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201. Dalton’s law of partial pressures states that:
ⓐ. The total pressure of a gas mixture equals the product of individual gas pressures.
ⓑ. The total pressure of a gas mixture equals the sum of pressures each gas would exert if present alone at the same volume and temperature.
ⓒ. Each gas in a mixture exerts zero pressure because others cancel it.
ⓓ. The total pressure depends only on the heaviest gas present.
Correct Answer: The total pressure of a gas mixture equals the sum of pressures each gas would exert if present alone at the same volume and temperature.
Explanation: According to Dalton, $P_{\text{total}} = P_1 + P_2 + P_3 + \dots$. Each component gas behaves independently and contributes partial pressure according to its mole fraction. Options A, C, and D are incorrect interpretations.
202. A 2 L container has 0.50 mol of $O_2$ and 0.50 mol of $N_2$ at 300 K. The total pressure is 24.6 atm. What is the partial pressure of each gas?
ⓐ. $O_2 = 12.3$ atm, $N_2 = 12.3$ atm
ⓑ. $O_2 = 24.6$ atm, $N_2 = 0$ atm
ⓒ. $O_2 = 6.15$ atm, $N_2 = 18.45$ atm
ⓓ. $O_2 = 8.0$ atm, $N_2 = 16.6$ atm
Correct Answer: $O_2 = 12.3$ atm, $N_2 = 12.3$ atm
Explanation: Total moles = 1.0. Each gas contributes half the moles, so mole fraction = 0.50. Partial pressure = mole fraction × total pressure. $P_{O_2} = P_{N_2} = 0.50 \times 24.6 = 12.3$ atm.
203. If a gas mixture contains 3 mol of $H_2$, 1 mol of $O_2$, and 2 mol of $N_2$, and total pressure is 6 atm, what is the partial pressure of $H_2$?
ⓐ. 1 atm
ⓑ. 2 atm
ⓒ. 3 atm
ⓓ. 4 atm
Correct Answer: 4 atm
Explanation: Total moles = 6. Mole fraction of $H_2 = 3/6 = 0.5$. So partial pressure = 0.5 × 6 atm = 3 atm. Wait—correction: 3 mol $H_2$ / (3+1+2 = 6 mol) = 0.5. 0.5 × 6 atm = 3 atm. Correct answer is C. 3 atm.
204. A mixture of gases contains 2 mol $He$ and 1 mol $Ar$. If the total pressure is 9 atm, the partial pressure of helium is:
ⓐ. 3 atm
ⓑ. 4.5 atm
ⓒ. 6 atm
ⓓ. 9 atm
Correct Answer: 6 atm
Explanation: Total moles = 3. Mole fraction of $He = 2/3$. So partial pressure = $(2/3) \times 9 = 6$ atm.
205. Why does Dalton’s law hold true for ideal gases?
ⓐ. Because molecules of different gases occupy different spaces
ⓑ. Because molecules of different gases have infinite volume
ⓒ. Because ideal gases do not interact, so each exerts pressure independently
ⓓ. Because gases at high pressure behave ideally
Correct Answer: Because ideal gases do not interact, so each exerts pressure independently
Explanation: In an ideal gas, there are no intermolecular forces, so each gas molecule collides with container walls as if other gases were absent. Therefore, partial pressures add up.
206. A gas mixture of 5 mol $N_2$ and 3 mol $O_2$ exerts a total pressure of 4 atm. What is the partial pressure of oxygen?
ⓐ. 1 atm
ⓑ. 1.5 atm
ⓒ. 2 atm
ⓓ. 2.5 atm
Correct Answer: 1.5 atm
Explanation: Total moles = 8. Mole fraction of $O_2 = 3/8 = 0.375$. Partial pressure = 0.375 × 4 atm = 1.5 atm.
207. In collecting gas over water, the measured pressure includes:
ⓐ. Only the gas pressure
ⓑ. Only vapor pressure of water
ⓒ. Sum of gas pressure and vapor pressure of water
ⓓ. Neither gas nor vapor pressure
Correct Answer: Sum of gas pressure and vapor pressure of water
Explanation: When gas is collected over water, it is saturated with water vapor. Thus the observed pressure is $P_{\text{gas}} + P_{\text{H2O}}$. The actual dry gas pressure = observed pressure – vapor pressure of water.
208. A sample of hydrogen gas is collected over water at 25°C. If total pressure is 760 mmHg and vapor pressure of water is 24 mmHg, what is the pressure of hydrogen?
ⓐ. 24 mmHg
ⓑ. 736 mmHg
ⓒ. 760 mmHg
ⓓ. 784 mmHg
Correct Answer: 736 mmHg
Explanation: Dry hydrogen pressure = total pressure – water vapor pressure = 760 – 24 = 736 mmHg.
209. A mixture of 2 mol $CO_2$, 2 mol $O_2$, and 4 mol $N_2$ exerts 1 atm total pressure. The partial pressure of nitrogen is:
ⓐ. 0.25 atm
ⓑ. 0.40 atm
ⓒ. 0.50 atm
ⓓ. 0.80 atm
Correct Answer: 0.50 atm
Explanation: Total moles = 8. Mole fraction of $N_2 = 4/8 = 0.5$. Partial pressure = 0.5 × 1 atm = 0.50 atm.
210. Dalton’s law can be combined with the ideal gas equation to give:
ⓐ. $P_{\text{total}}V = n_{\text{total}}RT$
ⓑ. $P_{\text{total}}V = (n_1 + n_2)RT$
ⓒ. $P_{\text{total}} = P_1 + P_2 + P_3 \dots$
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Dalton’s law adds partial pressures: $P_{\text{total}} = P_1 + P_2 + P_3 \dots$. Using ideal gas law for each gas, total equation becomes $P_{\text{total}}V = (n_1+n_2+n_3 \dots)RT = n_{\text{total}}RT$. Thus, all statements are correct.
211. A 10 L container holds 0.5 mol of $O_2$ and 1.0 mol of $N_2$ at 300 K. What is the total pressure exerted by the mixture?
ⓐ. 1.23 atm
ⓑ. 2.46 atm
ⓒ. 3.69 atm
ⓓ. 4.92 atm
Correct Answer: 2.46 atm
Explanation: Total moles = 1.5. Using $P = \dfrac{nRT}{V}$, with $R = 0.0821 \, L\,atm\,K^{-1}mol^{-1}$:
$P = \dfrac{(1.5)(0.0821)(300)}{10} \approx 2.46 \, atm$. The total pressure is the sum of partial pressures of $O_2$ and $N_2$.
212. In the above question, what is the partial pressure of oxygen?
ⓐ. 0.82 atm
ⓑ. 1.23 atm
ⓒ. 1.64 atm
ⓓ. 2.46 atm
Correct Answer: 0.82 atm
Explanation: Mole fraction of $O_2 = 0.5/1.5 = 0.333$. Partial pressure $P_{O_2} = \chi_{O_2} \times P_{total} = 0.333 \times 2.46 \approx 0.82 \, atm$.
213. Which formula relates partial pressure of a gas to its mole fraction?
ⓐ. $P_i = n_iRT$
ⓑ. $P_i = \dfrac{n_i}{V}$
ⓒ. $P_i = \chi_i \times P_{total}$
ⓓ. $P_i = P_{total} – \chi_i$
Correct Answer: $P_i = \chi_i \times P_{total}$
Explanation: Mole fraction $\chi_i = \dfrac{n_i}{n_{total}}$. Multiplying this fraction by the total pressure gives the partial pressure of the component gas.
214. A gas mixture contains 2 mol $H_2$, 1 mol $O_2$, and 1 mol $N_2$. If the total pressure is 8 atm, what is the partial pressure of hydrogen?
ⓐ. 2 atm
ⓑ. 3 atm
ⓒ. 4 atm
ⓓ. 5 atm
Correct Answer: 4 atm
Explanation: Total moles = 4. Mole fraction of $H_2 = 2/4 = 0.5$. Partial pressure $= 0.5 \times 8 = 4 \, atm$.
215. Why does Dalton’s law fail for real gases at high pressure?
ⓐ. Because gases stop colliding with container walls
ⓑ. Because intermolecular forces and finite molecular volume cannot be neglected
ⓒ. Because molecules disappear at high pressure
ⓓ. Because temperature becomes zero at high pressure
Correct Answer: Because intermolecular forces and finite molecular volume cannot be neglected
Explanation: Dalton’s law assumes ideal behavior where gases do not interact. At high pressure and low temperature, intermolecular attractions and finite size of molecules cause deviations.
216. A mixture of 2 mol $CH_4$, 3 mol $CO_2$, and 5 mol $N_2$ exerts 9 atm total pressure. The partial pressure of $CO_2$ is:
ⓐ. 1 atm
ⓑ. 2.7 atm
ⓒ. 3 atm
ⓓ. 5 atm
Correct Answer: 2.7 atm
Explanation: Total moles = 10. Mole fraction of $CO_2 = 3/10 = 0.3$. Partial pressure = 0.3 × 9 = 2.7 atm.
217. When gas is collected over water, why must the vapor pressure of water be subtracted from the total pressure?
ⓐ. Because vapor pressure adds to the mass of gas
ⓑ. Because measured pressure is the sum of dry gas and water vapor pressure
ⓒ. Because water vapor reduces the number of gas molecules
ⓓ. Because water vapor exerts no pressure
Correct Answer: Because measured pressure is the sum of dry gas and water vapor pressure
Explanation: The gas collected is saturated with water vapor. Hence, total pressure = pressure of dry gas + vapor pressure of water. To find the dry gas pressure, subtract vapor pressure of water.
218. A mixture of gases at 27°C and 1 atm contains 0.4 mol of $H_2$ and 0.6 mol of $N_2$. What is the partial pressure of nitrogen?
ⓐ. 0.2 atm
ⓑ. 0.4 atm
ⓒ. 0.6 atm
ⓓ. 0.8 atm
Correct Answer: 0.6 atm
Explanation: Mole fraction of $N_2 = 0.6/1.0 = 0.6$. Partial pressure $= 0.6 \times 1 = 0.6 \, atm$.
219. Which of the following statements is correct about Dalton’s law?
ⓐ. It applies only to identical gases.
ⓑ. It is valid for non-reacting gases.
ⓒ. It holds true for reacting gas mixtures as well.
ⓓ. It is applicable only at high pressures.
Correct Answer: It is valid for non-reacting gases.
Explanation: Dalton’s law works when gases do not chemically react with each other. For reacting mixtures, the composition changes, so the law cannot be directly applied.
220. In scuba diving, the composition of compressed air cylinders must consider Dalton’s law because:
ⓐ. Partial pressures of gases affect solubility in blood, influencing safety
ⓑ. Mole fractions remain constant with depth
ⓒ. Total pressure is independent of depth
ⓓ. Oxygen exerts zero partial pressure under water
Correct Answer: Partial pressures of gases affect solubility in blood, influencing safety
Explanation: At high pressures underwater, nitrogen dissolves more in blood due to higher partial pressure, risking decompression sickness. Oxygen partial pressure must also be regulated to avoid toxicity. This application directly follows Dalton’s law.
221. Graham’s law of diffusion states that the rate of diffusion of a gas is:
ⓐ. Directly proportional to the square root of its molar mass
ⓑ. Inversely proportional to the square root of its molar mass
ⓒ. Directly proportional to its molar mass
ⓓ. Independent of molar mass
Correct Answer: Inversely proportional to the square root of its molar mass
Explanation: Graham’s law:
$$
r \propto \frac{1}{\sqrt{M}}
$$
where $r$ is diffusion rate and $M$ is molar mass. Lighter gases diffuse faster, heavier gases slower.
222. If gas A has molar mass 4 and gas B has molar mass 64, the ratio of their diffusion rates ($r_A:r_B$) is:
ⓐ. 1:4
ⓑ. 1:2
ⓒ. 4:1
ⓓ. 8:1
Correct Answer: 4:1
Explanation: By Graham’s law:
$$
\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4
$$
So gas A diffuses 4 times faster than gas B.
223. Hydrogen diffuses four times faster than which of the following gases?
ⓐ. Oxygen ($M=32$)
ⓑ. Carbon dioxide ($M=44$)
ⓒ. Nitrogen ($M=28$)
ⓓ. Methane ($M=16$)
Correct Answer: Oxygen ($M=32$)
Explanation: Hydrogen’s molar mass = 2. According to Graham’s law, if H₂ diffuses 4 times faster, other gas’s molar mass should be 16 times greater (since $r_H/r_{gas} = 4$). $M_{gas} = 2 \times 16 = 32$, i.e., O₂.
224. Which pair of gases will diffuse at approximately the same rate?
ⓐ. $O_2 (M=32)$ and $CH_4 (M=16)$
ⓑ. $CO (M=28)$ and $N_2 (M=28)$
ⓒ. $CO_2 (M=44)$ and $Cl_2 (M=71)$
ⓓ. $H_2 (M=2)$ and $He (M=4)$
Correct Answer: $CO (M=28)$ and $N_2 (M=28)$
Explanation: Since both have nearly equal molar mass, they diffuse at nearly the same rate. Other pairs differ significantly in molar masses.
225. If 1 L of hydrogen escapes from a small hole in 10 minutes, how much oxygen will escape in the same time under identical conditions?
ⓐ. 0.25 L
ⓑ. 2.50 L
ⓒ. 1.0 L
ⓓ. 0.50 L
Correct Answer: 0.50 L
Explanation: $ \dfrac{r_H}{r_O} = \sqrt{\dfrac{M_O}{M_H}} = \sqrt{\dfrac{32}{2}} = \sqrt{16} = 4$. Thus H₂ diffuses 4 times faster. Oxygen volume = $1/4 = 0.25$. Wait: Correction—Given 1 L H₂ → O₂ diffuses 1/√16 = 1/4 as fast. So volume of O₂ = $1.0/4 = 0.25$ L. Correct answer is A. 0.25 L.
226. Which property of gases is explained by Graham’s law?
ⓐ. Liquefaction
ⓑ. Compressibility
ⓒ. Diffusion and effusion
ⓓ. Critical temperature
Correct Answer: Diffusion and effusion
Explanation: Graham’s law describes how gas molecules spread (diffusion) or pass through a small hole (effusion). Rate is inversely proportional to square root of molar mass.
227. The ratio of diffusion rates of NH₃ (M = 17) and HCl (M = 36.5) is approximately:
ⓐ. 1:1
ⓑ. 2.5:1
ⓒ. 2:1
ⓓ. 1.5:1
Correct Answer: 1.5:1
Explanation:
$$
\frac{r_{NH_3}}{r_{HCl}} = \sqrt{\frac{36.5}{17}} \approx \sqrt{2.15} \approx 1.46 \approx 1.5
$$
So NH₃ diffuses about 1.5 times faster than HCl.
228. When ammonia and hydrogen chloride gases diffuse toward each other, a white ring of $NH_4Cl$ forms closer to HCl because:
ⓐ. HCl is lighter and diffuses faster
ⓑ. NH₃ is lighter and diffuses faster
ⓒ. Both diffuse equally
ⓓ. NH₃ does not diffuse at all
Correct Answer: NH₃ is lighter and diffuses faster
Explanation: NH₃ (M=17) diffuses faster than HCl (M=36.5). Hence, the point of meeting (white ring) is closer to the slower HCl end of the tube.
229. If gas A diffuses twice as fast as gas B, the ratio of their molar masses is:
ⓐ. 1:2
ⓑ. 4:1
ⓒ. 2:1
ⓓ. 1:4
Correct Answer: 1:4
Explanation: Graham’s law: $r_A/r_B = \sqrt{M_B/M_A} = 2$. Squaring gives $M_B/M_A = 4$. Thus $M_A:M_B = 1:4$.
230. Which real-life application uses Graham’s law of diffusion?
ⓐ. Oxygen entering blood through alveoli in lungs
ⓑ. High pressure cooking in a cooker
ⓒ. Melting of ice at 0°C
ⓓ. Expansion of metals on heating
Correct Answer: Oxygen entering blood through alveoli in lungs
Explanation: Diffusion of gases across membranes (like $O_2$ and $CO_2$ exchange in lungs) follows Graham’s law. Lighter molecules diffuse faster through biological membranes, which is critical in respiration.
231. According to Graham’s law, if two gases diffuse at the same rate, then:
ⓐ. They must have equal densities
ⓑ. They must have equal molar masses
ⓒ. They must have equal volumes
ⓓ. They must have equal temperatures
Correct Answer: They must have equal molar masses
Explanation: Rate of diffusion depends only on molar mass (at same T and P). If diffusion rates are equal, molar masses are equal, regardless of density, volume, or temperature.
232. The rate of diffusion of helium is how many times greater than that of methane ($CH_4$, M = 16)?
ⓐ. 30
ⓑ. 16
ⓒ. 8
ⓓ. 2
Correct Answer: 2
Explanation: $r_{He}/r_{CH_4} = \sqrt{M_{CH_4}/M_{He}} = \sqrt{16/4} = \sqrt{4} = 2$. So helium diffuses twice as fast as methane.
233. In an experiment, 100 mL of hydrogen diffuses in 25 seconds. How long will it take for the same volume of oxygen ($M=32$) to diffuse under identical conditions?
ⓐ. 25 s
ⓑ. 50 s
ⓒ. 70 s
ⓓ. 75 s
Correct Answer: 75 s
Explanation: Time is inversely proportional to rate. $t_O/t_H = r_H/r_O = \sqrt{M_O/M_H} = \sqrt{32/2} = \sqrt{16} = 4$. So $t_O = 25 \times 4 = 100 \, s$. Wait—check: 100 mL of H₂ in 25 s. For O₂, $r_H/r_O = \sqrt{32/2} = 4$. So O₂ diffuses 4 times slower → takes 4 × 25 = 100 s. Correct answer: D. 100 s.
234. Which of the following gases will diffuse fastest at the same conditions?
ⓐ. $O_2$ (M=32)
ⓑ. $CO_2$ (M=44)
ⓒ. $CH_4$ (M=16)
ⓓ. $Cl_2$ (M=71)
Correct Answer: $CH_4$ (M=16)
Explanation: Lighter molar mass means faster diffusion. $CH_4$ has the lowest molar mass among given gases, so it diffuses fastest.
235. Which relation is correct for diffusion rates of gases A and B?
ⓐ. $\dfrac{r_A}{r_B} = \dfrac{M_A}{M_B}$
ⓑ. $\dfrac{r_A}{r_B} = \sqrt{\dfrac{M_B}{M_A}}$
ⓒ. $\dfrac{r_A}{r_B} = \dfrac{1}{\sqrt{M_A+M_B}}$
ⓓ. $\dfrac{r_A}{r_B} = \dfrac{T_A}{T_B}$
Correct Answer: $\dfrac{r_A}{r_B} = \sqrt{\dfrac{M_B}{M_A}}$
Explanation: Graham’s law defines relative diffusion rates by square root of inverse molar masses. None of the other options reflect the correct dependence.
236. Ammonia gas ($M=17$) diffuses through a porous pot in 40 s. How long would hydrogen ($M=2$) take under the same conditions?
ⓐ. 14 s
ⓑ. 20 s
ⓒ. 28 s
ⓓ. 40 s
Correct Answer: 14 s
Explanation: $t_H/t_{NH_3} = \sqrt{M_H/M_{NH_3}} = \sqrt{2/17} \approx 0.34$. $ t_H = 0.34 \times 40 = 13.6 \approx 14 \, s$.
237. Which gas diffuses about 1.6 times faster than chlorine ($Cl_2$, M=71)?
ⓐ. Oxygen (M=32)
ⓑ. Ammonia (M=17)
ⓒ. Hydrogen (M=2)
ⓓ. Methane (M=16)
Correct Answer: Oxygen (M=32)
Explanation: $r_{O_2}/r_{Cl_2} = \sqrt{71/32} \approx 1.49 \approx 1.5$. This matches closest to 1.6, so O₂ diffuses \~1.5 times faster than chlorine.
238. Effusion of a gas refers to:
ⓐ. Gas escaping through a large opening into another container
ⓑ. Gas spreading slowly through liquid medium
ⓒ. Gas passing through a very small hole without collisions
ⓓ. Gas molecules colliding with each other at high pressure
Correct Answer: Gas passing through a very small hole without collisions
Explanation: Effusion is the process of gas molecules escaping one container to another through an extremely tiny hole without intermolecular collisions. Diffusion refers to spreading due to random molecular motion.
239. The diffusion rate of $CO$ (M=28) compared to $CO_2$ (M=44) is:
ⓐ. 0.80
ⓑ. 1.0
ⓒ. 1.25
ⓓ. 1.50
Correct Answer: 1.25
Explanation: $r_{CO}/r_{CO_2} = \sqrt{44/28} = \sqrt{1.57} \approx 1.25$. Thus CO diffuses \~25% faster than CO₂.
240. In an experiment, two gases diffuse through the same pinhole. If gas A takes 36 s and gas B takes 64 s, the ratio of their molar masses ($M_A:M_B$) is:
ⓐ. 1:2
ⓑ. 9:16
ⓒ. 16:9
ⓓ. 81:256
Correct Answer: 81:256
Explanation: Time taken is proportional to $\sqrt{M}$. So $(t_A/t_B)^2 = M_A/M_B$. Substituting: $ (36/64)^2 = (0.5625)^2 = 0.316 \approx 9/28$. Wait—check: $M_A/M_B = (36/64)^2 = (0.5625)^2 = 0.316$. Correction: let’s recalc— $(36/64)^2 = (9/16)^2 = 81/256 \approx 0.316$. So ratio = 81:256.
241. Which of the following is not an assumption of the kinetic theory of gases?
ⓐ. Gas molecules are in continuous random motion.
ⓑ. The volume of individual gas molecules is negligible compared to the volume of the container.
ⓒ. Molecules exert attractive and repulsive forces on each other at all times.
ⓓ. Collisions between molecules and container walls are perfectly elastic.
Correct Answer: Molecules exert attractive and repulsive forces on each other at all times.
Explanation: Kinetic theory assumes ideal gases have no intermolecular forces; molecules act independently. Options A, B, and D are correct assumptions. Deviations occur in real gases, but for ideal gases intermolecular forces are neglected.
242. According to kinetic theory, pressure of a gas arises due to:
ⓐ. Attractive forces between molecules
ⓑ. Collisions of gas molecules with the walls of the container
ⓒ. Weight of the gas molecules
ⓓ. Vibrations of gas molecules in fixed positions
Correct Answer: Collisions of gas molecules with the walls of the container
Explanation: Pressure is the macroscopic effect of microscopic elastic collisions of molecules against the walls. Each collision exerts a force; the sum over time and area produces measurable pressure.
243. Which assumption of kinetic theory explains why gases have neither definite shape nor definite volume?
ⓐ. Gas molecules are far apart with negligible volume.
ⓑ. Gas molecules are in constant random motion.
ⓒ. Collisions are perfectly elastic.
ⓓ. No intermolecular attractions exist.
Correct Answer: Gas molecules are far apart with negligible volume.
Explanation: Because intermolecular spaces are huge compared to molecular size, gases can expand indefinitely and occupy the shape and volume of their container.
244. Which assumption of kinetic theory explains the compressibility of gases?
ⓐ. Gas molecules have negligible volume compared to container volume.
ⓑ. Gas molecules exert strong attractive forces.
ⓒ. Gas molecules lose energy on collision.
ⓓ. Gas molecules move in fixed paths.
Correct Answer: Gas molecules have negligible volume compared to container volume.
Explanation: Since most of the container volume is empty space, gases can be compressed significantly. Liquids and solids cannot be compressed as much because particles are closely packed.
245. Which of the following is assumed to remain constant during molecular collisions in kinetic theory?
ⓐ. Total kinetic energy
ⓑ. Velocity of each molecule
ⓒ. Direction of each molecule
ⓓ. Volume of container
Correct Answer: Total kinetic energy
Explanation: Kinetic theory assumes collisions are perfectly elastic. Energy may redistribute between molecules, but the total kinetic energy remains constant, conserving energy at the microscopic level.
246. Which assumption of kinetic theory accounts for the diffusion of gases?
ⓐ. Gas molecules have definite volume.
ⓑ. Gas molecules are in constant random motion.
ⓒ. Gas molecules exert strong forces of attraction.
ⓓ. Gas molecules lose energy on collision.
Correct Answer: Gas molecules are in constant random motion.
Explanation: Continuous random motion allows gas molecules to spread out and mix with others spontaneously. Diffusion occurs without external force because of this molecular motion.
247. According to kinetic theory, the average kinetic energy of gas molecules is directly proportional to:
ⓐ. Absolute pressure
ⓑ. Absolute temperature
ⓒ. Volume of the gas
ⓓ. Molar mass of the gas
Correct Answer: Absolute temperature
Explanation: The assumption is $KE_{avg} = \tfrac{3}{2} k_B T$. This shows energy depends only on temperature, not on pressure, volume, or molar mass.
248. Why does kinetic theory assume intermolecular forces are negligible in gases?
ⓐ. Because molecules are too small to interact
ⓑ. Because molecules are widely separated compared to their size
ⓒ. Because temperature is always high
ⓓ. Because molecules have zero mass
Correct Answer: Because molecules are widely separated compared to their size
Explanation: The large intermolecular spaces in gases mean attractive forces are negligible under normal conditions. This assumption simplifies gas behavior into ideal gas laws.
249. Which of the following correctly describes motion of gas molecules according to kinetic theory?
ⓐ. Molecules move in straight lines until colliding with other molecules or container walls.
ⓑ. Molecules move in circular paths under constant acceleration.
ⓒ. Molecules vibrate around fixed positions.
ⓓ. Molecules oscillate harmonically with zero collisions.
Correct Answer: Molecules move in straight lines until colliding with other molecules or container walls.
Explanation: Between collisions, molecules are unaffected by forces and move uniformly in straight lines. Collisions change speed and direction but are perfectly elastic.
250. Which assumption of kinetic theory best explains why gases have very low densities compared to liquids and solids?
ⓐ. Gas molecules exert no attractive forces.
ⓑ. Gas molecules have negligible volume compared to container.
ⓒ. Gas molecules are in constant motion.
ⓓ. Gas molecules collide elastically.
Correct Answer: Gas molecules have negligible volume compared to container.
Explanation: Since the actual volume occupied by molecules is tiny relative to container volume, the mass per unit volume (density) is low. Liquids and solids have molecules tightly packed, leading to higher density.
251. In the kinetic theory of gases, the pressure exerted by a gas on the container walls is due to:
ⓐ. Gravitational pull on gas molecules
ⓑ. Elastic collisions of molecules with the walls
ⓒ. Attractive forces between gas molecules
ⓓ. Vibrations of molecules in fixed positions
Correct Answer: Elastic collisions of molecules with the walls
Explanation: Pressure is the macroscopic manifestation of microscopic collisions. Each collision transfers momentum to the wall. Since collisions are perfectly elastic, total kinetic energy is conserved, and the continuous bombardment produces uniform pressure.
252. Which expression represents the pressure of an ideal gas derived from kinetic theory?
ⓐ. $P = \dfrac{1}{3} \rho \overline{c^2}$
ⓑ. $P = \rho \overline{c}$
ⓒ. $P = \dfrac{2}{3} \rho \overline{c}$
ⓓ. $P = \dfrac{1}{2} \rho \overline{c^2}$
Correct Answer: $P = \dfrac{1}{3} \rho \overline{c^2}$
Explanation: The derivation gives:
$$
P = \frac{1}{3}\rho \overline{c^2}
$$
where $\rho$ = mass density of the gas and $\overline{c^2}$ = mean square velocity. The factor $1/3$ arises from considering three-dimensional motion.
253. In the pressure derivation, why is the factor $\tfrac{1}{3}$ included in the equation?
ⓐ. To account for intermolecular attractions
ⓑ. To consider motion in three dimensions (x, y, z)
ⓒ. To balance energy loss in collisions
ⓓ. To match experimental data only
Correct Answer: To consider motion in three dimensions (x, y, z)
Explanation: The velocity squared is distributed equally among three orthogonal directions. Thus,
$$
\overline{c^2} = \overline{c_x^2} + \overline{c_y^2} + \overline{c_z^2} = 3\overline{c_x^2}
$$
leading to the factor $1/3$ in the pressure equation.
254. Which step is essential in deriving the kinetic theory equation for pressure?
ⓐ. Considering intermolecular potential energy
ⓑ. Calculating change in momentum of a molecule after collision with wall
ⓒ. Assuming molecules form a lattice
ⓓ. Including external gravitational force
Correct Answer: Calculating change in momentum of a molecule after collision with wall
Explanation: A molecule of mass $m$ and velocity $c_x$ collides elastically with the wall. Change in momentum = $2mc_x$. The rate of change of momentum (force) leads directly to pressure when divided by wall area.
255. If $N$ molecules of mass $m$ are in a volume $V$, which expression gives pressure?
ⓐ. $P = \dfrac{Nm}{V} \overline{c}$
ⓑ. $P = \dfrac{1}{2} Nm \overline{c^2}$
ⓒ. $P = \dfrac{3V}{Nm} \overline{c^2}$
ⓓ. $P = \dfrac{Nm}{3V} \overline{c^2}$
Correct Answer: $P = \dfrac{Nm}{3V} \overline{c^2}$
Explanation: In kinetic theory, pressure depends on number density and mean square speed. $\rho = \dfrac{Nm}{V}$. Hence, $P = \dfrac{1}{3} \rho \overline{c^2} = \dfrac{Nm}{3V}\overline{c^2}$.
256. Which relation connects pressure and average kinetic energy of molecules?
ⓐ. $PV = \tfrac{1}{2}Nm\overline{c^2}$
ⓑ. $PV = \tfrac{1}{3}Nm\overline{c^2}$
ⓒ. $PV = \tfrac{2}{3} N \overline{E_k}$
ⓓ. $PV = N\overline{E_k}$
Correct Answer: $PV = \tfrac{2}{3} N \overline{E_k}$
Explanation: From kinetic theory,
$$
\overline{E_k} = \frac{1}{2}m\overline{c^2}, \quad PV = \frac{1}{3}Nm\overline{c^2} = \frac{2}{3}N\overline{E_k}
$$
This links macroscopic quantities (P, V) with microscopic molecular kinetic energy.
257. Which of the following is the correct dimension of the term $\rho \overline{c^2}$ in the pressure equation?
ⓐ. $ML^{-1}T^{-2}$
ⓑ. $ML^0T^0$
ⓒ. $ML^{-2}T^{-1}$
ⓓ. $M^0L^0T^0$
Correct Answer: $ML^{-1}T^{-2}$
Explanation: $\rho$ (density) has dimensions $ML^{-3}$. $\overline{c^2}$ has $L^2T^{-2}$. Multiplying gives $ML^{-1}T^{-2}$, which are dimensions of pressure (force per unit area).
258. In deriving pressure, why is the motion of molecules along y and z directions considered?
ⓐ. To include intermolecular attractions
ⓑ. To ensure isotropy of gas motion
ⓒ. To simplify equations by ignoring collisions
ⓓ. To account for external fields
Correct Answer: To ensure isotropy of gas motion
Explanation: Gases are isotropic; molecules are equally likely to move in all directions. Therefore, pressure contribution is equally shared among x, y, z axes, leading to the factor $1/3$.
259. Which of the following correctly relates root mean square speed $c_{rms}$ to pressure equation?
ⓐ. $P = \dfrac{1}{3}\rho c_{rms}$
ⓑ. $P = \dfrac{1}{2}\rho c_{rms}^3$
ⓒ. $P = \rho c_{rms}^2$
ⓓ. $P = \dfrac{1}{3}\rho c_{rms}^2$
Correct Answer: $P = \dfrac{1}{3}\rho c_{rms}^2$
Explanation: RMS speed is defined by $c_{rms} = \sqrt{\overline{c^2}}$. Substituting into the pressure expression gives:
$$
P = \frac{1}{3}\rho c_{rms}^2
$$
260. Which important conclusion arises from the derivation of the pressure equation?
ⓐ. Gas pressure depends only on volume of molecules
ⓑ. Gas pressure depends directly on molecular kinetic energy and density
ⓒ. Gas pressure is independent of molecular motion
ⓓ. Gas pressure is proportional to molar mass only
Correct Answer: Gas pressure depends directly on molecular kinetic energy and density
Explanation: The equation $P = \tfrac{1}{3}\rho \overline{c^2}$ shows that pressure depends on how many molecules exist per unit volume (density) and how fast they move (kinetic energy). Thus pressure is a direct macroscopic result of microscopic motion.
261. According to kinetic theory, the average kinetic energy of a molecule of an ideal gas is:
ⓐ. Directly proportional to pressure
ⓑ. Inversely proportional to temperature
ⓒ. Directly proportional to absolute temperature
ⓓ. Independent of temperature
Correct Answer: Directly proportional to absolute temperature
Explanation: The relation is
$$
\overline{E_k} = \tfrac{3}{2}k_B T
$$
where $k_B$ is Boltzmann constant and $T$ is absolute temperature. Thus, molecular kinetic energy increases linearly with $T$, not with pressure or volume.
262. The average translational kinetic energy per mole of a gas is given by:
ⓐ. $\tfrac{1}{2}RT$
ⓑ. $\tfrac{3}{2}RT$
ⓒ. $RT$
ⓓ. $\tfrac{5}{2}RT$
Correct Answer: $\tfrac{3}{2}RT$
Explanation: Multiplying molecular kinetic energy ($\tfrac{3}{2}k_BT$) by Avogadro’s number ($N_A$) gives:
$$
E_{avg/mol} = \tfrac{3}{2} N_A k_B T = \tfrac{3}{2}RT
$$
This is valid for translational degrees of freedom in an ideal monatomic gas.
263. If the temperature of a gas is doubled, its average kinetic energy becomes:
ⓐ. Half of original
ⓑ. Same as before
ⓒ. Twice of original
ⓓ. Four times of original
Correct Answer: Twice of original
Explanation: Since $\overline{E_k} \propto T$, doubling the absolute temperature doubles the average kinetic energy. It does not become four times because the relationship is linear, not quadratic.
264. Which constant directly connects molecular kinetic energy with absolute temperature?
ⓐ. Universal gas constant (R)
ⓑ. Boltzmann constant ($k_B$)
ⓒ. Planck’s constant (h)
ⓓ. Avogadro’s number ($N_A$)
Correct Answer: Boltzmann constant ($k_B$)
Explanation: For a single molecule, $\overline{E_k} = \tfrac{3}{2}k_B T$. For one mole, this becomes $\tfrac{3}{2}RT$, since $R = N_A k_B$. Thus, $k_B$ is the fundamental link.
265. At what temperature will the average kinetic energy of oxygen molecules be twice that at 300 K?
ⓐ. 150 K
ⓑ. 300 K
ⓒ. 450 K
ⓓ. 600 K
Correct Answer: 600 K
Explanation: $\overline{E_k} \propto T$. To double kinetic energy, temperature must be doubled. Thus, at 600 K, oxygen molecules have twice the kinetic energy they had at 300 K.
266. The average kinetic energy of molecules in 1 mol of gas at 27°C is approximately:
ⓐ. $3.7 \, kJ$
ⓑ. $5.0 \, kJ$
ⓒ. $6.2 \, kJ$
ⓓ. $3.4 \, kJ$
Correct Answer: $3.7 \, kJ$
Explanation: $\overline{E_k/mol} = \tfrac{3}{2}RT$. With $T=300 K$, $R=8.314\, J\,mol^{-1}K^{-1}$:
$\overline{E_k/mol} = \tfrac{3}{2}(8.314)(300) \approx 3741 J = 3.7 \, kJ$.
267. Which of the following gases will have the same average kinetic energy at 300 K?
ⓐ. H₂, O₂, and CO₂
ⓑ. He and Ne only
ⓒ. Only gases of same molar mass
ⓓ. None of the above
Correct Answer: H₂, O₂, and CO₂
Explanation: Average kinetic energy depends only on temperature, not on molar mass. At the same absolute temperature, all gases have identical average kinetic energy per molecule.
268. Which relation connects pressure, volume, and average kinetic energy of molecules?
ⓐ. $PV = N \overline{E_k}$
ⓑ. $PV = \tfrac{1}{2}N \overline{E_k}$
ⓒ. $PV = \tfrac{2}{3}N \overline{E_k}$
ⓓ. $PV = \tfrac{3}{2}N \overline{E_k}$
Correct Answer: $PV = \tfrac{2}{3}N \overline{E_k}$
Explanation: From kinetic theory:
$$
PV = \tfrac{1}{3}Nm\overline{c^2} = \tfrac{2}{3}N\overline{E_k}
$$
This establishes the macroscopic–microscopic connection.
269. If the rms speed of oxygen molecules at 300 K is $c$, what will be their rms speed at 1200 K?
ⓐ. $c$
ⓑ. $c/2$
ⓒ. $4c$
ⓓ. $2c$
Correct Answer: $2c$
Explanation: $c_{rms} = \sqrt{\dfrac{3RT}{M}}$. Since $c_{rms} \propto \sqrt{T}$, quadrupling the temperature increases rms speed by factor $\sqrt{4} = 2$.
270. Which statement best summarizes the relationship between kinetic energy and temperature?
ⓐ. Higher molecular mass gives higher kinetic energy at same temperature.
ⓑ. Kinetic energy of all gases is identical at the same absolute temperature.
ⓒ. Kinetic energy depends on pressure as well as temperature.
ⓓ. Lighter gases have less kinetic energy at given temperature.
Correct Answer: Kinetic energy of all gases is identical at the same absolute temperature.
Explanation: At a given temperature, $\overline{E_k} = \tfrac{3}{2}k_B T$ applies universally. While lighter gases move faster (higher velocity), their kinetic energy per molecule equals that of heavier gases at the same T.
271. Which of the following relations between molecular speeds is correct for an ideal gas?
ⓐ. $u_{rms} < u_{avg} < u_{mp}$
ⓑ. $u_{rms} = u_{avg} = u_{mp}$
ⓒ. $u_{avg} < u_{mp} < u_{rms}$
ⓓ. $u_{mp} < u_{avg} < u_{rms}$
Correct Answer: $u_{mp} < u_{avg} < u_{rms}$
Explanation: For Maxwell–Boltzmann distribution,
$$
u_{mp} = \sqrt{\frac{2RT}{M}}, \quad u_{avg} = \sqrt{\frac{8RT}{\pi M}}, \quad u_{rms} = \sqrt{\frac{3RT}{M}}
$$
Numerically, $u_{mp} < u_{avg} < u_{rms}$.
272. The expression for the most probable speed ($u_{mp}$) is:
ⓐ. $\sqrt{\frac{3RT}{M}}$
ⓑ. $\sqrt{\frac{8RT}{\pi M}}$
ⓒ. $\sqrt{\frac{2RT}{M}}$
ⓓ. $\sqrt{\frac{RT}{M}}$
Correct Answer: $\sqrt{\frac{2RT}{M}}$
Explanation: Derived from the maximum of the Maxwell–Boltzmann distribution curve, the most probable speed is $u_{mp} = \sqrt{2RT/M}$. This is the speed possessed by the largest fraction of molecules.
273. The average speed of gas molecules is given by:
ⓐ. $\sqrt{\frac{3RT}{M}}$
ⓑ. $\sqrt{\frac{RT}{2M}}$
ⓒ. $\sqrt{\frac{2RT}{M}}$
ⓓ. $\sqrt{\frac{8RT}{\pi M}}$
Correct Answer: $\sqrt{\frac{8RT}{\pi M}}$
Explanation: Average speed is obtained by integrating the Maxwell–Boltzmann speed distribution. The result is:
$$
u_{avg} = \sqrt{\frac{8RT}{\pi M}}
$$
274. The root mean square (RMS) speed is defined as:
ⓐ. Arithmetic mean of molecular speeds
ⓑ. Square root of average of squared speeds
ⓒ. Speed of maximum number of molecules
ⓓ. Reciprocal of average speed
Correct Answer: Square root of average of squared speeds
Explanation: RMS speed:
$$
u_{rms} = \sqrt{\overline{c^2}} = \sqrt{\frac{3RT}{M}}
$$
It is always greater than both average and most probable speeds.
275. If the molar mass of oxygen is 32 g/mol, what is the RMS speed at 300 K?
ⓐ. 272 m/s
ⓑ. 345 m/s
ⓒ. 461 m/s
ⓓ. 540 m/s
Correct Answer: 461 m/s
Explanation:
$$
u_{rms} = \sqrt{\frac{3RT}{M}}
$$
$R = 8.314\, J\,mol^{-1}K^{-1}, T=300\,K, M=0.032\,kg/mol$.
$u_{rms} = \sqrt{\dfrac{3(8.314)(300)}{0.032}} \approx 461\,m/s$.
276. At the same temperature, which gas has the highest RMS speed?
ⓐ. Helium (M=4)
ⓑ. Oxygen (M=32)
ⓒ. Nitrogen (M=28)
ⓓ. Carbon dioxide (M=44)
Correct Answer: Helium (M=4)
Explanation: $u_{rms} \propto 1/\sqrt{M}$. Lighter gases move faster at the same temperature. Helium, having the lowest molar mass, has the highest RMS speed.
277. If hydrogen ($M=2$) has an RMS speed of 1840 m/s at a given temperature, what is the RMS speed of oxygen ($M=32$) at the same temperature?
ⓐ. 460 m/s
ⓑ. 920 m/s
ⓒ. 1840 m/s
ⓓ. 3680 m/s
Correct Answer: 460 m/s
Explanation: $\dfrac{u_{rms,H_2}}{u_{rms,O_2}} = \sqrt{\dfrac{M_{O_2}}{M_{H_2}}} = \sqrt{\dfrac{32}{2}} = 4$.
Thus $u_{rms,O_2} = 1840/4 = 460 \, m/s$.
278. Which of the following correctly matches the gas speeds?
ⓐ. $u_{mp} = \sqrt{\tfrac{2RT}{M}}$
ⓑ. $u_{avg} = \sqrt{\tfrac{8RT}{\pi M}}$
ⓒ. $u_{rms} = \sqrt{\tfrac{3RT}{M}}$
ⓓ. All of the above
Correct Answer: All of the above
Explanation: All three are standard formulas derived from Maxwell–Boltzmann distribution. They differ slightly in magnitude but have the same temperature and molar mass dependence.
279. For nitrogen gas at 300 K, the three molecular speeds are approximately:
ⓐ. $u_{mp} = 422$ m/s, $u_{avg} = 467$ m/s, $u_{rms} = 515$ m/s
ⓑ. $u_{mp} = 515$ m/s, $u_{avg} = 422$ m/s, $u_{rms} = 467$ m/s
ⓒ. $u_{mp} = 467$ m/s, $u_{avg} = 515$ m/s, $u_{rms} = 422$ m/s
ⓓ. All are equal at 300 K
Correct Answer: $u_{mp} = 422$ m/s, $u_{avg} = 467$ m/s, $u_{rms} = 515$ m/s
Explanation: For $N_2 (M=28)$ at 300 K, using formulas gives:
$u_{mp} \approx 422$ m/s,
$u_{avg} \approx 467$ m/s,
$u_{rms} \approx 515$ m/s.
The order confirms $u_{mp} < u_{avg} < u_{rms}$.
280. Which conclusion can be drawn from the three types of molecular speeds?
ⓐ. They are identical at all temperatures.
ⓑ. They differ slightly but all increase with temperature.
ⓒ. They decrease as temperature increases.
ⓓ. They depend only on pressure, not temperature.
Correct Answer: They differ slightly but all increase with temperature.
Explanation: All three (most probable, average, RMS) speeds depend on temperature ($\sqrt{T}$). They are not equal but close in value. As $T$ rises, molecules move faster, increasing all three speeds.
281. Why do real gases deviate from ideal behavior at high pressure?
ⓐ. Because molecular volume becomes negligible
ⓑ. Because intermolecular attractions are insignificant
ⓒ. Because molecular volume cannot be ignored compared to container volume
ⓓ. Because gas molecules stop colliding with walls
Correct Answer: Because molecular volume cannot be ignored compared to container volume
Explanation: At high pressure, molecules are forced close together, so the finite volume of molecules is no longer negligible. This reduces the free volume available for motion, causing deviation from $PV = nRT$.
282. Why do real gases deviate from ideal behavior at low temperature?
ⓐ. Because molecular kinetic energy becomes very high
ⓑ. Because intermolecular attractive forces become significant
ⓒ. Because pressure increases with cooling
ⓓ. Because volume becomes infinite
Correct Answer: Because intermolecular attractive forces become significant
Explanation: At low temperature, molecules move slower, allowing intermolecular attractions (van der Waals forces) to dominate. These reduce pressure exerted on walls, causing deviation from ideal predictions.
283. Which gas behaves most ideally under ordinary conditions?
ⓐ. CO₂
ⓑ. NH₃
ⓒ. H₂
ⓓ. SO₂
Correct Answer: H₂
Explanation: Hydrogen has very weak intermolecular forces and small molecular size. Gases with stronger intermolecular attractions (like NH₃, SO₂, CO₂) show larger deviations.
284. Which assumption of kinetic theory is mainly violated when real gases deviate from ideal behavior?
ⓐ. Collisions are perfectly elastic
ⓑ. Intermolecular forces are negligible
ⓒ. Motion of molecules is random
ⓓ. Molecular volume is negligible
Correct Answer: Intermolecular forces are negligible
Explanation: The ideal gas assumption that molecules do not attract or repel each other breaks down under high pressure and low temperature, leading to deviations.
285. Which gas shows the greatest deviation from ideal behavior?
ⓐ. He
ⓑ. H₂
ⓒ. CO₂
ⓓ. N₂
Correct Answer: CO₂
Explanation: CO₂ has larger molecular mass and stronger van der Waals forces compared to lighter gases. Thus, it deviates significantly from ideality, especially near liquefaction conditions.
286. What is the compressibility factor $Z$ for an ideal gas?
ⓐ. 0
ⓑ. Less than 1
ⓒ. Greater than 1
ⓓ. 1
Correct Answer: 1
Explanation: Compressibility factor is defined as $Z = \dfrac{PV}{nRT}$. For an ideal gas, $PV = nRT$, so $Z = 1$. For real gases, $Z$ may be greater than or less than 1 depending on conditions.
287. For a real gas, $Z < 1$ indicates:
ⓐ. Gas is more compressible than ideal
ⓑ. Gas is less compressible than ideal
ⓒ. Gas has no intermolecular forces
ⓓ. Gas behaves exactly like an ideal gas
Correct Answer: Gas is more compressible than ideal
Explanation: When $Z < 1$, attractive forces dominate, reducing pressure compared to ideal prediction, making the gas easier to compress.
288. For a real gas, $Z > 1$ indicates:
ⓐ. Attractive forces dominate
ⓑ. Volume of molecules is negligible
ⓒ. Gas is more compressible than ideal
ⓓ. Repulsive forces dominate
Correct Answer: Repulsive forces dominate
Explanation: When repulsions (finite molecular volume effects) exceed attractions, measured pressure is higher than ideal, giving $Z > 1$. The gas behaves less compressibly.
289. Which condition favors ideal behavior most strongly?
ⓐ. High pressure and low temperature
ⓑ. Low pressure and high temperature
ⓒ. Low pressure and low temperature
ⓓ. High pressure and high temperature
Correct Answer: Low pressure and high temperature
Explanation: At low pressure, intermolecular distances are large, reducing interactions. At high temperature, molecular kinetic energy dominates over attractions. Together, these conditions make gases behave nearly ideally.
290. Why does ammonia gas deviate strongly from ideal behavior?
ⓐ. Because of very low molar mass
ⓑ. Because it is non-polar
ⓒ. Because it forms hydrogen bonds
ⓓ. Because its molecules are spherical
Correct Answer: Because it forms hydrogen bonds
Explanation: Ammonia is polar and can form hydrogen bonds, leading to strong intermolecular attractions. This makes its behavior deviate from the ideal gas law, especially at lower temperatures.
291. The compressibility factor $Z$ is defined as:
ⓐ. $Z = \dfrac{nRT}{PV}$
ⓑ. $Z = \dfrac{V}{nRT}$
ⓒ. $Z = \dfrac{P}{RT}$
ⓓ. $Z = \dfrac{PV}{nRT}$
Correct Answer: $Z = \dfrac{PV}{nRT}$
Explanation: Compressibility factor measures deviation of real gases from ideal behavior. For an ideal gas, $PV=nRT$ so $Z=1$. For real gases, $Z$ may be greater or less than 1 depending on molecular interactions.
292. If $Z = 1$ for a gas, it means:
ⓐ. Gas shows strong deviation from ideality
ⓑ. Gas behaves exactly like an ideal gas
ⓒ. Gas is incompressible
ⓓ. Gas has no molar volume
Correct Answer: Gas behaves exactly like an ideal gas
Explanation: At $Z = 1$, measured $PV$ matches the ideal gas law prediction. In practice, only at low pressure and high temperature do real gases approach $Z=1$.
293. For a real gas, $Z < 1$ indicates:
ⓐ. Dominance of repulsive forces
ⓑ. Infinite compressibility
ⓒ. No molecular interaction
ⓓ. Dominance of attractive forces
Correct Answer: Dominance of attractive forces
Explanation: When intermolecular attractions dominate, gas molecules exert less pressure than predicted by the ideal gas law, giving $Z < 1$. The gas is more compressible.
294. For a real gas, $Z > 1$ indicates:
ⓐ. Dominance of attractive forces
ⓑ. Dominance of repulsive forces
ⓒ. Ideal gas behavior
ⓓ. Zero volume of molecules
Correct Answer: Dominance of repulsive forces
Explanation: Repulsive forces (due to finite molecular size) raise the pressure compared to ideal gas predictions, leading to $Z > 1$. The gas becomes less compressible.
295. At very low pressure, the compressibility factor of a real gas approaches:
ⓐ. 0
ⓑ. Less than 1
ⓒ. Greater than 1
ⓓ. 1
Correct Answer: 1
Explanation: At low pressure, molecular interactions are negligible because molecules are far apart. The gas behaves ideally, and $Z$ approaches 1.
296. At high pressure, the compressibility factor of most real gases is:
ⓐ. Exactly 1
ⓑ. Less than 1
ⓒ. Greater than 1
ⓓ. Equal to zero
Correct Answer: Greater than 1
Explanation: At high pressure, molecular volume becomes significant, and repulsive forces dominate, giving $Z > 1$.
297. Which of the following gases shows the largest deviation in compressibility factor due to hydrogen bonding?
ⓐ. $O_2$
ⓑ. $N_2$
ⓒ. $CO_2$
ⓓ. $NH_3$
Correct Answer: $NH_3$
Explanation: Ammonia strongly hydrogen bonds, giving significant deviations from ideality, especially at moderate pressures and temperatures, reflected in compressibility factor values.
298. The value of $Z$ for CO₂ at moderate pressures is often less than 1. This indicates:
ⓐ. Attractive forces dominate at this range
ⓑ. Repulsive forces dominate at this range
ⓒ. Gas behaves ideally
ⓓ. Volume of molecules is negligible
Correct Answer: Attractive forces dominate at this range
Explanation: When $Z<1$, intermolecular attractions reduce the effective pressure, making the gas more compressible than an ideal gas.
299. Which expression connects compressibility factor with molar volume?
ⓐ. $Z = \dfrac{PV_m}{RT}$
ⓑ. $Z = \dfrac{RT}{PV_m}$
ⓒ. $Z = \dfrac{P}{V_mRT}$
ⓓ. $Z = \dfrac{n}{PV_m}$
Correct Answer: $Z = \dfrac{PV_m}{RT}$
Explanation: Since molar volume $V_m = \dfrac{V}{n}$, substituting into $Z = PV/nRT$ gives $Z = PV_m/RT$. This form is often used to analyze deviations.
300. Which of the following best explains why compressibility factor is useful?
ⓐ. It directly measures gas density.
ⓑ. It quantifies deviation of a real gas from ideal behavior.
ⓒ. It eliminates the gas constant from equations.
ⓓ. It only applies to monatomic gases.
Correct Answer: It quantifies deviation of a real gas from ideal behavior.
Explanation: $Z$ shows whether a gas is more or less compressible than predicted by the ideal gas law. It provides insight into the relative importance of intermolecular attractions and repulsions under different conditions.
This section features Class 11 Chemistry MCQs – Chapter 5: States of Matter (Part 3). Based on the NCERT/CBSE Class 11 Chemistry syllabus, this chapter develops deeper insight into the properties of gases and liquids. It is a must for board exams and a high-scoring chapter for competitive exams like JEE and NEET. The entire chapter consists of 494 MCQs with step-by-step solutions, divided into 5 sections. Here in Part 3, you will find another 100 MCQs focusing on topics such as Van der Waals equation of real gases, liquefaction of gases, critical temperature, critical pressure, critical volume, and the concept of isotherms. Questions also highlight deviation of real gases from ideal behavior, compressibility curves, and their applications in practical chemistry. These concepts form the backbone of numerical problem-solving in both board and competitive exams.
- 👉 Total MCQs in this chapter: 494.
- 👉 This page contains: Third set of 100 solved MCQs with answers.
- 👉 Helpful for exam-oriented practice and competitive preparation.
- 👉 To access other chapters, subjects, or classes, use the navigation buttons above.
- 👉 To continue practicing, move to the Part 4 button above.