if you mistakenly used 0.25 mol instead of 0.50 mol. Answer choice C (22.4 L) is the volume of 1.0 mol of gas at STP; this error occurs if you assume all 50.0 g becomes gas, forgetting that only the CO₂ portion (44 g/mol of the original 100 g/mol) contributes to gas volume. Answer choice D (44.8 L) suggests using 2.0 mol, possibly from incorrectly doubling somewhere in your calculation. Remember: at STP, always use 22.4 L/mol as your conversion factor, and pay careful attention to mole ratios from balanced equations—they're your bridge between mass and volume.
An unknown gas has a mass of 1.64 g and occupies a volume of 1.00 L at a pressure of 745 torr and a temperature of 27 °C. What is the approximate molar mass of the gas? (R = 0.0821 L·atm/mol·K, 1 atm = 760 torr)
Answer: 40 g/mol When you encounter a gas problem asking for molar mass, you're dealing with a combination of the ideal gas law and the definition of molar mass. The key insight is connecting PV = nRT with the relationship between moles, mass, and molar mass. Start by converting your given values to standard units. The pressure needs to be in atmospheres: 745 torr ÷ 760 torr/atm = 0.980 atm. The temperature must be in Kelvin: 27°C + 273 = 300 K. Using the ideal gas law PV = nRT, solve for moles: n = PV/RT = (0.980 atm)(1.00 L)/(0.0821 L·atm/mol·K)(300 K) = 0.0398 mol. Since molar mass = mass/moles, you get: MM = 1.64 g ÷ 0.0398 mol = 41.2 g/mol, which rounds to approximately 40 g/mol. Answer choice (C) 40 g/mol is correct. Choice (A) 4.0 g/mol would result from a calculation error, likely forgetting to convert temperature to Kelvin or making an order-of-magnitude mistake. Choice (B) 28 g/mol suggests using incorrect pressure conversion or arithmetic errors in the final division. Choice (D) 80 g/mol indicates doubling the correct answer, possibly from using the wrong gas constant or pressure units. For DAT gas problems, always double-check your unit conversions first—temperature to Kelvin and pressure to atmospheres when using R = 0.0821. Set up PV = nRT to find moles, then use molar mass = given mass ÷ calculated moles.
Boyle's law describes the relationship between pressure and volume, while Charles's law describes the relationship between volume and temperature. These laws are special cases of the ideal gas law under which respective conditions?
Answer: Constant n and T for Boyle's; Constant n and P for Charles's. When you encounter gas law questions, remember that Boyle's and Charles's laws are simplified versions of the ideal gas law ( $PV = nRT$ ) that apply when certain variables remain constant. Boyle's law states that pressure and volume are inversely proportional: $P_1V_1 = P_2V_2$ . This relationship only holds when temperature and the amount of gas remain constant. If temperature changed, the gas would expand or contract due to thermal effects, confounding the pressure-volume relationship you're trying to observe. Charles's law describes the direct relationship between volume and temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ . This works only when pressure and the amount of gas stay constant. If pressure varied, it would independently affect the volume, masking the pure temperature effect. Looking at the answer choices: Choice A correctly identifies that Boyle's law requires constant n (amount of gas) and T (temperature), while Charles's law requires constant n and P (pressure). Choice B incorrectly suggests Boyle's law needs constant volume, which is impossible since Boyle's law specifically examines how volume changes. Choice C swaps the conditions—it assigns Charles's law conditions to Boyle's law and vice versa. Choice D incorrectly states that Boyle's law requires constant temperature AND pressure, which would make volume completely fixed, and that Charles's law needs constant volume and temperature, which would contradict the law's purpose. Remember: each gas law isolates one relationship by holding all other variables constant except the two being studied.
Hydrogen gas is collected over water at 25 °C. The total pressure of the gas mixture in the collection tube is 755 torr. If the vapor pressure of water at 25 °C is 24 torr, what is the partial pressure of the hydrogen gas?
Answer: 731 torr When you encounter a gas collection problem involving water, you're dealing with Dalton's Law of Partial Pressures. The key insight is that the total pressure measured includes both your desired gas AND water vapor pressure. In this setup, hydrogen gas bubbles through water and gets collected. The total pressure reading (755 torr) represents the sum of hydrogen's partial pressure plus the water vapor that naturally evaporates into the collection space. Since water vapor pressure depends only on temperature, you can look up that at 25°C, water exerts 24 torr of pressure. Using Dalton's Law: $P_{total} = P_{H_2} + P_{H_2O}$ Solving for hydrogen's partial pressure: $P_{H_2} = P_{total} - P_{H_2O} = 755 \text{ torr} - 24 \text{ torr} = 731 \text{ torr}$ Choice A (24 torr) gives you only the water vapor pressure, not the hydrogen. Choice C (755 torr) incorrectly assumes the total pressure equals the hydrogen pressure, ignoring water vapor entirely. Choice D (779 torr) represents the common error of adding the water vapor pressure instead of subtracting it. The correct answer is B (731 torr). Remember this pattern: in gas-over-water collection problems, always subtract the water vapor pressure from the total pressure to find your target gas's partial pressure. The water vapor pressure depends only on temperature and can be found in reference tables.
A container holds a mixture of helium and argon gases. If the mole fraction of helium is 0.25 and its partial pressure is 150 torr, what is the total pressure inside the container?
Answer: 600 torr When you encounter gas mixture problems involving mole fractions and partial pressures, you're working with Dalton's Law of Partial Pressures. The key relationship is that partial pressure equals mole fraction times total pressure: $P_i = X_i \times P_{total}$ . Given that helium has a mole fraction of 0.25 and a partial pressure of 150 torr, you can solve for total pressure by rearranging the equation: $P_{total} = \frac{P_{He}}{X_{He}} = \frac{150 \text{ torr}}{0.25} = 600 \text{ torr}$ . This confirms answer C is correct. Let's examine why the other options are wrong. Answer A (37.5 torr) results from incorrectly multiplying the partial pressure by the mole fraction (150 × 0.25), which would give you a component of total pressure, not the total itself. Answer B (113 torr) comes from subtracting the helium's partial pressure from 150 torr and dividing by something arbitrary—there's no theoretical basis for this calculation. Answer D (750 torr) might result from adding 150 torr to 600 torr, mistakenly thinking you need to add the helium's contribution to the total rather than recognizing that partial pressure is already included in the total. Remember this pattern for the DAT: in gas mixture problems, partial pressure is always a fraction of total pressure determined by mole fraction. If you know any two of these three values (partial pressure, mole fraction, total pressure), you can calculate the third. Always check that your partial pressures are smaller than your total pressure.
Under which of the following conditions does a real gas most closely approximate the behavior of an ideal gas?
Answer: Low pressure and high temperature When you encounter questions about real gas behavior versus ideal gas behavior, focus on the two key assumptions of ideal gas theory: gas particles have no volume and no intermolecular forces. Real gases deviate from ideal behavior when these assumptions break down. At high temperature and low pressure, real gases behave most like ideal gases. High temperature gives gas particles enough kinetic energy that intermolecular attractions (like van der Waals forces) become negligible compared to their motion. Low pressure means the gas particles are spread far apart, making their actual volume insignificant compared to the container volume. Under these conditions, the gas particles act like point masses with no interactions—exactly what ideal gas theory assumes. Looking at the wrong answers: Choice B (low pressure, low temperature) gets the pressure right but fails because low temperature reduces particle kinetic energy, making intermolecular attractions more significant. Choice C (high pressure, low temperature) combines both problematic conditions—high pressure forces particles closer together where their actual volume matters, while low temperature again emphasizes intermolecular forces. Choice D (high pressure, high temperature) fixes the temperature issue but creates problems with high pressure, where particle volume becomes significant and particles are forced into closer proximity. Study tip for the DAT: Remember "Hot and Spread Out" for ideal gas conditions. High temperature overcomes intermolecular forces, while low pressure keeps particles spread apart. This combination minimizes both major sources of deviation from ideal behavior—a concept that frequently appears on standardized exams.
If the absolute temperature of a sample of an ideal gas is quadrupled, the average kinetic energy of its molecules will
Answer: increase by a factor of four. When you encounter questions about ideal gases and temperature, think about the kinetic molecular theory, which directly relates molecular motion to temperature. The average kinetic energy of gas molecules is given by the equation $KE_{avg} = \frac{3}{2}kT$ , where k is Boltzmann's constant and T is absolute temperature. This shows a direct proportional relationship between kinetic energy and temperature. If the absolute temperature is quadrupled (multiplied by 4), then: $KE_{new} = \frac{3}{2}k(4T) = 4 \times \frac{3}{2}kT = 4 \times KE_{original}$ Therefore, the average kinetic energy increases by a factor of four, making C correct. Looking at the wrong answers: A suggests the kinetic energy only doubles when temperature quadruples, which ignores the direct proportionality shown in the kinetic energy equation. B claims the energy increases by a factor of sixteen, which would only occur if kinetic energy were proportional to temperature squared—but it's not. D states that kinetic energy remains unchanged if volume is constant, but this confuses kinetic energy with other gas properties; kinetic energy depends only on temperature, not volume. Remember this key relationship: average kinetic energy of ideal gas molecules is directly proportional to absolute temperature. Volume, pressure, and amount of gas don't affect this relationship. When you see temperature changes in kinetic theory questions, apply the direct proportionality rule—if temperature increases by a factor of n, kinetic energy increases by the same factor of n.
Which of the following gas samples would occupy the largest volume at Standard Temperature and Pressure (STP)?
Answer: 16 g of He (Molar Mass ≈ 4 g/mol) When you encounter gas volume problems at STP, remember that equal numbers of moles of any gas occupy equal volumes. This means you need to convert the given masses to moles first, then determine which sample has the most moles. To find moles, use the formula: moles = mass ÷ molar mass. Let's calculate for each option: For option A: $\frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol of O₂}$ For option B: $\frac{16 \text{ g}}{16 \text{ g/mol}} = 1.0 \text{ mol of CH₄}$ For option C: $\frac{16 \text{ g}}{4 \text{ g/mol}} = 4.0 \text{ mol of He}$ For option D: $\frac{16 \text{ g}}{64 \text{ g/mol}} = 0.25 \text{ mol of SO₂}$ Since helium has the lowest molar mass, 16 grams represents the greatest number of moles (4.0 mol), making C correct. Option A is wrong because oxygen's higher molar mass (32 g/mol) means fewer moles per gram. Option B gives you exactly 1 mole, which is less than helium's 4 moles. Option D represents the trap of choosing the gas with the highest molar mass—SO₂ actually gives you the fewest moles and therefore the smallest volume. Remember this pattern: when comparing equal masses of different gases, the one with the lowest molar mass will always occupy the largest volume at STP. Look for the lightest gas when masses are equal.
The 'b' parameter in the van der Waals equation, $(P +$ $an^{2}$ $/$ $V^{2}$ $)(V - nb) = nRT$ , is a correction for the volume of the gas particles. Which of the following noble gases would be expected to have the largest 'b' value?
Answer: Kr When you encounter van der Waals equation problems, focus on what each parameter physically represents. The 'b' parameter corrects for the finite volume that gas particles themselves occupy - it represents the excluded volume per mole of gas particles. Since noble gas atoms are roughly spherical, the 'b' value directly correlates with atomic size. Larger atoms occupy more space and exclude more volume from being available to other particles. Looking at the periodic trends, atomic radius increases as you move down a group due to additional electron shells. Among these noble gases, Kr (krypton) is the largest atom. It sits in the fourth period with four electron shells, making it significantly larger than the others. This larger size means each Kr atom excludes more volume, resulting in the highest 'b' value. Choice A (He) is incorrect because helium has the smallest atomic radius of all noble gases, with only one electron shell, giving it the lowest 'b' value. Choice C (Ar) is wrong because argon, while larger than He and Ne, is still smaller than Kr since it's in the third period. Choice D (Ne) is incorrect because neon, in the second period, has a much smaller atomic radius than Kr. For DAT questions about gas behavior, remember that van der Waals corrections always relate to real molecular properties: 'a' corrects for intermolecular forces, while 'b' corrects for molecular size. When comparing atoms in the same group, larger atoms mean larger correction factors.
Consider the complete combustion of propane: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g). If all gases are measured at the same temperature and pressure, what volume of oxygen is required to completely react with 2.0 L of propane?
Answer: 10.0 L When you encounter gas stoichiometry problems, remember that gases at the same temperature and pressure have volumes directly proportional to their mole ratios. This means you can use the balanced equation coefficients as volume ratios. Looking at the balanced equation: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g), the coefficient ratio shows that 1 mole of propane reacts with 5 moles of oxygen. Since all gases are at the same conditions, this translates to a 1:5 volume ratio. If you start with 2.0 L of propane, you can set up the proportion: $\frac{1 \text{ L C₃H₈}}{5 \text{ L O₂}} = \frac{2.0 \text{ L C₃H₈}}{x \text{ L O₂}}$ Solving: $x = 2.0 \text{ L} × 5 = 10.0 \text{ L O₂}$ This confirms answer choice D is correct. Let's examine why the other options are wrong: Choice A (2.0 L) assumes a 1:1 ratio, ignoring the stoichiometric coefficients entirely. Choice B (5.0 L) represents the volume of oxygen needed for just 1.0 L of propane, not 2.0 L. Choice C (6.0 L) might result from incorrectly adding the propane volume to some calculation, but has no basis in the stoichiometry. Study tip: For gas stoichiometry at constant temperature and pressure, always use the balanced equation coefficients directly as volume ratios. This avoids unnecessary conversions through moles and makes calculations much faster on the DAT.
A gas mixture contains 2.0 moles of O₂, 3.0 moles of N₂, and 1.0 mole of Ar. If the total pressure of the mixture is 2.4 atm, what is the partial pressure of N₂?
Answer: 1.2 atm When you encounter gas mixture problems, you're dealing with Dalton's Law of Partial Pressures, which states that each gas in a mixture contributes to the total pressure proportionally to its mole fraction. To find the partial pressure of N₂, you need to calculate its mole fraction first. The total moles in the mixture is: 2.0 mol O₂ + 3.0 mol N₂ + 1.0 mol Ar = 6.0 mol total. The mole fraction of N₂ is: $\chi_{N_2} = \frac{3.0 \text{ mol N}_2}{6.0 \text{ mol total}} = 0.5$ The partial pressure equals the mole fraction times the total pressure: $P_{N_2} = \chi_{N_2} \times P_{total} = 0.5 \times 2.4 \text{ atm} = 1.2 \text{ atm}$ Looking at the wrong answers: Choice A (0.4 atm) represents the partial pressure of Ar, which has 1.0 mole out of 6.0 total moles. Choice B (0.8 atm) represents the partial pressure of O₂, which has 2.0 moles out of 6.0 total moles. Choice D (1.6 atm) might result from incorrectly using 3.0 moles divided by total pressure instead of using mole fraction properly. The correct answer is C (1.2 atm). Study tip: Always set up partial pressure problems in two steps: first calculate the mole fraction, then multiply by total pressure. This systematic approach prevents mix-ups between the different gases in the mixture and ensures you're using the correct proportions.
A rigid steel tank containing an ideal gas is heated from 200 K to 400 K. If the initial pressure was 150 kPa, what is the final pressure?
Answer: 300 kPa When you encounter gas problems 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 remains constant. Since the steel tank is rigid, the volume cannot change. For an ideal gas at constant volume, the relationship is: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ Setting up the calculation with the given values: - Initial pressure (P₁) = 150 kPa - Initial temperature (T₁) = 200 K - Final temperature (T₂) = 400 K Solving for final pressure: $P_2 = P_1 \times \frac{T_2}{T_1} = 150 \text{ kPa} \times \frac{400 \text{ K}}{200 \text{ K}} = 150 \times 2 = 300 \text{ kPa}$ This confirms answer C is correct. Looking at the wrong answers: A) 75 kPa represents an inverse relationship error—as if pressure decreased when temperature increased, which violates Gay-Lussac's Law. B) 150 kPa suggests no pressure change occurred, ignoring the temperature effect entirely. D) 600 kPa results from incorrectly multiplying the pressure by the temperature ratio (400/200) twice, or from other calculation errors. Remember this pattern: in rigid containers, doubling the absolute temperature doubles the pressure. Always use Kelvin temperatures for gas law calculations, and watch for the direct proportionality between pressure and temperature when volume is constant—this relationship appears frequently on science exams.
The van der Waals equation, $(P +$ $an^{2}$ $/$ $V^{2}$ $)(V - nb) = nRT$ , corrects the ideal gas law for real gas behavior. The 'a' parameter in the equation primarily accounts for...
Answer: the attractive intermolecular forces between gas molecules. When you encounter the van der Waals equation on the DAT, focus on understanding what each correction term addresses in real gas behavior. The ideal gas law assumes gas molecules have no volume and don't interact with each other, but real gases deviate from this idealized behavior. The van der Waals equation modifies the ideal gas law with two correction terms. The $a$ parameter appears in the pressure correction term $(P + \frac{an^2}{V^2})$ . This term accounts for the fact that gas molecules do attract each other through weak intermolecular forces (like van der Waals forces). These attractions effectively reduce the pressure that gas molecules exert on container walls compared to what the ideal gas law predicts. The $\frac{an^2}{V^2}$ term compensates by adding back this "lost" pressure. Choice A correctly identifies that the $a$ parameter accounts for attractive intermolecular forces. Choice B incorrectly suggests inelastic collisions—the van der Waals model still assumes elastic collisions. Choice C confuses the two correction terms; the finite volume of gas molecules is actually corrected by the $b$ parameter in the $(V - nb)$ term, not the $a$ parameter. Choice D is wrong because molecular velocity is already accounted for in the $nRT$ portion of the equation through temperature. Remember this pattern: in van der Waals corrections, $a$ = attractions (pressure correction) and $b$ = molecular bulk/volume (volume correction). This mnemonic will help you quickly identify what each parameter represents.

Start this quiz