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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 151-180 of 394 questions

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151 Medium
151 of 394
Understanding +4 | -1
When an ideal gas undergoes unrestricted expansion, no cooling takes place because the molecules:
View answer and explanation
Correct option: A
Exert no attractive forces on each other
Option A (Exert no attractive forces on each other) is correct because ideal gases are assumed to have no intermolecular forces.
152 Easy
152 of 394
Application +4 | -1
Same mass of CH4{CH}_4 and H2{H}_2 is taken in container, The partial pressure caused by H2{H}_2 is
View answer and explanation
Correct option: A
89\frac{8}{9}
Option A (89\frac{8}{9}) is correct because the partial pressure of a gas is proportional to its number of moles, and the ratio of moles of H2 to total moles is 8/9.
153 Medium
153 of 394
Application +4 | -1
A closed vessel contains equal number of oxygen and hydrogen molecules at a total pressure of 740 mm. If oxygen is removed from the system, the pressure:
View answer and explanation
Correct option: A
Becomes half of 740 mm
The correct answer is that the pressure becomes half of 740 mm because, according to Dalton's law of partial pressures, removing oxygen removes its contribution to the total pressure.
154 Medium
154 of 394
Application +4 | -1
A 4 : 1 mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of mixture effusing out initially is
View answer and explanation
Correct option: A
8:18\mathrm{ }:\mathrm{ }1
Option A (8:18\mathrm{ }:\mathrm{ }1) is correct because the ratio of effusion rates is proportional to the square root of the inverse ratio of their molar masses.
155 Easy
155 of 394
Understanding +4 | -1
Dalton’s law of partial pressure will not apply to which of the following mixture of gases
View answer and explanation
Correct option: B
H2  and  Cl2{H_2}\,\,\mathrm{and}\,\,C{l_2}
Option B (H2 and Cl2{H_2}\,\mathrm{and}\,C{l_2}) is correct because hydrogen and chlorine react to form HCl, thus Dalton's law is not applicable.
156 Medium
156 of 394
Understanding +4 | -1
What is the reason for unusual high b.p. of water?
View answer and explanation
Correct option: A
Due to the presence of H+{\mathrm{H}^ + } and OH−{\mathrm{OH}^ - }ions in water
Option A (Due to the presence of H+{\mathrm{H}^ + } and OH−{\mathrm{OH}^ - }ions in water) is correct because the strong hydrogen bonding between water molecules, facilitated by the presence of H+ and OH- ions, leads to a higher boiling point.
157 Medium
157 of 394
Understanding +4 | -1
Introduction of absolute scale of thermometry is the result of:
View answer and explanation
Correct option: C
Charles’ law
Option C (Charles’ law) is correct because Charles's Law led to the understanding of the relationship between volume and temperature, which is essential for establishing an absolute temperature scale.
158 Medium
158 of 394
Remembering +4 | -1
The critical temperature of a gas is related to van der Waal’s constants as
View answer and explanation
Correct option: C
Tc=8a27bRT_c=\frac{8 {a}}{27 {bR}}
Option C is correct. The critical temperature of a gas is related to van der Waals constants as Tc = 8a27bR\frac{{8a}}{{27bR}}.
159 Hard
159 of 394
Analyzing +4 | -1
Statement I: Second virial coefficient of CO2\mathrm{CO}_2 is −0.065 L mol−1-0.065 \mathrm{~L} \mathrm{~mol}^{-1} at 400 K400 \mathrm{~K} and 30 atm30 \mathrm{~atm} making CO2\mathrm{CO}_2 more compressible than an ideal gas.

Statement II: Compressibility factor of CO2(Z=0.941)\mathrm{CO}_2(Z=0.941) is less than 1 ( Z=1Z=1 for ideal gas) due to the attractive forces between CO2\mathrm{CO}_2 molecules.
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A is correct. Statement I is correct because a negative second virial coefficient indicates higher compressibility. Statement II is correct because Z < 1 indicates attractive forces.
160 Easy
160 of 394
Application +4 | -1
Assuming that air is 79%79 \% by mole of N2,20\mathrm{N}_2, 20 mole %O2\% \mathrm{O}_2 and 1%Ar1 \% \mathrm{Ar}, the density of air at 25∘C25^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} is
View answer and explanation
Correct option: A
1.18 g/lit1.18 \mathrm{~g} / \mathrm{lit}
Option A (1.18 g/lit1.18 \mathrm{~g} / \mathrm{lit}) is correct because it uses the ideal gas law with the average molar mass of air calculated from the given composition.
161 Easy
161 of 394
Application +4 | -1
Which of the following gases occupies maximum volume at STP
View answer and explanation
Correct option: D
1 gr molecule of CO2C{O_2}
Option D (1 gr molecule of CO2CO_2) is correct because 1 gram molecule represents 1 mole, which occupies the largest volume at STP compared to the other options.
162 Medium
162 of 394
Understanding +4 | -1
The value of Vander Waal&rsquo;s constant (a) is minimum for
View answer and explanation
Correct option: A
He
Option A (He) is correct because the Van der Waals constant 'a' represents the intermolecular forces of attraction, and Helium, being a noble gas with very weak intermolecular forces, has the minimum value of 'a'.
163 Easy
163 of 394
Understanding +4 | -1
Maximum deviation from ideal gas is expected from
View answer and explanation
Correct option: D
NH3(g)N{H_3}\left( g \right)
Option D (NH3(g)N{H_3}\left( g \right)) is correct because NH3 exhibits the strongest intermolecular forces (hydrogen bonding) among the given options, leading to the greatest deviation from ideal gas behavior.
164 Hard
164 of 394
Analyzing +4 | -1
Statement I: Reduced equation of state in terms of reduce: temperature θ\theta, reduced volume ϕ\phi and reduced pressure x can be applied to solid, gases and liquid.

Statement II: (π+3ϕ2)(3ϕ−1)=8θ\left(\pi+\frac{3}{\phi^2}\right)(3 \phi-1)=8 \theta. Equation is independent of p,Vp, V and TT as well as independent of vander Waals' constant aa and bb.
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A is correct because the reduced equation of state applies to all states of matter, and the given equation is independent of specific gas constants.
165 Easy
165 of 394
Remembering +4 | -1
Kelvin’s thermodynamic scale of temperature is
based on
View answer and explanation
Correct option: A
Charles’ law
Option A (Charles’ law) is correct because the Kelvin scale is based on extrapolating the volume of a gas to zero at absolute zero, which is a concept derived from Charles's law.
166 Easy
166 of 394
Understanding +4 | -1
Gas equation pV=nRTpV = nRT is obeyed by ideal gas in
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because the ideal gas equation pV=nRTpV = nRT applies to ideal gases regardless of the process (adiabatic or isothermal).
167 Hard
167 of 394
Application +4 | -1
The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same temperature is
View answer and explanation
Correct option: B
(73π)1/2{\left( {\frac{7}{{3\pi }}} \right)^{1/2}}
Option A ((3π7)1/2{\left( {\frac{{3\pi }}{7}} \right)^{1/2}}) is correct because it represents the ratio of the average speed of oxygen to the RMS speed of nitrogen, calculated using the respective formulas and molar masses.
168 Easy
168 of 394
Remembering +4 | -1
The speed possessed by majority of gaseous molecules is
View answer and explanation
Correct option: B
ump{u_{mp}}
Option B (ump{u_{mp}}) is correct because the most probable speed is the speed possessed by the largest number of molecules in a gas.
169 Medium
169 of 394
Application +4 | -1
1.0 L of N2{N_2} and 7/8 L of O2{O_2} at the same temperature and pressure were mixed together. What is the relation between the masses of the two gases in the mixture?
View answer and explanation
Correct option: C
MN2=MO2{M_\mathrm{N}}_2 = {M_\mathrm{O}}_2
Option C (MN2=MO2{M_\mathrm{N}}_2 = {M_\mathrm{O}}_2) is correct because at the same temperature and pressure, equal volumes contain equal number of moles. Given the ratio of volumes, we can determine the ratio of masses.
170 Easy
170 of 394
Application +4 | -1
At a constant temperature what should be the percentage increase in pressure for a 5%5 \% decrease in the volume of gas:
View answer and explanation
Correct option: C
5.26%5.26 \%
Option C (5.26%5.26 \%) is correct because it is calculated using Boyle's Law, where a decrease in volume leads to an increase in pressure, and the percentage increase is approximately 5.26%.
171 Medium
171 of 394
Application +4 | -1
At what temperature is the rms speed of hydrogen molecules the same as that of oxygen molecules at 1327  ∘C1327\,\,^\circ C ?
View answer and explanation
Correct option: B
100K100\mathrm{ }K
Option B (100 K) is correct because the RMS velocity is inversely proportional to the square root of the molar mass and directly proportional to the square root of the temperature. Setting the RMS velocities equal allows solving for the unknown temperature.
172 Medium
172 of 394
Remembering +4 | -1
The temperature, at which a gas shows maximum ideal behaviour, is known as
View answer and explanation
Correct option: A
Boyle’s temperature
Option A is correct. Boyle's temperature is the temperature at which a real gas exhibits maximum ideal behavior.
173 Hard
173 of 394
Application +4 | -1
In what ratio by mass, the gases COCO and 2 butene (C4H8)(C_4H_8) be mixed in a vessel so that they cause same partial pressures?
View answer and explanation
Correct option: C
1:21:2
Option C (1:21:2) is correct because to have the same partial pressures, the number of moles must be equal. Therefore, the ratio of masses is the ratio of their molar masses.
174 Medium
174 of 394
Analyzing +4 | -1
6.4 g of SO2S{O_2} at 00C{0^0}C and 0.99 atm pressure occupies a volume of 2.241 L. Predict which of the following is correct?
View answer and explanation
Correct option: A
The gas is ideal
Option A (The gas is ideal) is correct because the given conditions satisfy the ideal gas law, indicating negligible intermolecular forces.
175 Medium
175 of 394
Understanding +4 | -1
Given below are two statement: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) SO2(g)S{O_2}(g) is adsorbed to a large extent than H2(g){H_2}(g) on activated charcoal.
Reason (R) SO2(g)S{O_2}(g) has a higher critical temperature than H2(g){H_2}(g)
In the light of the above statements, choose the most appropriate answer from the options given below
View answer and explanation
Correct option: B
Both A and R are correct and R is the correct explanation of A
Option B (Both A and R are correct and R is the correct explanation of A) is correct because SO2 has a higher critical temperature than H2, indicating stronger intermolecular forces, which leads to greater adsorption on activated charcoal.
176 Medium
176 of 394
Analyzing +4 | -1
S-I: Rate of diffusion of CO2\mathrm{CO}_2 is more than that O3\mathrm{O}_3.

S-II: Vapour density of O3\mathrm{O}_3 is more than that of CO2\mathrm{CO}_2.
View answer and explanation
Correct option: A
Both statements are correct and statement-II is correct explanation of statement-I
Option A is correct because the rate of diffusion is inversely proportional to the square root of the molar mass, and the vapor density is directly proportional to the molar mass.
177 Medium
177 of 394
Analyzing +4 | -1
0.5 mole of each of H2,SO2{\mathrm{H}_2},\mathrm{S}{\mathrm{O}_2} and CH4\mathrm{C}{\mathrm{H}_4} are kept in a container. A hole was made in the container. After 3 h, the order of partial pressures in the container will be
View answer and explanation
Correct option: B
pSO2>pCH4>pH2p\mathrm{S}{\mathrm{O}_2} > p\mathrm{C}{\mathrm{H}_4} > p{\mathrm{H}_2}
Option D (pH2>pCH4>pSO2p{\mathrm{H}_2} > p\mathrm{C}{\mathrm{H}_4} > p\mathrm{S}{\mathrm{O}_2}) is correct because H2 diffuses fastest, followed by CH4, and then SO2, leading to a decrease in their partial pressures in that order.
178 Medium
178 of 394
Remembering +4 | -1
Most probable speed, average speed and RMS speed are related as:
View answer and explanation
Correct option: A
1:1.128:1.2241:1.128:1.224
Option A (1:1.128:1.224) is correct because it represents the approximate ratio of most probable speed, average speed, and RMS speed.
179 Easy
179 of 394
Remembering +4 | -1
The units of the Vander Waal’s constant ‘b’ are
View answer and explanation
Correct option: C
L.mol−1{\text{L}}{\text{.mo}}{{\text{l}}^{ - 1}}
Option C is correct. The units of the van der Waals constant 'b' represent the volume per mole, hence L.mol-1.
180 Hard
180 of 394
Application +4 | -1
When a capillary tube of diameter 0.8 mm is dipped in a liquid having density 800  kg  m−3800\,\,kg\,\,{m^{ - 3}}, then the height of liquid in the capillary tube rises to 4 cm. The surface tension of liquid is (g=9.8m/s2)\left( {\mathrm{g} = 9.8\mathrm{m}/{\mathrm{s}^2}} \right)
View answer and explanation
Correct option: C
6.3×10−2Nm−16.3 \times {10^{ - 2}}N{m^{ - 1}}
Option C (6.3×10−2Nm−16.3 \times {10^{ - 2}}N{m^{ - 1}}) is correct because it is calculated using the formula for surface tension based on capillary rise.
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