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101. Why does water have a much higher boiling point than hydrogen sulfide ($H_2S$)?
ⓐ. Water is lighter, so it requires more energy to boil
ⓑ. Water forms extensive hydrogen bonds, which require large thermal energy to break
ⓒ. Hydrogen sulfide has stronger intermolecular forces than water
ⓓ. Both have equal boiling points because they are hydrides
Correct Answer: Water forms extensive hydrogen bonds, which require large thermal energy to break
Explanation: Water molecules form strong intermolecular hydrogen bonds due to high electronegativity of oxygen. Breaking these requires more energy, raising the boiling point. $H_2S$ lacks hydrogen bonding and is held together mainly by weak dipole–dipole and dispersion forces, hence it boils at much lower temperatures.
102. Which factor causes ionic solids like NaCl to have very high melting points?
ⓐ. Presence of weak dispersion forces
ⓑ. Strong electrostatic forces between oppositely charged ions
ⓒ. Hydrogen bonding between ions
ⓓ. Dipole–dipole forces between ions
Correct Answer: Strong electrostatic forces between oppositely charged ions
Explanation: In ionic solids, large Coulombic forces hold ions tightly in a rigid lattice. Enormous energy is required to overcome these interactions, giving high melting points. Hydrogen bonding and dipole–dipole do not apply to ionic crystals.
103. Why does methane ($CH_4$) have a very low boiling point?
ⓐ. Strong hydrogen bonding
ⓑ. Presence of dipole–dipole interactions
ⓒ. Only weak London dispersion forces hold the molecules together
ⓓ. High molecular mass of carbon
Correct Answer: Only weak London dispersion forces hold the molecules together
Explanation: Methane is nonpolar and cannot form hydrogen bonds or dipole–dipole interactions. Its molecules interact only through temporary induced dipoles (dispersion forces), which are weak, hence very low boiling point.
104. Which factor increases both melting and boiling points of a substance?
ⓐ. Strong intermolecular forces
ⓑ. High vapor pressure
ⓒ. Low molecular mass
ⓓ. High thermal energy
Correct Answer: Strong intermolecular forces
Explanation: Stronger intermolecular forces require more energy input to separate molecules, raising both melting and boiling points. High vapor pressure (B) actually indicates weak intermolecular forces, while molecular mass (C) and thermal energy (D) are not the primary determinants.
105. Why do nonpolar molecules with larger molar mass generally have higher boiling points than smaller nonpolar molecules?
ⓐ. They form hydrogen bonds with each other
ⓑ. They experience stronger London dispersion forces due to more electrons and higher polarizability
ⓒ. They are heavier, so harder to vaporize
ⓓ. They exist in ionic form at higher mass
Correct Answer: They experience stronger London dispersion forces due to more electrons and higher polarizability
Explanation: As molecular size and electron cloud increase, temporary dipoles become stronger, enhancing London dispersion forces. This raises boiling points in larger nonpolar molecules like pentane compared to methane.
106. Which property best explains why iodine ($I_2$) has a higher melting point than fluorine ($F_2$)?
ⓐ. Iodine forms hydrogen bonds
ⓑ. Iodine has stronger London dispersion forces due to more electrons
ⓒ. Fluorine molecules have stronger ion–dipole interactions
ⓓ. Fluorine has a higher molecular mass
Correct Answer: Iodine has stronger London dispersion forces due to more electrons
Explanation: Iodine molecules are larger and more polarizable, producing stronger London dispersion forces than small fluorine molecules. Hence, iodine is solid at room temperature, while fluorine is a gas.
107. Which statement best explains why salt (NaCl) raises the boiling point of water?
ⓐ. It increases hydrogen bonding in water
ⓑ. It decreases intermolecular forces in water
ⓒ. It introduces ion–dipole interactions, making it harder for water molecules to escape into vapor
ⓓ. It decreases density of water
Correct Answer: It introduces ion–dipole interactions, making it harder for water molecules to escape into vapor
Explanation: Dissolved NaCl interacts strongly with water molecules via ion–dipole forces, reducing the number of “free” water molecules available to evaporate. Thus, higher temperature (boiling point elevation) is needed to vaporize water.
108. Why do molecular solids like naphthalene have relatively low melting points compared to ionic solids?
ⓐ. They have weak dispersion forces holding the molecules together
ⓑ. They contain strong covalent networks
ⓒ. They are stabilized by metallic bonding
ⓓ. They have hydrogen bonding in the lattice
Correct Answer: They have weak dispersion forces holding the molecules together
Explanation: Molecular solids are held by weak van der Waals forces, which require less energy to overcome. Ionic solids, in contrast, have strong Coulombic attractions requiring high energy to melt.
109. Why does graphite have a very high melting point?
ⓐ. Because it has hydrogen bonding between layers
ⓑ. Because it has strong covalent bonding within layers forming a giant covalent structure
ⓒ. Because it has weak dispersion forces only
ⓓ. Because of ion–dipole forces in the lattice
Correct Answer: Because it has strong covalent bonding within layers forming a giant covalent structure
Explanation: In graphite, carbon atoms are covalently bonded in hexagonal layers. Breaking this requires breaking covalent bonds, which demands high energy. Although layers are weakly held by dispersion forces, the intralayer bonds dominate, giving a very high melting point.
110. Why do metals typically have high melting and boiling points?
ⓐ. Because of strong delocalized metallic bonding between cations and sea of electrons
ⓑ. Because of weak London forces between atoms
ⓒ. Because they form dipole–dipole interactions
ⓓ. Because metallic atoms have low density
Correct Answer: Because of strong delocalized metallic bonding between cations and sea of electrons
Explanation: Metallic bonding involves electrostatic attraction between positively charged metal ions and a delocalized electron cloud. This bonding is strong and requires significant thermal energy to break, resulting in high melting and boiling points.
111. Boyle’s law can be stated as:
ⓐ. At constant pressure, volume is directly proportional to temperature
ⓑ. At constant temperature, pressure is directly proportional to volume
ⓒ. At constant temperature, pressure is inversely proportional to volume
ⓓ. At constant volume, pressure is directly proportional to temperature
Correct Answer: At constant temperature, pressure is inversely proportional to volume
Explanation: Boyle’s law describes the relationship between pressure and volume of a gas at constant temperature: $P \propto \tfrac{1}{V}$. When one increases, the other decreases, keeping $PV = k$ constant for a fixed amount of gas.
112. Which of the following mathematical forms correctly represents Boyle’s law?
ⓐ. $PV = k$
ⓑ. $\dfrac{P}{V} = k$
ⓒ. $\dfrac{V}{T} = k$
ⓓ. $P = kT$
Correct Answer: $PV = k$
Explanation: For a fixed mass of gas at constant temperature, the product of pressure and volume remains constant. Option B is incorrect because pressure is not directly proportional to volume. Options C and D describe other gas laws.
113. A sample of gas occupies 4.0 L at 2 atm pressure. What will be its volume at 1 atm, keeping temperature constant?
ⓐ. 2.0 L
ⓑ. 4.0 L
ⓒ. 6.0 L
ⓓ. 8.0 L
Correct Answer: 8.0 L
Explanation: By Boyle’s law: $P_1V_1 = P_2V_2$. Substituting: $(2)(4.0) = (1)(V_2)$. So $V_2 = 8.0 \, L$. Hence volume doubles when pressure is halved at constant temperature.
114. The graph of Boyle’s law (P vs V) at constant temperature is:
ⓐ. A straight line passing through the origin
ⓑ. A hyperbola
ⓒ. A parabola opening upward
ⓓ. A horizontal straight line
Correct Answer: A hyperbola
Explanation: Since $P \propto \tfrac{1}{V}$, the graph of P against V is a rectangular hyperbola. A plot of P versus 1/V would give a straight line.
115. For a gas at constant temperature, when volume decreases by half, pressure:
ⓐ. Remains unchanged
ⓑ. Doubles
ⓒ. Reduces to half
ⓓ. Becomes zero
Correct Answer: Doubles
Explanation: Boyle’s law shows that $P \propto 1/V$. If volume becomes $\tfrac{1}{2} V$, then pressure becomes $2P$. Thus halving volume doubles the pressure.
116. A gas is kept at 300 K. Its pressure increases from 1 atm to 5 atm. If initial volume is 10 L, what is the final volume?
ⓐ. 2.0 L
ⓑ. 10.0 L
ⓒ. 50.0 L
ⓓ. 5.0 L
Correct Answer: 2.0 L
Explanation: Using $P_1V_1 = P_2V_2$: $(1)(10) = (5)(V_2)$. So $V_2 = 2.0 \, L$. This shows that increasing pressure five times reduces volume five times at constant T.
117. Which of the following experimental setups best verifies Boyle’s law?
ⓐ. Heating gas in a sealed container and measuring pressure
ⓑ. Cooling gas at constant volume and measuring temperature
ⓒ. Trapping air in a J-tube and varying mercury levels to change volume and pressure
ⓓ. Compressing ice into liquid
Correct Answer: Trapping air in a J-tube and varying mercury levels to change volume and pressure
Explanation: In Boyle’s law experiments, a J-shaped glass tube filled partly with mercury traps air in one limb. Varying mercury height changes pressure, and corresponding volume changes can be measured, confirming $PV = \text{constant}$.
118. If the pressure of a fixed mass of gas is plotted against its reciprocal of volume (1/V), the graph is:
ⓐ. Parabola
ⓑ. Straight line through the origin
ⓒ. Hyperbola
ⓓ. Circle
Correct Answer: Straight line through the origin
Explanation: Since $P \propto 1/V$, plotting P against 1/V gives a straight line through the origin. The slope equals the constant $k$.
119. Boyle’s law is valid under which condition?
ⓐ. Low pressure and high temperature
ⓑ. Constant temperature
ⓒ. Constant volume
ⓓ. Extremely high pressure and low temperature
Correct Answer: Constant temperature
Explanation: Boyle’s law applies only when temperature is held constant for a fixed amount of gas. At very high pressure or low temperature, deviations occur due to real gas behavior.
120. Which real-life application depends on Boyle’s law?
ⓐ. Capillary rise in liquids
ⓑ. Functioning of lungs during breathing
ⓒ. Heat conduction in solids
ⓓ. Melting of ice at 0°C
Correct Answer: Functioning of lungs during breathing
Explanation: During inhalation, lung volume increases, causing pressure inside lungs to drop, drawing in air. During exhalation, lung volume decreases, pressure rises, and air is expelled. This inverse pressure–volume relationship is a direct application of Boyle’s law.
121. A gas has volume 6.0 L at 3 atm pressure. What will be its pressure if the volume changes to 2.0 L at constant temperature?
ⓐ. 1 atm
ⓑ. 3 atm
ⓒ. 6 atm
ⓓ. 9 atm
Correct Answer: 9 atm
Explanation: Using Boyle’s law: $P_1V_1 = P_2V_2$. Substituting values: $(3)(6) = P_2 (2)$. So $P_2 = 18/2 = 9 \, \text{atm}$. Pressure increases threefold when volume decreases to one-third.
122. A gas sample initially at 1 atm and 500 mL is compressed isothermally to 250 mL. The final pressure is:
ⓐ. 0.5 atm
ⓑ. 1 atm
ⓒ. 2 atm
ⓓ. 4 atm
Correct Answer: 2 atm
Explanation: Boyle’s law: $P_1V_1 = P_2V_2$. $1 \times 500 = P_2 \times 250$. Hence $P_2 = 500/250 = 2 \, \text{atm}$. Pressure doubles when volume halves.
123. Which of the following pairs of quantities always remain constant in Boyle’s law?
ⓐ. Pressure and temperature
ⓑ. Volume and number of moles
ⓒ. Temperature and number of moles
ⓓ. Pressure and volume
Correct Answer: Temperature and number of moles
Explanation: Boyle’s law holds for a fixed amount of gas at constant temperature. Pressure and volume vary inversely, but the condition for validity is constant T and n.
124. The constant obtained in Boyle’s law, $k = PV$, depends on:
ⓐ. Nature of the gas and its amount only
ⓑ. Pressure and volume values chosen
ⓒ. Whether the gas is compressed or expanded
ⓓ. The color of the container
Correct Answer: Nature of the gas and its amount only
Explanation: $k$ is a constant for a given mass of gas at fixed temperature. It depends on number of moles and temperature, not on arbitrary pressure or volume values.
125. Which curve best represents isothermal compression of a gas?
ⓐ. Isobaric line (horizontal)
ⓑ. Isochoric line (vertical)
ⓒ. Hyperbolic curve in PV graph
ⓓ. Straight line slope upward in P–V graph
Correct Answer: Hyperbolic curve in PV graph
Explanation: At constant temperature, Boyle’s law gives $P \propto 1/V$. Thus, the curve is a rectangular hyperbola in the P–V plane.
126. A balloon filled with air at 1 atm has a volume of 2.5 L. If the balloon is squeezed until volume becomes 1.25 L, the new pressure inside will be:
ⓐ. 0.5 atm
ⓑ. 1 atm
ⓒ. 2 atm
ⓓ. 2.5 atm
Correct Answer: 2 atm
Explanation: Using $P_1V_1 = P_2V_2$: $1 \times 2.5 = P_2 \times 1.25$. Thus $P_2 = 2 \, \text{atm}$.
127. Boyle’s law fails at:
ⓐ. Very high temperature and low pressure
ⓑ. Low temperature and high pressure
ⓒ. Room temperature and moderate pressure
ⓓ. Any condition of constant T
Correct Answer: Low temperature and high pressure
Explanation: Under these conditions, real gases deviate significantly from ideal behavior because intermolecular forces and molecular volume cannot be neglected. Boyle’s law applies best at high T and low P.
128. Which of the following statements is correct for Boyle’s law?
ⓐ. Pressure is directly proportional to volume at constant T
ⓑ. Pressure is inversely proportional to volume at constant T
ⓒ. Pressure is independent of volume
ⓓ. Pressure increases with temperature at constant V
Correct Answer: Pressure is inversely proportional to volume at constant T
Explanation: This is the fundamental statement of Boyle’s law. $PV = \text{constant}$ shows the inverse relation.
129. A given gas occupies 200 mL at 740 mmHg. What volume will it occupy at 760 mmHg, assuming constant T?
ⓐ. 194.7 mL
ⓑ. 200 mL
ⓒ. 220 mL
ⓓ. 740 mL
Correct Answer: 194.7 mL
Explanation: Boyle’s law: $P_1V_1 = P_2V_2$. $740 \times 200 = 760 \times V_2$. $V_2 = (148000 / 760) = 194.7 \, mL$. The slight rise in pressure decreases the volume a little.
130. Which practical device works on the principle of Boyle’s law?
ⓐ. Refrigerator
ⓑ. Hydraulic press
ⓒ. Syringe
ⓓ. Mercury thermometer
Correct Answer: Syringe
Explanation: In a syringe, pulling the piston increases volume, decreasing pressure, so liquid is drawn in. Pushing the piston decreases volume, increasing pressure, expelling liquid. This is a direct application of Boyle’s law.
131. Charles’ law states that for a fixed mass of gas at constant pressure:
ⓐ. Volume is directly proportional to pressure
ⓑ. Volume is inversely proportional to temperature
ⓒ. Volume is directly proportional to absolute temperature
ⓓ. Volume is constant regardless of temperature
Correct Answer: Volume is directly proportional to absolute temperature
Explanation: Charles’ law says $V \propto T$ when pressure and number of moles are constant. This means that as the temperature of a gas increases (in Kelvin), molecules move faster and occupy more space, so the volume expands. If the gas is cooled, volume contracts proportionally. Options A and B describe Boyle’s law, not Charles’.
132. The mathematical expression for Charles’ law is:
ⓐ. $V \times T = k$
ⓑ. $\dfrac{V}{T} = k$
ⓒ. $V = kP$
ⓓ. $\dfrac{P}{V} = k$
Correct Answer: $\dfrac{V}{T} = k$
Explanation: For a fixed pressure and number of moles, the ratio of volume to absolute temperature remains constant. Here, $V$ is the volume, $T$ is absolute temperature (in Kelvin), and $k$ is a constant.
133. A gas has a volume of 500 mL at 27°C. What will be its volume at 127°C if pressure is constant?
ⓐ. 250 mL
ⓑ. 500 mL
ⓒ. 666.7 mL
ⓓ. 750 mL
Correct Answer: 666.7 mL
Explanation: Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300 \,K$, $T_2 = 127 + 273 = 400 \,K$. Using Charles’ law: $V_1/T_1 = V_2/T_2$. So $500/300 = V_2/400$. $V_2 = (500 \times 400)/300 = 666.7 \, mL$.
134. Which of the following graphs represents Charles’ law?
ⓐ. Pressure vs Volume (hyperbola)
ⓑ. Volume vs Temperature in °C (straight line not through origin)
ⓒ. Volume vs Temperature in Kelvin (straight line through origin)
ⓓ. Pressure vs 1/Volume (straight line through origin)
Correct Answer: Volume vs Temperature in Kelvin (straight line through origin)
Explanation: Charles’ law gives $V \propto T$. When T is in Kelvin, the graph is a straight line through the origin. If plotted in °C, the line cuts the x-axis at –273°C, the absolute zero.
135. Why is temperature in Kelvin used in Charles’ law instead of Celsius?
ⓐ. Because Celsius is not scientific
ⓑ. Because gas molecules stop moving at 0 K, making it the true zero of temperature
ⓒ. Because Kelvin values are smaller than Celsius
ⓓ. Because Kelvin is more convenient for calculations only
Correct Answer: Because gas molecules stop moving at 0 K, making it the true zero of temperature
Explanation: At absolute zero (0 K = –273.15°C), the volume of an ideal gas becomes zero. Hence, Kelvin is the correct scale since it is directly proportional to molecular kinetic energy. Celsius would not show a proportional relationship.
136. A gas sample has volume 2.0 L at 300 K. What will be its volume at 450 K if pressure remains constant?
ⓐ. 1.33 L
ⓑ. 2.0 L
ⓒ. 4.5 L
ⓓ. 3.0 L
Correct Answer: 3.0 L
Explanation: Using $V_1/T_1 = V_2/T_2$: $2.0/300 = V_2/450$. Thus $V_2 = (2.0 \times 450)/300 = 3.0 \, L$. Volume increases with temperature at constant pressure.
137. What happens to the volume of a gas if temperature is reduced to half at constant pressure?
ⓐ. Volume becomes half
ⓑ. Volume becomes double
ⓒ. Volume remains constant
ⓓ. Volume becomes zero
Correct Answer: Volume becomes half
Explanation: Since $V \propto T$, halving the absolute temperature halves the volume. For example, if temperature drops from 400 K to 200 K, volume decreases by the same factor.
138. Charles’ law can be explained using kinetic theory because:
ⓐ. Increasing temperature decreases molecular collisions
ⓑ. Increasing temperature increases average kinetic energy, so molecules occupy larger volume to keep pressure constant
ⓒ. Decreasing temperature reduces intermolecular forces completely
ⓓ. Molecules disappear at high temperature
Correct Answer: Increasing temperature increases average kinetic energy, so molecules occupy larger volume to keep pressure constant
Explanation: Higher temperature makes molecules move faster, colliding more frequently and with greater force. To maintain constant pressure, volume expands, which balances the collision frequency with container walls.
139. A balloon has a volume of 2.5 L at 20°C. At what temperature will it occupy 5.0 L, assuming pressure is constant?
ⓐ. 293 K
ⓑ. 386 K
ⓒ. 586 K
ⓓ. 313 K
Correct Answer: 586 K
Explanation: Convert initial T to Kelvin: $T_1 = 20 + 273 = 293 \,K$. Using $V_1/T_1 = V_2/T_2$: $2.5/293 = 5.0/T_2$. So $T_2 = (5.0 \times 293)/2.5 = 586 \,K$.
140. Which everyday example best demonstrates Charles’ law?
ⓐ. Boiling of water at 100°C
ⓑ. Expansion of air in a hot-air balloon as it is heated
ⓒ. Compression of air in a syringe
ⓓ. Melting of ice into water
Correct Answer: Expansion of air in a hot-air balloon as it is heated
Explanation: In a hot-air balloon, heating the air increases its temperature, so volume expands at nearly constant pressure. The balloon rises because expanded hot air is less dense than cooler outside air. This directly illustrates Charles’ law.
141. If the volume of a gas is 600 mL at 300 K, what will be the volume at 600 K at constant pressure?
ⓐ. 300 mL
ⓑ. 600 mL
ⓒ. 900 mL
ⓓ. 1200 mL
Correct Answer: 1200 mL
Explanation: By Charles’ law $V_1/T_1 = V_2/T_2$. Substituting values: $600/300 = V_2/600$. $V_2 = (600 \times 600)/300 = 1200 \, \text{mL}$. Thus doubling the temperature doubles the volume.
142. What would happen to the volume of a gas at constant pressure if its absolute temperature approaches zero?
ⓐ. Volume becomes very large
ⓑ. Volume becomes zero
ⓒ. Volume remains constant
ⓓ. Volume becomes negative
Correct Answer: Volume becomes zero
Explanation: According to Charles’ law, $V \propto T$. At absolute zero (0 K), the kinetic energy of molecules is zero, so theoretically, volume becomes zero. In reality, real gases liquefy before reaching 0 K.
143. Which of the following equations is derived from Charles’ law?
ⓐ. $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$
ⓑ. $P_1V_1 = P_2V_2$
ⓒ. $P \propto T$
ⓓ. $V \times T = k$
Correct Answer: $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$
Explanation: The proportionality $V \propto T$ can be expressed as $V/T = k$. Comparing two states gives $V_1/T_1 = V_2/T_2$. Option B is Boyle’s law, C is Gay-Lussac’s law, and D is incorrect form.
144. A gas has a volume of 4.0 L at 27°C. What will be its volume at 127°C, keeping pressure constant?
ⓐ. 4.0 L
ⓑ. 5.33 L
ⓒ. 8.0 L
ⓓ. 16.0 L
Correct Answer: 5.33 L
Explanation: Convert temperatures to Kelvin: $T_1 = 27+273=300\,K$, $T_2 = 127+273=400\,K$. Using Charles’ law: $V_1/T_1 = V_2/T_2$. $4.0/300 = V_2/400$. $V_2 = (4.0 \times 400)/300 = 5.33 \, L$.
145. Which of the following gases will show exact proportionality of volume with temperature under Charles’ law?
ⓐ. Real gases at low temperature
ⓑ. Real gases at high pressure
ⓒ. Ideal gases under all conditions
ⓓ. Real gases at very low pressure and high temperature
Correct Answer: Ideal gases under all conditions
Explanation: Charles’ law is strictly valid only for ideal gases, which assume no intermolecular forces and negligible volume. Real gases approximate this behavior best at low pressure and high temperature, where deviations are minimal.
146. Why does a football appear deflated when taken out on a cold winter morning?
ⓐ. Pressure increases at low temperature
ⓑ. Volume decreases as temperature decreases, reducing internal pressure and elasticity
ⓒ. Intermolecular forces disappear at low temperature
ⓓ. Air escapes through pores at night
Correct Answer: Volume decreases as temperature decreases, reducing internal pressure and elasticity
Explanation: According to Charles’ law, cooling reduces temperature, which decreases the volume of air inside the ball at constant pressure. This lowers internal pressure and the ball appears deflated.
147. A sample of gas occupies 750 mL at 27°C. At what temperature will it occupy 1000 mL at constant pressure?
ⓐ. 127°C
ⓑ. 363 K
ⓒ. 400 K
ⓓ. 100°C
Correct Answer: 400 K
Explanation: Convert to Kelvin: $T_1 = 27+273=300 \,K$. $V_1/T_1 = V_2/T_2$. $750/300 = 1000/T_2$. $T_2 = (1000 \times 300)/750 = 400 \,K$. Thus required temperature is 400 K (127°C).
148. The intercept of the extrapolated Volume vs Temperature (°C) graph of a gas is:
ⓐ. 0°C
ⓑ. –100°C
ⓒ. 273.15°C
ⓓ. –273.15°C
Correct Answer: –273.15°C
Explanation: The graph of volume against Celsius temperature extrapolates to zero volume at –273.15°C, which corresponds to 0 K, the absolute zero temperature. This is the foundation of the Kelvin scale.
149. If a balloon expands from 2.0 L at 300 K to 3.0 L, what is the final temperature at constant pressure?
ⓐ. 200 K
ⓑ. 300 K
ⓒ. 450 K
ⓓ. 600 K
Correct Answer: 450 K
Explanation: Using Charles’ law: $V_1/T_1 = V_2/T_2$. $2.0/300 = 3.0/T_2$. $T_2 = (3.0 \times 300)/2.0 = 450 K$. This shows direct proportionality between volume and temperature.
150. Which real-life application illustrates Charles’ law?
ⓐ. Working of pressure cooker
ⓑ. Syringe expelling medicine
ⓒ. Rising of hot-air balloons
ⓓ. Melting of ice in water
Correct Answer: Rising of hot-air balloons
Explanation: Hot air balloons rise because heating the air increases molecular kinetic energy, which increases the gas volume at constant pressure. The expanded hot air inside the balloon is less dense than surrounding cold air, providing lift. This is a classic demonstration of Charles’ law.
151. Gay Lussac’s law states that for a fixed mass of gas at constant volume:
ⓐ. Pressure is directly proportional to absolute temperature
ⓑ. Pressure is inversely proportional to absolute temperature
ⓒ. Pressure is directly proportional to volume
ⓓ. Pressure is inversely proportional to volume
Correct Answer: Pressure is directly proportional to absolute temperature
Explanation: Gay Lussac’s law says $P \propto T$ when volume and moles remain constant. As temperature increases, molecules move faster, colliding more forcefully with container walls, increasing pressure.
152. The mathematical form of Gay Lussac’s law is:
ⓐ. $\dfrac{P}{T} = k$
ⓑ. $P \times T = k$
ⓒ. $\dfrac{P}{V} = k$
ⓓ. $PV = nRT$
Correct Answer: $\dfrac{P}{T} = k$
Explanation: For a fixed volume of gas, the ratio of pressure to absolute temperature is constant. This is the simplest form of Gay Lussac’s law. Option D is the ideal gas law.
153. A gas has pressure of 2 atm at 300 K. What will be its pressure at 450 K if volume is constant?
ⓐ. 1.33 atm
ⓑ. 2.0 atm
ⓒ. 3.0 atm
ⓓ. 4.5 atm
Correct Answer: 3.0 atm
Explanation: Using $P_1/T_1 = P_2/T_2$: $2/300 = P_2/450$. So $P_2 = (2 \times 450)/300 = 3.0 \, \text{atm}$. Pressure rises as temperature increases.
154. Which graph represents Gay Lussac’s law correctly?
ⓐ. P vs T in Kelvin: straight line through origin
ⓑ. P vs V: hyperbola
ⓒ. P vs 1/V: straight line through origin
ⓓ. P vs T in Celsius: straight line cutting axis at –273°C
Correct Answer: P vs T in Kelvin: straight line through origin
Explanation: Gay Lussac’s law gives $P \propto T$ (in Kelvin). So the graph is a straight line through origin. If plotted in °C, the line cuts temperature axis at –273°C.
155. A sealed container of gas has pressure 5 atm at 400 K. What will be the pressure at 200 K?
ⓐ. 2.5 atm
ⓑ. 10 atm
ⓒ. 1 atm
ⓓ. 5 atm
Correct Answer: 2.5 atm
Explanation: Using $P_1/T_1 = P_2/T_2$: $5/400 = P_2/200$. So $P_2 = (5 \times 200)/400 = 2.5 \, \text{atm}$. Lower temperature reduces molecular motion, halving the pressure.
156. Why must temperature always be measured in Kelvin when applying Gay Lussac’s law?
ⓐ. Because Celsius values are not integers
ⓑ. Because pressure is directly proportional to absolute temperature (Kelvin scale), not Celsius scale
ⓒ. Because Kelvin values are always positive
ⓓ. Because Celsius is not a scientific unit
Correct Answer: Because pressure is directly proportional to absolute temperature (Kelvin scale), not Celsius scale
Explanation: At –273.15°C (0 K), molecular motion stops, and pressure becomes zero. Using Celsius scale would not reflect the true proportionality. Therefore, absolute temperature in Kelvin must be used.
157. Which real-life example demonstrates Gay Lussac’s law?
ⓐ. Hot-air balloon rising in the sky
ⓑ. A sealed aerosol can bursting when heated
ⓒ. Expansion of liquid in a thermometer
ⓓ. Capillary rise of water in a tube
Correct Answer: A sealed aerosol can bursting when heated
Explanation: At constant volume, heating increases molecular motion, raising pressure until the can bursts. This is a direct application of Gay Lussac’s law.
158. If a gas exerts 760 mmHg pressure at 273 K, what pressure will it exert at 546 K at constant volume?
ⓐ. 380 mmHg
ⓑ. 760 mmHg
ⓒ. 1250 mmHg
ⓓ. 1520 mmHg
Correct Answer: 1520 mmHg
Explanation: By $P_1/T_1 = P_2/T_2$: $760/273 = P_2/546$. So $P_2 = (760 \times 546)/273 = 1520 \, \text{mmHg}$. Doubling the temperature doubles the pressure.
159. Why does the pressure inside a car tire increase when the car is driven for a long distance?
ⓐ. Because air escapes into the tire
ⓑ. Because tire expands and adds more air molecules
ⓒ. Because friction heats the air, raising its temperature and hence its pressure at constant volume
ⓓ. Because intermolecular forces increase with driving
Correct Answer: Because friction heats the air, raising its temperature and hence its pressure at constant volume
Explanation: The rolling tire and road friction heat the air inside, raising its temperature. At constant volume of the tire, Gay Lussac’s law predicts pressure increases directly with temperature.
160. Which of the following conditions makes Gay Lussac’s law least applicable?
ⓐ. Very high temperature and low pressure
ⓑ. Room temperature and moderate pressure
ⓒ. Low temperature and high pressure
ⓓ. Constant volume conditions
Correct Answer: Low temperature and high pressure
Explanation: Real gases deviate from ideal behavior at low temperature and high pressure because intermolecular forces and finite molecular size cannot be ignored. Thus, under such conditions, Gay Lussac’s law does not hold accurately.
161. A steel cylinder contains oxygen gas at 10 atm and 300 K. If the cylinder is heated to 600 K without changing volume, what will be the new pressure?
ⓐ. 5 atm
ⓑ. 10 atm
ⓒ. 20 atm
ⓓ. 30 atm
Correct Answer: 20 atm
Explanation: By Gay Lussac’s law, $P_1/T_1 = P_2/T_2$. Substituting: $10/300 = P_2/600$. So $P_2 = (10 \times 600)/300 = 20 \, \text{atm}$. Doubling temperature doubles pressure at constant volume.
162. A gas exerts a pressure of 2.5 atm at 350 K. What will be its pressure at 175 K at constant volume?
ⓐ. 1.25 atm
ⓑ. 2.5 atm
ⓒ. 3.5 atm
ⓓ. 5.0 atm
Correct Answer: 1.25 atm
Explanation: Using $P_1/T_1 = P_2/T_2$. Substituting: $2.5/350 = P_2/175$. So $P_2 = (2.5 \times 175)/350 = 1.25 \, \text{atm}$. Halving the temperature halves the pressure.
163. The graph of Gay Lussac’s law (Pressure vs Temperature) is:
ⓐ. A parabola
ⓑ. A hyperbola
ⓒ. A straight line through origin (Kelvin scale)
ⓓ. A horizontal line
Correct Answer: A straight line through origin (Kelvin scale)
Explanation: Since $P \propto T$ (in Kelvin), the graph is a straight line through the origin. If temperature is expressed in Celsius, the line cuts the axis at –273°C.
164. Which device demonstrates Gay Lussac’s law in everyday life?
ⓐ. Pressure cooker whistle releasing steam
ⓑ. Hot-air balloon rising
ⓒ. Mercury thermometer
ⓓ. Hydraulic lift
Correct Answer: Pressure cooker whistle releasing steam
Explanation: In a sealed pressure cooker, heating raises the temperature of steam. At constant volume, pressure increases until the safety valve (whistle) releases excess pressure. This is a direct consequence of Gay Lussac’s law.
165. A gas sample at constant volume has a pressure of 400 mmHg at 200 K. What will be the pressure at 400 K?
ⓐ. 200 mmHg
ⓑ. 400 mmHg
ⓒ. 600 mmHg
ⓓ. 800 mmHg
Correct Answer: 800 mmHg
Explanation: Using $P_1/T_1 = P_2/T_2$. Substituting: $400/200 = P_2/400$. So $P_2 = (400 \times 400)/200 = 800 \, \text{mmHg}$. Pressure doubles as temperature doubles.
166. Why does heating a sealed glass container sometimes cause it to shatter?
ⓐ. Heating reduces molecular motion, lowering pressure
ⓑ. Heating increases molecular motion, raising pressure beyond the container’s strength
ⓒ. Heating increases intermolecular forces sharply
ⓓ. Heating reduces the number of molecules
Correct Answer: Heating increases molecular motion, raising pressure beyond the container’s strength
Explanation: Gay Lussac’s law states that at constant volume, pressure increases with temperature. If the internal pressure exceeds the container’s mechanical limit, the container can burst.
167. Which condition is essential for the validity of Gay Lussac’s law?
ⓐ. Constant pressure
ⓑ. Constant volume
ⓒ. Variable moles of gas
ⓓ. High pressure and low temperature
Correct Answer: Constant volume
Explanation: The law holds for a fixed mass of gas at constant volume. Any change in volume alters the relation, so constant V is the key condition.
168. A gas occupies 5 L at 750 mmHg and 300 K. What pressure will it exert at 450 K, keeping volume constant?
ⓐ. 500 mmHg
ⓑ. 750 mmHg
ⓒ. 850.5 mmHg
ⓓ. 1125 mmHg
Correct Answer: 1125 mmHg
Explanation: By $P_1/T_1 = P_2/T_2$: $750/300 = P_2/450$. $P_2 = (750 \times 450)/300 = 1125 \, \text{mmHg}$.
169. Why does the pressure inside a sealed aerosol can increase when exposed to sunlight?
ⓐ. Number of gas molecules increases
ⓑ. Volume of can expands significantly
ⓒ. Temperature rise increases kinetic energy, raising pressure at constant volume
ⓓ. Gas molecules lose kinetic energy
Correct Answer: Temperature rise increases kinetic energy, raising pressure at constant volume
Explanation: When sunlight heats the can, the gas molecules gain energy and collide more forcefully with container walls. At constant volume, pressure increases proportionally to temperature, as per Gay Lussac’s law.
170. At what temperature will the pressure of a gas become zero, if it obeyed Gay Lussac’s law perfectly?
ⓐ. –10°C
ⓑ. –200°C
ⓒ. 273.15°C
ⓓ. –273.15°C
Correct Answer: –273.15°C
Explanation: Extrapolation of the $P$ vs $T$ (°C) line shows pressure reaches zero at –273.15°C. This corresponds to absolute zero (0 K), where molecular motion theoretically ceases.
171. Avogadro’s law states that:
ⓐ. At constant pressure and temperature, the volume of a gas is inversely proportional to its moles.
ⓑ. At constant pressure and temperature, the volume of a gas is directly proportional to its moles.
ⓒ. At constant pressure and temperature, the volume of a gas is directly proportional to its pressure.
ⓓ. At constant pressure and temperature, the volume of a gas is directly proportional to its temperature.
Correct Answer: At constant pressure and temperature, the volume of a gas is directly proportional to its moles.
Explanation: Avogadro’s law shows that $V \propto n$ when pressure and temperature are constant. This means doubling the number of moles doubles the volume under the same conditions. Options A, C, and D describe other laws (Boyle’s, Gay Lussac’s, Charles’).
172. The mathematical form of Avogadro’s law is:
ⓐ. $V = kT$
ⓑ. $V = kn$
ⓒ. $V = kP$
ⓓ. $V = k$
Correct Answer: $V = kn$
Explanation: For a gas sample at constant P and T, volume is proportional to the number of moles: $V = kn$. Here, $k$ is a proportionality constant depending on P and T.
173. According to Avogadro’s law, equal volumes of gases at the same temperature and pressure contain:
ⓐ. Equal number of molecules
ⓑ. Equal number of atoms
ⓒ. Equal masses
ⓓ. Equal densities
Correct Answer: Equal number of molecules
Explanation: Avogadro’s hypothesis states that at the same T and P, equal volumes of different gases contain the same number of molecules. The mass and density may vary depending on molar mass, but molecule count remains equal.
174. At STP, the molar volume of an ideal gas is:
ⓐ. 11.2 L
ⓑ. 12.4 L
ⓒ. 1.0 L
ⓓ. 22.4 L
Correct Answer: 22.4 L
Explanation: One mole of an ideal gas occupies 22.4 L at 0°C (273 K) and 1 atm. This value comes directly from Avogadro’s law and the ideal gas equation.
175. Which of the following is an application of Avogadro’s law?
ⓐ. Determining molecular weights of gases
ⓑ. Measuring gas diffusion rates
ⓒ. Calculating osmotic pressure of solutions
ⓓ. Explaining critical temperature of gases
Correct Answer: Determining molecular weights of gases
Explanation: Since equal volumes of gases at the same T and P contain equal molecules, comparing masses of equal volumes gives relative molecular masses. This was historically key in deducing molecular formulas.
176. If 2 L of $H_2$ gas reacts completely with 1 L of $O_2$ gas at the same T and P, the volume of $H_2O$ vapor formed will be:
ⓐ. 1 L
ⓑ. 3 L
ⓒ. 5 L
ⓓ. 2 L
Correct Answer: 2 L
Explanation: The reaction is $2H_2 + O_2 \to 2H_2O$. According to Avogadro’s law, volume ratios are the same as mole ratios. 2 volumes of $H_2$ react with 1 volume of $O_2$ to produce 2 volumes of $H_2O$.
177. Which statement is true about Avogadro’s number ($6.022 \times 10^{23}$)?
ⓐ. It is the number of atoms in 1 g of any element.
ⓑ. It is the number of molecules in 22.4 L of any gas at STP.
ⓒ. It is the number of protons in 1 mole of hydrogen.
ⓓ. It is the number of neutrons in 1 mole of oxygen.
Correct Answer: It is the number of molecules in 22.4 L of any gas at STP.
Explanation: One mole of a gas at STP occupies 22.4 L and contains Avogadro’s number of molecules. It is not defined in terms of protons, neutrons, or grams, but in terms of particles per mole.
178. If 1 mole of $N_2$ gas occupies 22.4 L at STP, how many liters will 2 moles occupy at the same conditions?
ⓐ. 11.2 L
ⓑ. 22.4 L
ⓒ. 44.8 L
ⓓ. 66.0 L
Correct Answer: 44.8 L
Explanation: By Avogadro’s law, doubling the number of moles doubles the volume under constant T and P. Thus, 2 moles occupy 44.8 L.
179. Which experiment confirmed Avogadro’s hypothesis?
ⓐ. Rutherford’s gold foil experiment
ⓑ. Gay Lussac’s law of combining volumes
ⓒ. Millikan’s oil drop experiment
ⓓ. Boyle’s J-tube experiment
Correct Answer: Gay Lussac’s law of combining volumes
Explanation: Gay Lussac observed simple whole-number ratios of gas volumes in reactions. Avogadro explained this by stating equal volumes contain equal numbers of molecules, thus linking volume ratios to mole ratios.
180. Which real-life example demonstrates Avogadro’s law?
ⓐ. A balloon inflates more when more air is blown into it
ⓑ. A pressure cooker whistle blowing due to high steam pressure
ⓒ. Ice melting into water
ⓓ. Alcohol evaporating at room temperature
Correct Answer: A balloon inflates more when more air is blown into it
Explanation: Blowing more air increases the number of moles (n) of gas inside. At constant temperature and pressure, volume increases proportionally. This is Avogadro’s law in action.
181. If 3 L of nitrogen gas contains $7.5 \times 10^{22}$ molecules at constant T and P, how many molecules will 6 L of nitrogen contain?
ⓐ. $3.75 \times 10^{22}$
ⓑ. $7.5 \times 10^{22}$
ⓒ. $1.5 \times 10^{23}$
ⓓ. $3.0 \times 10^{23}$
Correct Answer: $1.5 \times 10^{23}$
Explanation: By Avogadro’s law, volume is directly proportional to number of molecules (or moles). Doubling volume doubles the number of molecules. $7.5 \times 10^{22} \times 2 = 1.5 \times 10^{23}$.
182. At the same temperature and pressure, which of the following gases will have the same volume if they contain equal number of molecules?
ⓐ. $H_2$ and $O_2$
ⓑ. $CO_2$ and $CH_4$
ⓒ. $N_2$ and $Cl_2$
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Avogadro’s law states equal volumes at same T and P contain equal numbers of molecules, regardless of their identity. So any gases with equal molecules will occupy equal volumes, irrespective of molar mass.
183. One mole of oxygen gas at STP occupies 22.4 L. What volume will 0.5 mol of oxygen occupy under the same conditions?
ⓐ. 11.2 L
ⓑ. 22.4 L
ⓒ. 33.6 L
ⓓ. 44.8 L
Correct Answer: 11.2 L
Explanation: Volume is proportional to moles. Halving the moles halves the volume. $22.4 \div 2 = 11.2$ L.
184. Which of the following mixtures obeys Avogadro’s law most accurately?
ⓐ. Real gases at low pressure and high temperature
ⓑ. Real gases at high pressure and low temperature
ⓒ. Liquids at room temperature
ⓓ. Solids at room temperature
Correct Answer: Real gases at low pressure and high temperature
Explanation: Avogadro’s law is valid for ideal gases. Real gases behave ideally at low pressure (where intermolecular forces are negligible) and high temperature (where kinetic energy dominates).
185. If 5 L of hydrogen reacts with 2.5 L of oxygen at constant T and P, what is the volume of steam formed?
ⓐ. 2.5 L
ⓑ. 6 L
ⓒ. 7.5 L
ⓓ. 5 L
Correct Answer: 5 L
Explanation: Reaction: $2H_2 + O_2 \rightarrow 2H_2O$. Volume ratio = 2:1:2. If 5 L of $H_2$ reacts with 2.5 L of $O_2$, it forms 5 L of $H_2O$ vapor.
186. How many liters of nitrogen gas at STP contain $3.011 \times 10^{23}$ molecules?
ⓐ. 5.6 L
ⓑ. 11.2 L
ⓒ. 22.4 L
ⓓ. 44.8 L
Correct Answer: 11.2 L
Explanation: One mole ($6.022 \times 10^{23}$ molecules) occupies 22.4 L at STP. Half a mole has $3.011 \times 10^{23}$ molecules and volume = 22.4 ÷ 2 = 11.2 L.
187. Two flasks of equal volume contain equal masses of hydrogen and oxygen at the same T and P. Which statement is correct?
ⓐ. They contain equal number of molecules.
ⓑ. The flask with oxygen contains fewer molecules.
ⓒ. The flask with hydrogen contains fewer molecules.
ⓓ. Both contain equal moles.
Correct Answer: The flask with oxygen contains fewer molecules.
Explanation: Equal mass does not mean equal moles. Since oxygen has higher molar mass (32 g/mol) compared to hydrogen (2 g/mol), the same mass has fewer moles, hence fewer molecules.
188. Which expression correctly represents Avogadro’s law?
ⓐ. $\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$
ⓑ. $P_1V_1 = P_2V_2$
ⓒ. $\dfrac{P}{T} = k$
ⓓ. $V \propto \dfrac{1}{T}$
Correct Answer: $\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$
Explanation: For a given gas at constant T and P, the ratio of volume to number of moles is constant. Options B, C, and D represent Boyle’s, Gay Lussac’s, and an incorrect relation, respectively.
189. Which of the following is NOT consistent with Avogadro’s law?
ⓐ. Volume is directly proportional to number of molecules.
ⓑ. Molar volume of different gases is equal at the same T and P.
ⓒ. Volume is directly proportional to molecular mass.
ⓓ. Equal volumes of gases contain equal molecules at same T and P.
Correct Answer: Volume is directly proportional to molecular mass.
Explanation: Volume does not depend on molecular mass but on number of molecules. Light gases and heavy gases both occupy the same volume if they have equal moles at the same conditions.
190. Which real-life example illustrates Avogadro’s law?
ⓐ. The volume of a balloon increases as more air is blown in at same T and P.
ⓑ. A pressure cooker builds up steam pressure with heating.
ⓒ. Steam condensing into water on cooling.
ⓓ. Gas spreading out to fill a room.
Correct Answer: The volume of a balloon increases as more air is blown in at same T and P.
Explanation: Adding more air increases the number of moles inside. At constant pressure and temperature, volume expands proportionally. This is a straightforward demonstration of Avogadro’s law.
191. Which equation relates pressure, volume, temperature, and moles for an ideal gas?
ⓐ. $PV = k$
ⓑ. $V/T = nR$
ⓒ. $P/T = nR$
ⓓ. $PV = nRT$
Correct Answer: $PV = nRT$
Explanation: The ideal gas equation combines Boyle’s ($P \propto 1/V$), Charles’ ($V \propto T$), and Avogadro’s ($V \propto n$) laws into $PV=nRT$, where $R$ is the gas constant. It holds best at low pressure and high temperature for real gases. Options A, C, and D are incomplete relations that do not include all variables simultaneously.
192. Which value of $R$ should be used when $P$ is in atm and $V$ in liters?
ⓐ. $R = 8.314\, \text{J mol}^{-1}\text{K}^{-1}$
ⓑ. $R = 0.082057\, \text{L atm mol}^{-1}\text{K}^{-1}$
ⓒ. $R = 62.36\, \text{L Torr mol}^{-1}\text{K}^{-1}$
ⓓ. $R = 8.314\, \text{kPa L mol}^{-1}\text{K}^{-1}$
Correct Answer: $R = 0.082057\, \text{L atm mol}^{-1}\text{K}^{-1}$
Explanation: Units must be consistent. With $P$ in atm and $V$ in L, use $R=0.082057\,\text{L atm mol}^{-1}\text{K}^{-1}$. If using $P$ in kPa (or J for $PV$), options A or D are appropriate since $1\,\text{J} = 1\,\text{kPa·L}$. Option C is for pressure in Torr (mmHg).
193. How many moles of $O_2$ are in $2.46\,\text{L}$ at $0.987\,\text{atm}$ and $298\,\text{K}$?
ⓐ. $0.0497$ mol
ⓑ. $0.245$ mol
ⓒ. $0.123$ mol
ⓓ. $0.0993$ mol
Correct Answer: $0.0993$ mol
Explanation: Use $n=\dfrac{PV}{RT}$. Here $n=\dfrac{(0.987)(2.46)}{(0.082057)(298)}\approx\dfrac{2.428}{24.45}\approx 0.0993$ mol. Keeping consistent units (atm, L, K) is crucial; otherwise the numerical value of $R$ changes.
194. What volume will $0.250$ mol $N_2$ occupy at $1.00\,\text{atm}$ and $27^\circ\text{C}$?
ⓐ. $5.00\,\text{L}$
ⓑ. $6.15\,\text{L}$
ⓒ. $7.46\,\text{L}$
ⓓ. $22.4\,\text{L}$
Correct Answer: $6.15\,\text{L}$
Explanation: Convert $27^\circ\text{C}\to 300\,\text{K}$. $V=\dfrac{nRT}{P}=\dfrac{(0.250)(0.082057)(300)}{1.00}\approx 6.15\,\text{L}$. Option D is molar volume at STP for 1 mol, not for 0.25 mol at 300 K.
195. A gas has $n=0.100$ mol in a $2.00\,\text{L}$ rigid vessel at $350\,\text{K}$. What is $P$ in kPa?
ⓐ. $72.8\,\text{kPa}$
ⓑ. $101\,\text{kPa}$
ⓒ. $145.5\,\text{kPa}$
ⓓ. $290.9\,\text{kPa}$
Correct Answer: $145.5\,\text{kPa}$
Explanation: Use $R=8.314\,\text{kPa·L mol}^{-1}\text{K}^{-1}$. $P=\dfrac{nRT}{V}=\dfrac{(0.100)(8.314)(350)}{2.00}\approx 145.5\,\text{kPa}$. Using an $R$ with mismatched units would give an incorrect result.
196. What is the density of $CO_2$ at $27^\circ\text{C}$ and $750\,\text{mmHg}$? ($M=44.01\,\text{g mol}^{-1}$)
ⓐ. $0.89\,\text{g L}^{-1}$
ⓑ. $3.52\,\text{g L}^{-1}$
ⓒ. $2.20\,\text{g L}^{-1}$
ⓓ. $1.76\,\text{g L}^{-1}$
Correct Answer: $1.76\,\text{g L}^{-1}$
Explanation: Use $d=\dfrac{PM}{RT}$. Convert $P=750/760=0.9868\,\text{atm}$, $T=300\,\text{K}$, $R=0.082057$. $d=\dfrac{(0.9868)(44.01)}{(0.082057)(300)}\approx\dfrac{43.45}{24.62}\approx 1.76\,\text{g L}^{-1}$.
197. A gas has density $1.25\,\text{g L}^{-1}$ at STP. Its molar mass is approximately:
ⓐ. $22.4\,\text{g mol}^{-1}$
ⓑ. $28.0\,\text{g mol}^{-1}$
ⓒ. $32.0\,\text{g mol}^{-1}$
ⓓ. $44.0\,\text{g mol}^{-1}$
Correct Answer: $28.0\,\text{g mol}^{-1}$
Explanation: Rearranging $d=\dfrac{PM}{RT}\Rightarrow M=\dfrac{dRT}{P}$. At STP $P=1\,\text{atm}$, $T=273\,\text{K}$, $R=0.082057$: $M\approx 1.25\times 0.082057\times 273\approx 28.0\,\text{g mol}^{-1}$ (near $N_2$).
198. A $5.00\,\text{L}$ flask at $298\,\text{K}$ contains $0.20$ mol $N_2$ and $0.30$ mol $O_2$. What is the total pressure?
ⓐ. $1.22\,\text{atm}$
ⓑ. $4.90\,\text{atm}$
ⓒ. $3.67\,\text{atm}$
ⓓ. $2.45\,\text{atm}$
Correct Answer: $2.45\,\text{atm}$
Explanation: Total moles $n_{\text{tot}}=0.50$. $P=\dfrac{nRT}{V}=\dfrac{(0.50)(0.082057)(298)}{5.00}\approx \dfrac{12.23}{5.00}=2.45\,\text{atm}$. Each gas’s partial pressure equals its mole fraction times $P_{\text{total}}$.
199. How many liters of $CO_2$ at $1.00\,\text{atm}$ and $300\,\text{K}$ form when $10.0\,\text{g}$ $CaCO_3$ decomposes completely? $(CaCO_3\rightarrow CaO+CO_2)$
ⓐ. $1.23\,\text{L}$
ⓑ. $5.00\,\text{L}$
ⓒ. $3.69\,\text{L}$
ⓓ. $2.46\,\text{L}$
Correct Answer: $2.46\,\text{L}$
Explanation: Moles $CaCO_3=\dfrac{10.0}{100.09}\approx 0.100$ mol $\Rightarrow n(CO_2)=0.100$ mol. $V=\dfrac{nRT}{P}=\dfrac{(0.100)(0.082057)(300)}{1.00}\approx 2.46\,\text{L}$. Stoichiometry (1:1) links moles of $CO_2$ to $CaCO_3$.
200. Under which conditions does $PV=nRT$ best describe real gases?
ⓐ. 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 $P$, molecules are far apart (negligible attractions and volumes), and at high $T$, kinetic energy dominates over intermolecular forces. Thus real gases approach ideal behavior. High pressures/low temperatures magnify deviations due to attractions and finite molecular size.
You are now on Class 11 Chemistry MCQs – Chapter 5: States of Matter (Part 2). This chapter is a vital part of the NCERT/CBSE syllabus, providing the theoretical base of the gaseous and liquid state of matter. Questions from this chapter frequently appear in board exams, JEE, NEET, and state-level entrance exams. The complete chapter offers 494 MCQs with answers, divided into 5 structured parts for step-by-step preparation. This section brings you the second set of 100 MCQs, focusing on ideal gas equation (PV = nRT), derivation of gas laws, real gases, compressibility factor (Z), and deviations from ideal behavior. These MCQs also include applications of kinetic theory of gases, molecular velocity calculations (rms, average, most probable speed), and Graham’s law of diffusion. Such topics are not only essential for conceptual clarity but also heavily weighted in competitive exams.
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