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301. The van der Waals equation for real gases is:
ⓐ. $(P + \dfrac{a}{V^2})(V – b) = nRT$
ⓑ. $(P – \dfrac{a}{V^2})(V + b) = nRT$
ⓒ. $(P + \dfrac{a}{V})(V – b) = nRT$
ⓓ. $(P – \dfrac{a}{V})(V + b) = nRT$
Correct Answer: $(P + \dfrac{a}{V^2})(V – b) = nRT$
Explanation: Van der Waals introduced constants $a$ and $b$ to correct for intermolecular attractions and finite molecular volume, respectively. The correct modified gas equation is $(P + \tfrac{a}{V^2})(V – b) = nRT$.
302. In the van der Waals equation, the constant $a$ accounts for:
ⓐ. Finite size of molecules
ⓑ. Elastic nature of collisions
ⓒ. Gas compressibility at high pressures
ⓓ. Intermolecular attractive forces
Correct Answer: Intermolecular attractive forces
Explanation: Attractive forces reduce the effective pressure exerted on container walls. The term $\dfrac{a}{V^2}$ is added to the observed pressure to correct this.
303. In the van der Waals equation, the constant $b$ represents:
ⓐ. Pressure correction
ⓑ. Volume occupied by molecules
ⓒ. Temperature dependence of kinetic energy
ⓓ. Intermolecular attractions
Correct Answer: Volume occupied by molecules
Explanation: Gas molecules themselves occupy finite volume, so the free space is less than container volume. The correction $(V – b)$ accounts for this excluded volume.
304. For 1 mole of a gas, the van der Waals equation is:
ⓐ. $(P + \dfrac{a}{V^2})(V – b) = RT$
ⓑ. $(P + \dfrac{a}{V})(V – b) = RT$
ⓒ. $(P – \dfrac{a}{V^2})(V + b) = RT$
ⓓ. $(P + b)(V – a) = RT$
Correct Answer: $(P + \dfrac{a}{V^2})(V – b) = RT$
Explanation: For 1 mol, $n = 1$, so the general van der Waals equation simplifies to $(P + \tfrac{a}{V^2})(V – b) = RT$.
305. Why is $a/V^2$ added to pressure in the van der Waals equation?
ⓐ. To account for finite molecular volume
ⓑ. To correct for repulsive forces
ⓒ. To compensate for reduction in effective pressure due to attractions
ⓓ. To include compressibility factor
Correct Answer: To compensate for reduction in effective pressure due to attractions
Explanation: Intermolecular attractions reduce the frequency and intensity of wall collisions. Hence, the observed pressure is less than ideal. Adding $a/V^2$ corrects this.
306. Why is $V – b$ used in the van der Waals equation?
ⓐ. Because molecules can overlap in volume
ⓑ. Because actual free volume is less than container volume due to finite molecular size
ⓒ. Because intermolecular forces reduce pressure
ⓓ. Because temperature affects available volume
Correct Answer: Because actual free volume is less than container volume due to finite molecular size
Explanation: Molecules occupy a definite volume. The effective free volume for molecular motion is container volume minus excluded volume $b$.
307. Which condition makes the van der Waals equation reduce to the ideal gas law?
ⓐ. High pressure and low temperature
ⓑ. Low pressure and high temperature
ⓒ. Low temperature only
ⓓ. High pressure only
Correct Answer: Low pressure and high temperature
Explanation: At low pressure, $V$ is large, so $a/V^2 \to 0$, and $b$ is negligible compared to $V$. At high temperature, attractions are insignificant. Thus, the equation reduces to $PV = nRT$.
308. The van der Waals constant $a$ has which units (for 1 mol gas)?
ⓐ. L² atm mol⁻²
ⓑ. L atm mol⁻¹
ⓒ. L² atm mol⁻¹
ⓓ. L³ atm mol⁻²
Correct Answer: L³ atm mol⁻²
Explanation: Since the correction term is $a/V^2$ and has pressure units, $a$ must have units $P \times V^2 = (atm)(L^2) = L^3 atm mol⁻²$.
309. Which type of gas will have a higher value of $a$?
ⓐ. Non-polar gases like He and Ne
ⓑ. Polar gases like NH₃ and HCl
ⓒ. Noble gases only
ⓓ. Monatomic gases only
Correct Answer: Polar gases like NH₃ and HCl
Explanation: Polar gases experience strong intermolecular attractions (dipole–dipole, hydrogen bonding). Therefore, they have larger $a$ values compared to non-polar gases like He or Ne.
310. Which of the following gases will have the smallest value of $b$?
ⓐ. CO₂
ⓑ. SO₂
ⓒ. H₂
ⓓ. O₂
Correct Answer: H₂
Explanation: The constant $b$ corresponds to the excluded volume due to finite molecular size. Lighter and smaller molecules like hydrogen occupy very little space, hence have the smallest $b$ value.
311. Which is the main cause of deviation of real gases from ideal behavior at low temperature?
ⓐ. Increase in molecular volume
ⓑ. Decrease in molecular volume
ⓒ. Dominance of intermolecular attractive forces
ⓓ. Negligible intermolecular forces
Correct Answer: Dominance of intermolecular attractive forces
Explanation: At low temperature, molecules move slowly, and attractive forces reduce the pressure exerted on container walls. This leads to a deviation where compressibility factor $Z < 1$.
312. Why do gases deviate from ideality at high pressure?
ⓐ. Because molecular volume cannot be neglected
ⓑ. Because temperature decreases automatically
ⓒ. Because intermolecular attractions increase sharply
ⓓ. Because molecules stop moving
Correct Answer: Because molecular volume cannot be neglected
Explanation: At high pressure, molecules are forced close together, and the volume occupied by the molecules themselves becomes significant compared to container volume, causing deviations with $Z > 1$.
313. Which factor is more significant at low pressure?
ⓐ. Intermolecular attractions
ⓑ. Molecular volume
ⓒ. Shape of the container
ⓓ. Temperature of surroundings
Correct Answer: Intermolecular attractions
Explanation: At low pressure, the distance between molecules is large, so their actual volume is negligible. However, weak attractive forces still affect motion, leading to deviations below ideal predictions.
314. For a non-polar gas like helium, deviation from ideality is mainly due to:
ⓐ. Strong dipole–dipole forces
ⓑ. Hydrogen bonding
ⓒ. Finite molecular volume
ⓓ. Ionic interactions
Correct Answer: Finite molecular volume
Explanation: Helium has negligible attractions due to being non-polar. Its deviation arises at high pressures where finite molecular size (volume) reduces the free space available for motion.
315. For a polar gas like ammonia, deviation from ideal behavior is primarily due to:
ⓐ. Molecular volume
ⓑ. Hydrogen bonding and strong intermolecular attractions
ⓒ. High compressibility factor
ⓓ. Negligible attractive forces
Correct Answer: Hydrogen bonding and strong intermolecular attractions
Explanation: Ammonia molecules strongly attract each other through hydrogen bonding, especially at moderate pressure and low temperature. This makes measured pressure lower than predicted, giving $Z < 1$.
316. When repulsive forces dominate over attractive forces in a gas, what happens to compressibility factor $Z$?
ⓐ. $Z = 0$
ⓑ. $Z < 1$
ⓒ. $Z = 1$
ⓓ. $Z > 1$
Correct Answer: $Z > 1$
Explanation: Repulsive interactions (finite volume effect) increase the pressure compared to ideal gas predictions, leading to $Z > 1$.
317. Which of the following gases shows larger deviation due to molecular volume rather than intermolecular attraction?
ⓐ. Hydrogen
ⓑ. Oxygen
ⓒ. Carbon dioxide
ⓓ. Ammonia
Correct Answer: Hydrogen
Explanation: Light gases like H₂ have weak attractions, but at high pressures, the finite size of molecules dominates. Thus, deviations are mainly from molecular volume, not attractions.
318. Why does carbon dioxide show greater deviation than nitrogen at the same conditions?
ⓐ. Lower molar mass
ⓑ. Stronger intermolecular forces due to polarity and higher molecular mass
ⓒ. Negligible molecular size
ⓓ. Perfect elasticity of collisions
Correct Answer: Stronger intermolecular forces due to polarity and higher molecular mass
Explanation: CO₂ molecules are larger and exhibit stronger van der Waals forces than N₂, so CO₂ deviates more significantly from ideality.
319. Which of the following best describes the cause of deviation in van der Waals equation?
ⓐ. $a$ corrects for molecular attractions; $b$ corrects for molecular volume
ⓑ. $a$ corrects for molecular volume; $b$ corrects for molecular attractions
ⓒ. Both $a$ and $b$ correct for pressure
ⓓ. Both $a$ and $b$ correct for temperature
Correct Answer: $a$ corrects for molecular attractions; $b$ corrects for molecular volume
Explanation: In van der Waals’ model, pressure is increased by $a/V^2$ to account for attractions, and volume is reduced by $b$ to account for finite molecular size.
320. Which condition minimizes deviation from ideality caused by both intermolecular forces and molecular volume?
ⓐ. 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, molecules are far apart, reducing volume effects. At high temperature, kinetic energy dominates over attractions. These conditions minimize deviations, and gas behavior approaches ideal.
321. The critical temperature ($T_c$) of a gas is defined as:
ⓐ. The temperature above which a gas cannot be liquefied, no matter how much pressure is applied
ⓑ. The temperature below which a gas behaves ideally
ⓒ. The temperature at which intermolecular forces vanish
ⓓ. The temperature at which gas molecules stop moving
Correct Answer: The temperature above which a gas cannot be liquefied, no matter how much pressure is applied
Explanation: At and above the critical temperature, the kinetic energy of molecules overcomes intermolecular attractions completely, so no amount of pressure can condense the gas into a liquid.
322. The critical pressure ($P_c$) is defined as:
ⓐ. The minimum pressure required to liquefy a gas at any temperature
ⓑ. The pressure required to liquefy a gas at the critical temperature
ⓒ. The maximum pressure a gas can exert before liquefying at 0 K
ⓓ. The pressure at which gas molecules stop colliding
Correct Answer: The pressure required to liquefy a gas at the critical temperature
Explanation: $P_c$ is the pressure needed to liquefy a gas at its $T_c$. Below this temperature, gases can be liquefied at pressures lower than $P_c$.
323. The critical volume ($V_c$) is:
ⓐ. The maximum volume of gas at STP
ⓑ. The volume occupied by 1 mole of gas at the critical temperature and pressure
ⓒ. The minimum volume a gas can have at any temperature
ⓓ. The volume at which compressibility factor is zero
Correct Answer: The volume occupied by 1 mole of gas at the critical temperature and pressure
Explanation: $V_c$ is defined for 1 mol of gas at the critical point (when $T = T_c$, $P = P_c$). It is an experimentally measurable quantity.
324. The critical constants ($T_c, P_c, V_c$) are related to which equation of state?
ⓐ. Boyle’s law
ⓑ. Charles’ law
ⓒ. van der Waals equation
ⓓ. Dalton’s law
Correct Answer: van der Waals equation
Explanation: By differentiating van der Waals’ equation at the critical point, expressions for $T_c, P_c, V_c$ are derived in terms of $a$ and $b$, the van der Waals constants.
325. The relation between critical constants and van der Waals constants is:
ⓐ. $T_c = \dfrac{8a}{27Rb}, \; P_c = \dfrac{a}{27b^2}, \; V_c = 3b$
ⓑ. $T_c = \dfrac{a}{2Rb}, \; P_c = \dfrac{a}{2b^2}, \; V_c = 2b$
ⓒ. $T_c = \dfrac{a}{Rb}, \; P_c = \dfrac{a}{b^2}, \; V_c = b$
ⓓ. $T_c = \dfrac{27a}{8Rb}, \; P_c = \dfrac{27a}{b^2}, \; V_c = b/3$
Correct Answer: $T_c = \dfrac{8a}{27Rb}, \; P_c = \dfrac{a}{27b^2}, \; V_c = 3b$
Explanation: These relations are obtained by applying critical point conditions $\dfrac{\partial P}{\partial V}=0$ and $\dfrac{\partial^2 P}{\partial V^2}=0$ to the van der Waals equation.
326. Which gas has the highest critical temperature among the following?
ⓐ. He
ⓑ. O₂
ⓒ. CO₂
ⓓ. NH₃
Correct Answer: NH₃
Explanation: Gases with strong intermolecular attractions (like hydrogen bonding in NH₃) have higher critical temperatures. Helium has extremely low $T_c$ due to weak forces, while NH₃ is highest among the options.
327. Why is critical temperature important in liquefaction of gases?
ⓐ. It indicates when gases behave ideally
ⓑ. It defines the lowest temperature required for liquefaction
ⓒ. It determines the maximum temperature for applying pressure to liquefy gas
ⓓ. It is unrelated to liquefaction
Correct Answer: It determines the maximum temperature for applying pressure to liquefy gas
Explanation: Gases can only be liquefied below their critical temperature. Above $T_c$, increasing pressure alone cannot condense them.
328. For a van der Waals gas, the compressibility factor at the critical point is:
ⓐ. 0.277
ⓑ. 0.5
ⓒ. 0.75
ⓓ. 0.375
Correct Answer: 0.277
Explanation: At critical conditions, $Z_c = \dfrac{P_cV_c}{RT_c} = \dfrac{3}{8} = 0.375$. Wait correction → actual derived value is 0.375. Correct answer is D.
329. Which set of critical constants correctly describes CO₂?
ⓐ. $T_c = 304 \, K, \; P_c = 73 \, atm$
ⓑ. $T_c = 150 \, K, \; P_c = 34 \, atm$
ⓒ. $T_c = 500 \, K, \; P_c = 100 \, atm$
ⓓ. $T_c = 20 \, K, \; P_c = 5 \, atm$
Correct Answer: $T_c = 304 \, K, \; P_c = 73 \, atm$
Explanation: Experimentally, CO₂ has $T_c \approx 304 \, K$ and $P_c \approx 73 \, atm$. This is why CO₂ can be liquefied near room temperature under high pressure.
330. Which critical property determines whether a gas can be liquefied at room temperature (298 K)?
ⓐ. Critical pressure
ⓑ. Critical temperature
ⓒ. Critical volume
ⓓ. Compressibility factor
Correct Answer: Critical temperature
Explanation: If the gas’s critical temperature is above room temperature, it can be liquefied at 298 K by applying sufficient pressure. If $T_c < 298 K$, the gas cannot be liquefied at room temperature.
331. Andrews’ experiments on CO₂ demonstrated the concept of:
ⓐ. Boyle’s law
ⓑ. Critical temperature and liquefaction of gases
ⓒ. Graham’s law of diffusion
ⓓ. Dalton’s law of partial pressures
Correct Answer: Critical temperature and liquefaction of gases
Explanation: Andrews studied the behavior of CO₂ under varying pressures and temperatures. His work showed that gases can be liquefied by pressure only if the temperature is below a certain limit—the critical temperature.
332. In Andrews’ experiment, what happened to CO₂ when pressure was increased at a temperature below its critical temperature?
ⓐ. It could not be liquefied
ⓑ. It behaved as an ideal gas
ⓒ. It solidified directly
ⓓ. It condensed into liquid at a certain pressure
Correct Answer: It condensed into liquid at a certain pressure
Explanation: Below the critical temperature, increasing pressure compresses the gas until it suddenly liquefies, shown as a horizontal portion in Andrews’ isotherm.
333. In the isotherms of CO₂ obtained by Andrews, the horizontal flat portion represents:
ⓐ. Region of supercritical fluid
ⓑ. Phase where liquid and vapor coexist
ⓒ. Purely gaseous state
ⓓ. Purely liquid state
Correct Answer: Phase where liquid and vapor coexist
Explanation: The flat portion corresponds to constant pressure during liquefaction. Gas and liquid phases exist together until the gas is completely converted into liquid.
334. At temperatures above the critical temperature, Andrews found that:
ⓐ. CO₂ liquefies at very high pressure
ⓑ. CO₂ behaves as an ideal gas
ⓒ. CO₂ solidifies at constant volume
ⓓ. CO₂ cannot be liquefied by any amount of pressure
Correct Answer: CO₂ cannot be liquefied by any amount of pressure
Explanation: Above $T_c = 304 K$, kinetic energy of molecules is too high for attractive forces to condense them into liquid. Thus, only a supercritical fluid is formed.
335. What is the critical temperature of CO₂ observed in Andrews’ experiments?
ⓐ. 150 K
ⓑ. 273 K
ⓒ. 304 K
ⓓ. 500 K
Correct Answer: 304 K
Explanation: Andrews measured that CO₂ could only be liquefied below about 304 K. This temperature is now universally accepted as the critical temperature for CO₂.
336. In Andrews’ CO₂ isotherm, what does the portion beyond the horizontal line (on further compression) represent?
ⓐ. Complete gaseous state
ⓑ. Complete liquid state
ⓒ. Supercritical region
ⓓ. Solid CO₂ formation
Correct Answer: Complete liquid state
Explanation: Once all gas has condensed, further compression reduces only the volume of the liquid, so the curve rises steeply again.
337. What phenomenon did Andrews’ experiments directly demonstrate for the first time?
ⓐ. Gas diffusion through a membrane
ⓑ. The law of mass action
ⓒ. The continuity of gaseous and liquid states
ⓓ. Avogadro’s law
Correct Answer: The continuity of gaseous and liquid states
Explanation: Andrews showed that gas and liquid are not separate entities but two states of the same substance. Above the critical point, there is no distinction between them.
338. Which curve obtained by Andrews is called the “isotherm of critical temperature”?
ⓐ. A curve with a distinct flat horizontal portion
ⓑ. A curve with no horizontal portion but a point of inflection
ⓒ. A curve representing Boyle’s law
ⓓ. A curve at absolute zero
Correct Answer: A curve with no horizontal portion but a point of inflection
Explanation: At critical temperature, the isotherm shows an inflection point where first and second derivatives of pressure with respect to volume are zero. This distinguishes it from lower-temperature isotherms.
339. Why is the discovery of critical constants important from Andrews’ experiments?
ⓐ. It explained Avogadro’s law mathematically
ⓑ. It allowed prediction of gas compressibility
ⓒ. It provided conditions for liquefying gases
ⓓ. It showed gases cannot exist below 0°C
Correct Answer: It provided conditions for liquefying gases
Explanation: By identifying $T_c$, $P_c$, and $V_c$, Andrews’ work gave scientists exact parameters required for successful liquefaction, crucial for industries using oxygen, nitrogen, CO₂, etc.
340. The supercritical fluid region in Andrews’ isotherms lies:
ⓐ. Below the critical temperature
ⓑ. At high pressure but low temperature
ⓒ. Exactly at 0 K
ⓓ. Above the critical temperature and pressure
Correct Answer: Above the critical temperature and pressure
Explanation: When both T > Tc and P > Pc, the gas cannot be liquefied but exists in a supercritical fluid state, showing properties of both gases (diffusivity) and liquids (density).
341. Why is the critical temperature ($T_c$) important in the liquefaction of gases?
ⓐ. It defines the lowest temperature for liquefaction.
ⓑ. It defines the highest temperature at which liquefaction is possible by applying pressure.
ⓒ. It is the temperature where gas molecules stop moving.
ⓓ. It is unrelated to gas liquefaction.
Correct Answer: It defines the highest temperature at which liquefaction is possible by applying pressure.
Explanation: Above $T_c$, molecular kinetic energy is so high that intermolecular forces cannot bring molecules together into the liquid phase, regardless of pressure applied.
342. The critical pressure ($P_c$) of a gas is:
ⓐ. The minimum pressure needed to liquefy a gas at any temperature.
ⓑ. The pressure below which all gases become liquids.
ⓒ. The maximum pressure a gas can exert before liquefaction.
ⓓ. The pressure needed to liquefy a gas at the critical temperature.
Correct Answer: The pressure needed to liquefy a gas at the critical temperature.
Explanation: At $T_c$, gas can only liquefy if pressure equals or exceeds $P_c$. Below $T_c$, gases can liquefy at lower pressures.
343. The critical volume ($V_c$) refers to:
ⓐ. The volume occupied by 1 mol of a gas at STP.
ⓑ. The volume occupied by 1 mol of gas at the critical point.
ⓒ. The minimum volume below which no gas can exist.
ⓓ. The volume of liquid formed during liquefaction.
Correct Answer: The volume occupied by 1 mol of gas at the critical point.
Explanation: $V_c$ is defined for 1 mol of gas under $T_c$ and $P_c$. Together with $T_c$ and $P_c$, it characterizes the critical state.
344. Which relation connects $T_c$, $P_c$, and $V_c$ for a van der Waals gas?
ⓐ. $\dfrac{P_c V_c}{R T_c} = 1$
ⓑ. $\dfrac{P_c V_c}{R T_c} = 0$
ⓒ. $\dfrac{P_c V_c}{R T_c} = \dfrac{8}{27}$
ⓓ. $\dfrac{P_c V_c}{R T_c} = \dfrac{3}{8}$
Correct Answer: $\dfrac{P_c V_c}{R T_c} = \dfrac{3}{8}$
Explanation: This universal relation is derived from van der Waals’ equation. At the critical point, compressibility factor $Z_c = 3/8 = 0.375$, the same for all van der Waals gases.
345. Which property of gases can be predicted using critical constants?
ⓐ. Molar mass
ⓑ. Liquefaction behavior and conditions
ⓒ. Ionization potential
ⓓ. Atomic radius
Correct Answer: Liquefaction behavior and conditions
Explanation: By knowing $T_c$, $P_c$, and $V_c$, one can determine whether a gas can be liquefied at a given temperature and pressure. This is essential for industrial gas storage and transport.
346. Which gas is easiest to liquefy based on its critical temperature?
ⓐ. Helium ($T_c = 5.2 K$)
ⓑ. Oxygen ($T_c = 154.8 K$)
ⓒ. Carbon dioxide ($T_c = 304 K$)
ⓓ. Nitrogen ($T_c = 126.2 K$)
Correct Answer: Carbon dioxide ($T_c = 304 K$)
Explanation: Higher critical temperature means easier liquefaction because gases can be liquefied at or near room temperature. Helium is hardest since its $T_c$ is extremely low.
347. Why is knowledge of $T_c, P_c, V_c$ important in industry?
ⓐ. To calculate atomic structure
ⓑ. To optimize storage and liquefaction of gases like O₂, N₂, CO₂
ⓒ. To determine molecular weight directly
ⓓ. To identify isotopes of gases
Correct Answer: To optimize storage and liquefaction of gases like O₂, N₂, CO₂
Explanation: Industries rely on liquefied gases for refrigeration, transport, and medical uses. Knowledge of critical constants ensures efficient design of equipment.
348. Which of the following gases cannot be liquefied at room temperature (298 K) by applying pressure?
ⓐ. NH₃ ($T_c = 405 K$)
ⓑ. CO₂ ($T_c = 304 K$)
ⓒ. O₂ ($T_c = 154.8 K$)
ⓓ. SO₂ ($T_c = 430 K$)
Correct Answer: O₂ ($T_c = 154.8 K$)
Explanation: Oxygen’s critical temperature is much lower than 298 K, so it cannot be liquefied at room temperature regardless of pressure applied.
349. Which critical property primarily determines gas storage safety in pressurized cylinders?
ⓐ. $T_c$
ⓑ. Compressibility factor
ⓒ. $V_c$
ⓓ. $P_c$
Correct Answer: $P_c$
Explanation: The required pressure to liquefy or store a gas depends on critical pressure. Storage cylinders are designed to withstand pressures safely above this value.
350. At the critical point of a gas:
ⓐ. The gas and liquid phases become indistinguishable
ⓑ. The gas behaves ideally
ⓒ. The gas solidifies instantly
ⓓ. The gas becomes incompressible
Correct Answer: The gas and liquid phases become indistinguishable
Explanation: At $T_c, P_c, V_c$, there is no boundary between liquid and gas phases. The substance exists as a supercritical fluid, showing properties of both liquids (density) and gases (diffusion).
351. Vapour pressure of a liquid is defined as:
ⓐ. The pressure exerted by the liquid molecules on the container walls
ⓑ. The pressure exerted by vapour molecules in equilibrium with the liquid at a given temperature
ⓒ. The atmospheric pressure at which the liquid boils
ⓓ. The pressure due to dissolved gases in liquid
Correct Answer: The pressure exerted by vapour molecules in equilibrium with the liquid at a given temperature
Explanation: Vapour pressure is the equilibrium pressure of vapour above its liquid (or solid) phase at a given temperature. It depends only on the nature of the liquid and temperature, not on volume of liquid present.
352. Vapour pressure of a liquid increases with:
ⓐ. Increase in surface area
ⓑ. Increase in atmospheric pressure
ⓒ. Increase in volume of liquid
ⓓ. Increase in temperature
Correct Answer: Increase in temperature
Explanation: As temperature rises, more molecules acquire sufficient kinetic energy to escape into the vapour phase. This increases the number of vapour molecules striking container walls, raising vapour pressure.
353. The temperature at which vapour pressure of a liquid equals atmospheric pressure is called:
ⓐ. Melting point
ⓑ. Sublimation point
ⓒ. Critical temperature
ⓓ. Boiling point
Correct Answer: Boiling point
Explanation: Boiling occurs when vapour pressure of the liquid equals or exceeds external atmospheric pressure, allowing bubbles of vapour to form inside the liquid.
354. Which of the following liquids will have the highest vapour pressure at room temperature?
ⓐ. Water
ⓑ. Ether
ⓒ. Ethanol
ⓓ. Glycerol
Correct Answer: Ether
Explanation: Ether is highly volatile with weak intermolecular forces, so it evaporates readily and has high vapour pressure compared to strongly hydrogen-bonded liquids like water, ethanol, and glycerol.
355. Which factor does not affect the vapour pressure of a liquid?
ⓐ. Nature of the liquid (intermolecular forces)
ⓑ. Temperature
ⓒ. Volume of liquid present
ⓓ. Atmospheric pressure
Correct Answer: Volume of liquid present
Explanation: Vapour pressure is a characteristic property of a liquid at given T, independent of how much liquid is present, as long as some liquid remains to maintain equilibrium.
356. Liquids with stronger intermolecular forces will have:
ⓐ. Higher vapour pressure
ⓑ. Lower vapour pressure
ⓒ. Same vapour pressure as weaker liquids
ⓓ. Vapour pressure independent of forces
Correct Answer: Lower vapour pressure
Explanation: Stronger forces hold molecules in liquid phase more tightly, reducing their tendency to escape into vapour phase. Hence, vapour pressure is lower compared to weakly bonded liquids.
357. At 100°C, the vapour pressure of water is:
ⓐ. 100 mmHg
ⓑ. 373 mmHg
ⓒ. 760 mmHg
ⓓ. 1 mmHg
Correct Answer: 760 mmHg
Explanation: Normal boiling point of water is defined as the temperature at which its vapour pressure equals 1 atm (760 mmHg).
358. Which curve best represents the relation between vapour pressure and temperature?
ⓐ. Straight line with positive slope
ⓑ. Hyperbolic curve
ⓒ. Exponential curve
ⓓ. Horizontal line
Correct Answer: Exponential curve
Explanation: Clausius–Clapeyron equation shows vapour pressure increases exponentially with temperature, because more molecules can overcome intermolecular forces at higher T.
359. If vapour pressure of liquid A is greater than liquid B at the same temperature, then:
ⓐ. A is less volatile than B
ⓑ. A has weaker intermolecular forces than B
ⓒ. A has stronger intermolecular forces than B
ⓓ. A has higher boiling point than B
Correct Answer: A has weaker intermolecular forces than B
Explanation: Higher vapour pressure means molecules escape more easily, indicating weaker intermolecular attractions. Consequently, such liquids usually have lower boiling points.
360. Which equation relates vapour pressure and enthalpy of vaporization?
ⓐ. Boyle’s law
ⓑ. Clausius–Clapeyron equation
ⓒ. van der Waals equation
ⓓ. Dalton’s law
Correct Answer: Clausius–Clapeyron equation
Explanation: Clausius–Clapeyron equation is
$$
\ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_1} – \frac{1}{T_2} \right)
$$
It quantitatively relates vapour pressure to temperature and enthalpy of vaporization.
361. Surface tension of a liquid is defined as:
ⓐ. Force per unit length acting tangentially on the surface of a liquid at right angle to an imaginary line
ⓑ. Pressure difference between liquid and vapour phase
ⓒ. Force exerted by gravity on the liquid molecules
ⓓ. Resistance offered by liquid against flow
Correct Answer: Force per unit length acting tangentially on the surface of a liquid at right angle to an imaginary line
Explanation: Surface tension is caused by unbalanced intermolecular cohesive forces at the surface. It is expressed in $N/m$ or dyne/cm, and explains many liquid behaviors such as droplet formation.
362. Which liquid has the highest surface tension at room temperature?
ⓐ. Ethanol
ⓑ. Water
ⓒ. Benzene
ⓓ. Ether
Correct Answer: Water
Explanation: Due to extensive hydrogen bonding, water exhibits high surface tension (\~72 dyne/cm at 25 °C), greater than most common organic liquids.
363. Surface tension decreases with:
ⓐ. Decrease in temperature
ⓑ. Increase in temperature
ⓒ. Increase in intermolecular forces
ⓓ. Addition of non-volatile solute
Correct Answer: Increase in temperature
Explanation: With rising temperature, molecular kinetic energy increases, overcoming cohesive forces more easily. Thus, surface tension decreases as temperature rises, becoming zero at the critical temperature.
364. Which of the following substances lowers surface tension when added to water?
ⓐ. Salt
ⓑ. Soap
ⓒ. Sugar
ⓓ. Glycerol
Correct Answer: Soap
Explanation: Soap and detergents are surface-active agents (surfactants). They reduce surface tension of water by disrupting hydrogen bonds, enabling better spreading and cleaning action.
365. Why do small liquid drops form spherical shapes?
ⓐ. Because of gravitational forces
ⓑ. Because of surface tension minimizing surface area
ⓒ. Because of vapor pressure
ⓓ. Because of viscosity
Correct Answer: Because of surface tension minimizing surface area
Explanation: Surface tension tends to minimize surface area for a given volume. A sphere has the smallest surface area for a given volume, hence drops are spherical.
366. Which of the following applications is based on surface tension?
ⓐ. Rise of liquid in a capillary
ⓑ. Flow of current in a conductor
ⓒ. Diffusion of gases
ⓓ. Viscosity measurement
Correct Answer: Rise of liquid in a capillary
Explanation: In capillary rise, adhesive forces between liquid and solid, and cohesive forces within liquid molecules, combine with surface tension to pull liquid upward against gravity.
367. Surface tension is measured in SI units as:
ⓐ. $N \, m^{-2}$
ⓑ. $N \, m^{-1}$
ⓒ. $J \, mol^{-1}$
ⓓ. $Pa$
Correct Answer: $N \, m^{-1}$
Explanation: Since surface tension is force per unit length, its SI unit is newton per meter. It can also be expressed as energy per unit area ($J/m^2$), since surface energy is equivalent.
368. Why does adding alcohol to water decrease its surface tension?
ⓐ. Alcohol molecules form stronger hydrogen bonds
ⓑ. Alcohol molecules disrupt hydrogen bonding between water molecules
ⓒ. Alcohol evaporates quickly
ⓓ. Alcohol has higher density
Correct Answer: Alcohol molecules disrupt hydrogen bonding between water molecules
Explanation: Alcohols weaken the cohesive forces in water. This lowers surface tension, allowing easier spreading. This principle is used in industrial formulations.
369. Which phenomenon demonstrates surface tension in everyday life?
ⓐ. Walking of water striders on water surface
ⓑ. Expansion of gases on heating
ⓒ. Melting of ice into water
ⓓ. Solubility of salts in water
Correct Answer: Walking of water striders on water surface
Explanation: Insects like water striders exploit surface tension of water to stand and walk on it without sinking, due to the “skin-like” property of the water surface.
370. Surface tension becomes zero at:
ⓐ. 0 °C
ⓑ. Melting point of liquid
ⓒ. Critical temperature of liquid
ⓓ. Boiling point of liquid
Correct Answer: Critical temperature of liquid
Explanation: At critical temperature, distinction between liquid and gas phases disappears. Intermolecular forces are insufficient to maintain a surface, so surface tension becomes zero.
371. Viscosity of a liquid is defined as:
ⓐ. Resistance offered by a liquid to an increase in vapour pressure
ⓑ. Resistance offered by a liquid to flow due to internal friction
ⓒ. Tendency of a liquid to form spherical drops
ⓓ. Pressure exerted by liquid molecules on container walls
Correct Answer: Resistance offered by a liquid to flow due to internal friction
Explanation: Viscosity is a measure of internal resistance between adjacent layers of liquid moving at different velocities. It is expressed in poise (cgs) or $Pa \cdot s$ (SI).
372. Which of the following liquids has the highest viscosity at room temperature?
ⓐ. Water
ⓑ. Ethanol
ⓒ. Glycerol
ⓓ. Benzene
Correct Answer: Glycerol
Explanation: Glycerol has extensive hydrogen bonding between molecules, making it much thicker and more resistant to flow compared to water, ethanol, or benzene.
373. How does viscosity of a liquid change with temperature?
ⓐ. Increases with increase in temperature
ⓑ. First increases then decreases
ⓒ. Remains constant
ⓓ. Decreases with increase in temperature
Correct Answer: Decreases with increase in temperature
Explanation: At higher temperature, molecules gain kinetic energy, reducing cohesive forces and intermolecular friction. Hence, liquids flow more easily, decreasing viscosity.
374. How does viscosity of gases change with temperature?
ⓐ. Decreases with increase in temperature
ⓑ. Becomes zero
ⓒ. Remains constant
ⓓ. Increases with increase in temperature
Correct Answer: Increases with increase in temperature
Explanation: In gases, higher temperature means faster molecules, leading to more frequent momentum transfer between layers. This increases viscosity, opposite to liquids.
375. Which law relates viscous force to velocity gradient in liquids?
ⓐ. Boyle’s law
ⓑ. Charles’ law
ⓒ. Newton’s law of viscosity
ⓓ. Graham’s law
Correct Answer: Newton’s law of viscosity
Explanation: According to Newton’s law, viscous force is directly proportional to the area of liquid layers and velocity gradient, expressed as:
$$
F = \eta A \frac{dv}{dx}
$$
where $\eta$ is coefficient of viscosity.
376. The SI unit of coefficient of viscosity is:
ⓐ. $N \, m^{-2}$
ⓑ. $Pa \cdot s$ (or $kg \, m^{-1} \, s^{-1}$)
ⓒ. $N \, m^{-1}$
ⓓ. $J \, mol^{-1}$
Correct Answer: $Pa \cdot s$ (or $kg \, m^{-1} \, s^{-1}$)
Explanation: Since $\eta = \dfrac{F}{A \cdot (dv/dx)}$, its SI unit is $N \, s \, m^{-2} = Pa \cdot s$. In cgs, it is expressed in poise ($1 \, poise = 0.1 \, Pa \cdot s$).
377. Why does honey flow more slowly than water?
ⓐ. Honey has higher density only
ⓑ. Honey has stronger intermolecular forces and higher viscosity
ⓒ. Honey has lower vapour pressure
ⓓ. Honey has higher surface tension
Correct Answer: Honey has stronger intermolecular forces and higher viscosity
Explanation: Hydrogen bonding and sugar-water interactions give honey much higher viscosity compared to water, making it flow slowly.
378. Which factor does not directly affect viscosity of liquids?
ⓐ. Intermolecular forces
ⓑ. Temperature
ⓒ. Molar mass of molecules
ⓓ. External pressure
Correct Answer: External pressure
Explanation: Viscosity depends mainly on molecular structure, interactions, and temperature. External pressure has negligible effect on viscosity of liquids under normal conditions.
379. What is the effect of adding impurities like electrolytes on viscosity of water?
ⓐ. Always decreases viscosity
ⓑ. Always increases viscosity
ⓒ. Can either increase or decrease depending on solute
ⓓ. Has no effect at all
Correct Answer: Can either increase or decrease depending on solute
Explanation: Some solutes (like salts) increase viscosity by enhancing intermolecular attractions, while others (like alcohols) may decrease it by disrupting hydrogen bonding.
380. Which real-life application is based on the concept of viscosity?
ⓐ. Lubrication in machinery
ⓑ. Formation of rain drops
ⓒ. Capillary rise of liquids
ⓓ. Boiling of liquids at high altitude
Correct Answer: Lubrication in machinery
Explanation: Oils used in engines and machines must have appropriate viscosity to reduce friction and wear. Too high viscosity hinders flow; too low viscosity fails to provide proper lubrication.
381. Which of the following equations is used to measure viscosity of liquids using capillary flow?
ⓐ. Clausius–Clapeyron equation
ⓑ. Poiseuille’s equation
ⓒ. van der Waals equation
ⓓ. Dalton’s law
Correct Answer: Poiseuille’s equation
Explanation: Poiseuille’s equation describes laminar flow of liquid through a narrow tube:
$$
\eta = \frac{\pi r^4 \Delta P t}{8 V l}
$$
where $r$ = radius of tube, $l$ = length, $\Delta P$ = pressure difference, $t$ = time, and $V$ = volume collected.
382. According to Poiseuille’s equation, viscosity of a liquid is directly proportional to:
ⓐ. Flow rate of liquid
ⓑ. Pressure difference
ⓒ. Time taken for a fixed volume to flow
ⓓ. Radius of the tube raised to fourth power
Correct Answer: Time taken for a fixed volume to flow
Explanation: In an Ostwald viscometer, flow time is directly proportional to viscosity when same liquid volume is allowed to pass. Thus, comparing times for two liquids gives relative viscosities.
383. Relative viscosity of two liquids (1 and 2) using Ostwald viscometer is given by:
ⓐ. $\dfrac{\eta_1}{\eta_2} = \dfrac{t_1 d_1}{t_2 d_2}$
ⓑ. $\dfrac{\eta_1}{\eta_2} = \dfrac{t_1}{t_2}$
ⓒ. $\dfrac{\eta_1}{\eta_2} = \dfrac{d_1}{d_2}$
ⓓ. $\dfrac{\eta_1}{\eta_2} = \dfrac{t_2 d_1}{t_1 d_2}$
Correct Answer: $\dfrac{\eta_1}{\eta_2} = \dfrac{t_1 d_1}{t_2 d_2}$
Explanation: Viscosity depends on both flow time $t$ and density $d$. Hence, relative viscosity is given by the ratio of $t \times d$ terms for the two liquids.
384. What happens to the viscosity of liquids when strong intermolecular hydrogen bonding is present?
ⓐ. It decreases drastically
ⓑ. It increases significantly
ⓒ. It becomes independent of temperature
ⓓ. It remains equal to that of water
Correct Answer: It increases significantly
Explanation: Hydrogen bonding enhances cohesive forces, restricting molecular mobility. Liquids like glycerol and honey show high viscosity due to hydrogen bonding.
385. Why does viscosity of gases increase with temperature, unlike liquids?
ⓐ. Because gas molecules collide more frequently and transfer momentum
ⓑ. Because gas molecules form hydrogen bonds at higher T
ⓒ. Because gas density decreases
ⓓ. Because pressure increases with temperature
Correct Answer: Because gas molecules collide more frequently and transfer momentum
Explanation: In gases, higher temperature increases molecular velocity and collisions, enhancing momentum transfer between layers. Hence viscosity increases with temperature.
386. Which of the following liquids will show the highest decrease in viscosity with rise in temperature?
ⓐ. Water
ⓑ. Glycerol
ⓒ. Mercury
ⓓ. Ethanol
Correct Answer: Glycerol
Explanation: Strong hydrogen bonding in glycerol makes it very viscous at low temperatures. A rise in temperature breaks some hydrogen bonds, causing a sharp fall in viscosity.
387. The coefficient of viscosity is related to viscous drag force by the equation:
ⓐ. $F = \eta A \dfrac{dv}{dx}$
ⓑ. $F = \dfrac{\eta}{A} \dfrac{dx}{dv}$
ⓒ. $F = \dfrac{A}{\eta} \dfrac{dv}{dx}$
ⓓ. $F = \eta \dfrac{A}{dx}$
Correct Answer: $F = \eta A \dfrac{dv}{dx}$
Explanation: Newton’s law of viscosity states that viscous force $F$ is proportional to surface area $A$ and velocity gradient $\dfrac{dv}{dx}$. The constant of proportionality is the coefficient of viscosity $\eta$.
388. Unit of viscosity in the CGS system is:
ⓐ. Pascal second (Pa·s)
ⓑ. Newton second per meter squared
ⓒ. Poise (P)
ⓓ. Joule per second
Correct Answer: Poise (P)
Explanation: In the cgs system, viscosity is measured in poise (P), where $1\, P = 0.1\, Pa \cdot s$. Submultiples like centipoise (cP) are often used (water ≈ 1 cP at 20°C).
389. Which factor does not significantly influence the viscosity of a liquid?
ⓐ. Intermolecular attractions
ⓑ. Temperature
ⓒ. Molecular shape and size
ⓓ. Volume of liquid taken
Correct Answer: Volume of liquid taken
Explanation: Viscosity is a bulk property; it depends on molecular interactions, size, and temperature, but not on how much liquid is present in the sample.
390. Which industrial application directly depends on control of viscosity?
ⓐ. Refining of petroleum products and lubrication oils
ⓑ. Electroplating of metals
ⓒ. Electrolysis of aqueous solutions
ⓓ. Melting of alloys
Correct Answer: Refining of petroleum products and lubrication oils
Explanation: Correct viscosity ensures proper lubrication in engines and machinery. Petrochemical industries carefully adjust viscosity of fuels and lubricants for optimal flow and performance.
391. Which of the following best explains why detergents help in cleaning greasy clothes?
ⓐ. They increase the viscosity of water.
ⓑ. They lower the surface tension of water.
ⓒ. They increase the boiling point of water.
ⓓ. They neutralize the dirt chemically.
Correct Answer: They lower the surface tension of water.
Explanation: Detergents are surfactants that reduce hydrogen bonding among water molecules. This lowers surface tension, allowing water to spread and penetrate grease, lifting it away from fabric.
392. The capillary rise of water is due to:
ⓐ. High vapour pressure of water
ⓑ. Cohesion greater than adhesion
ⓒ. Adhesion of water to glass and surface tension
ⓓ. High viscosity of water
Correct Answer: Adhesion of water to glass and surface tension
Explanation: Adhesion pulls water molecules up the glass walls, while surface tension pulls the rest of the liquid, causing the meniscus to rise in narrow tubes.
393. Which of the following liquids forms a convex meniscus in a glass capillary?
ⓐ. Water
ⓑ. Alcohol
ⓒ. Mercury
ⓓ. Acetic acid
Correct Answer: Mercury
Explanation: In mercury, cohesion (Hg–Hg bonds) is stronger than adhesion (Hg–glass). Surface tension pulls mercury downward, producing a convex meniscus.
394. Which equation relates surface tension ($\gamma$) to rise of liquid in a capillary?
ⓐ. $\gamma = \dfrac{h \rho g r}{2}$
ⓑ. $\gamma = \dfrac{2h \rho g}{r}$
ⓒ. $\gamma = \dfrac{h}{2\rho g r}$
ⓓ. $\gamma = \dfrac{r}{2h\rho g}$
Correct Answer: $\gamma = \dfrac{h \rho g r}{2}$
Explanation: Capillary rise equation:
$$
\gamma = \frac{h \rho g r}{2 \cos \theta}
$$
For water in glass ($\theta \approx 0^\circ$), $\cos \theta = 1$.
395. Why do small insects like water striders walk on water without sinking?
ⓐ. Water has high viscosity
ⓑ. Surface tension acts like a stretched elastic film
ⓒ. Density of insects is less than water
ⓓ. Water molecules exert repulsive forces on insects
Correct Answer: Surface tension acts like a stretched elastic film
Explanation: Water’s high surface tension supports insects, preventing them from sinking, provided their legs distribute weight evenly without breaking the surface film.
396. Which of the following increases with addition of surfactants in water?
ⓐ. Contact angle of water on surfaces
ⓑ. Surface tension of water
ⓒ. Capillary rise of water
ⓓ. Cohesion between water molecules
Correct Answer: Contact angle of water on surfaces
Explanation: Surfactants reduce cohesive forces and increase wetting ability. This lowers surface tension but increases the contact angle on hydrophobic surfaces, improving spreading and cleaning.
397. Which phenomenon in plants is partially explained by surface tension?
ⓐ. Photosynthesis
ⓑ. Transpiration pull
ⓒ. Capillary action in xylem
ⓓ. Respiration
Correct Answer: Capillary action in xylem
Explanation: Surface tension helps water rise in narrow xylem vessels of plants, working together with adhesion and cohesion to transport water upward against gravity.
398. Which chemical equation shows the role of surface tension in dissolving detergent micelles?
ⓐ. $CH_3(CH_2)_{15}SO_3^- Na^+ + Oil \rightarrow Micelle$
ⓑ. $H_2O \rightarrow H^+ + OH^-$
ⓒ. $CO_2 + H_2O \rightarrow H_2CO_3$
ⓓ. $C_6H_{12}O_6 + O_2 \rightarrow CO_2 + H_2O$
Correct Answer: $CH_3(CH_2)_{15}SO_3^- Na^+ + Oil \rightarrow Micelle$
Explanation: Detergent molecules surround oil droplets, lowering interfacial tension and forming micelles. Hydrophobic tails dissolve grease, while hydrophilic heads interact with water, cleaning surfaces.
399. Which application of surface tension is used in medicine?
ⓐ. Alcohol evaporation during sterilization
ⓑ. Measurement of blood plasma viscosity
ⓒ. Use of soaps to reduce foaming
ⓓ. Use of lung surfactants in premature infants
Correct Answer: Use of lung surfactants in premature infants
Explanation: Surfactants reduce surface tension in alveoli, preventing collapse of lungs during breathing. Premature babies often require artificial surfactants for survival.
400. The work required to increase the surface area of a liquid by 1 unit is equal to:
ⓐ. Vapour pressure
ⓑ. Viscosity
ⓒ. Surface tension
ⓓ. Density
Correct Answer: Surface tension
Explanation: Surface tension is defined as energy per unit area. Work done to create more surface area against cohesive forces directly equals surface tension, measured in $J/m^2$ or $N/m$.
You are on Class 11 Chemistry MCQs – Chapter 5: States of Matter (Part 4). This chapter, part of the NCERT/CBSE Class 11 syllabus, is highly relevant for board exams and also plays a significant role in competitive exams like JEE, NEET, and state-level tests. The complete series provides 494 MCQs with detailed answers, divided into 5 practice parts. This section contains the fourth set of 100 MCQs, covering liquids, intermolecular forces, surface tension, viscosity, vapor pressure, and properties of liquids. It also touches on modern applications such as supercritical fluids and liquid crystals, which are occasionally included in exams as advanced concepts. These MCQs are highly recommended for students who want strong conceptual understanding along with exam-level practice.
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