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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 121-150 of 394 questions

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121 Hard
121 of 394
Application +4 | -1
Helium has the van der Waal's constant b=24 mL mol−1b=24 \mathrm{~mL} \mathrm{~mol}^{-1}. The molecular diameter of helium will be
View answer and explanation
Correct option: A
267 pm267 \mathrm{~pm}
Option A (267 pm267 \mathrm{~pm}) is correct because it is calculated using the relation between van der Waals constant 'b' and the molecular diameter.
122 Hard
122 of 394
Understanding +4 | -1
The equation, [Pr+3Vr2][3Vr−1]=8Tr:\left[ {{P_r} + \frac{3}{{V_r^2}}} \right]\left[ {3{V_r} - 1} \right] = 8{T_r}:
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because the given equation represents the law of corresponding states, which relates reduced pressure, volume, and temperature and provides better results at the boiling point.
123 Hard
123 of 394
Application +4 | -1
As the temperature is raised from 200C to 400C{20^0}C\,\mathrm{to}\,{40^0}C the average kinetic energy of neon atoms changes by a factor of which of the following?
View answer and explanation
Correct option: C
313/293\sqrt {313/293}
Option B (313/293\sqrt {313/293}) is correct because the average kinetic energy is directly proportional to the absolute temperature, and the ratio of the square roots of the temperatures in Kelvin gives the correct factor.
124 Medium
124 of 394
Application +4 | -1
The inversion temperature Ti( K)T_i({~K}) of hydrogen is (given van der Waal's constants aa and bb are 0.244 atm L mol−20.244 {~atm} {~L} {~mol}^{-2} and 0.027Lmol−10.027 {Lmol}^{-1} respectively)
View answer and explanation
Correct option: B
220
Option B (220) is correct because the inversion temperature is calculated using the formula Ti=2aRbT_i = \frac{{2a}}{{Rb}}, where a and b are van der Waals constants.
125 Medium
125 of 394
Application +4 | -1
The root mean square speed of the molecules of diatomic gas is uu. When the temperature is doubled, the molecules dissociates into two atoms. The new speed of the atom is:
View answer and explanation
Correct option: C
2u2u
Option C (2u2u) is correct because when the temperature is doubled and the diatomic molecule dissociates, the molar mass is halved, leading to a doubling of the RMS velocity.
126 Hard
126 of 394
Understanding +4 | -1
Statement I: In van der Waals' equation (p+n2aV2)(V−nb)=nRT\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T Pressure correction between molecules. (n2aV2)\left(\frac{n^2 a}{V^2}\right) is due to force of attraction

Statement II: Pressure of the gas is due to collisions of molecules against sides of the container
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
Option B is correct because both statements are true, but Statement II doesn't explain Statement I.
127 Hard
127 of 394
Application +4 | -1
If "A" joule kinetic energy possess by N\mathrm{N} number of O2\mathrm{O}_2 molecules at 150k150 \mathrm{k} and " 2 A2 \mathrm{~A} " joule kinetic energy possess by O2\mathrm{O}_2 molecules of another sample at 300k300 \mathrm{k}, Find how many molecules of O2\mathrm{O}_2 present in another sample.
View answer and explanation
Correct option: A
N\mathrm{N} molecules of O2\mathrm{O}_2
Option A (N molecules of O2) is correct because the number of molecules is directly proportional to the kinetic energy at a given temperature, and when both kinetic energy and temperature double, the number of molecules remains the same.
128 Medium
128 of 394
Understanding +4 | -1
Under identical condition of temperature and pressure, which among the following gases has lowest density?
View answer and explanation
Correct option: B
He
Option B (He) is correct because, under identical conditions, the gas with the lowest molar mass will have the lowest density according to the ideal gas law.
129 Medium
129 of 394
Remembering +4 | -1
Van der Waal found that two assumptions made by kinetic theory of gases were wrong, one of them is that gas molecules are
View answer and explanation
Correct option: C
Point particles without significant volume
Option D (Point particles without significant volume) is correct because Van der Waals' corrections addressed the assumption that gas molecules have negligible volume.
130 Hard
130 of 394
Application +4 | -1
If two molecules of A and B having mass 100 kg and 64 kg and rate of diffusion of A is 12×10−312 \times {10^{ - 3}} then what will be the rate of diffusion of B
View answer and explanation
Correct option: A
15×10−315 \times {10^{ - 3}}
Option A (15×10−315 \times {10^{ - 3}}) is correct because it applies Graham's Law of Diffusion to calculate the rate of diffusion of gas B based on the given information.
131 Medium
131 of 394
Analyzing +4 | -1
Assertion (A): The value of van der Waals' constant 'aa' is higher for ammonia than for nitrogen.

Reason (R): Intermolecular hydrogen bonding is present in ammonia.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the presence of intermolecular hydrogen bonding in ammonia leads to stronger attractive forces, resulting in a higher value for the van der Waals constant 'a'.
132 Medium
132 of 394
Analyzing +4 | -1
Which of the following statements on critical constants of gases are correct?

I. Larger the Tc/pcT_c / p_c value of a gas, larger would be the excluded volume.

II. Critial temperature (Tc)\left(T_c\right) of a gas is greater than its Boyle temperature (TB)\left(T_B\right).
III. At the critial point in the van der Waals' gas isotherm.

(∂p∂Vm)Tc=0\left(\frac{\partial p}{\partial V_m}\right)_{T_c}=0

Select the correct answer using the codes given below.
View answer and explanation
Correct option: B
Both I and III
Option B (Both I and III) is correct because statement I correctly relates Tc/Pc to excluded volume, and statement III correctly describes the condition at the critical point.
133 Medium
133 of 394
Application +4 | -1
2g2\mathrm{g} of hydrogen diffuses from a container in 10 minute. How many gram of oxygen would diffused through the same container in the same time under similar conditions?
View answer and explanation
Correct option: D
8 g8\,\mathrm{g}
Option D (8 g8\,\mathrm{g}) is correct because Graham's law of diffusion states that the rate of diffusion is inversely proportional to the square root of the molar mass.
134 Medium
134 of 394
Application +4 | -1
One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is:
View answer and explanation
Correct option: B
4  cal4\,\,\mathrm{cal}
Option B (4 cal4\,\mathrm{cal}) is correct because the molar specific heat at constant volume for a mixture is calculated using the weighted average of the individual molar specific heats.
135 Medium
135 of 394
Application +4 | -1
Equal volumes of two gases which don’t react together are confined in separate vessels. Their pressure is 100 mm and 300 mm of Hg respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture? (Temperature remains constant)
View answer and explanation
Correct option: D
200mm200\mathrm{ }mm
Option D (200mm200\mathrm{ }mm) is correct because the final pressure is the average of the initial pressures since the volumes are equal.
136 Medium
136 of 394
Analyzing +4 | -1
When air is blown to balloon (at constant temperature) its pressure and volume both increases. This violates:
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because Boyle's law, Charles' law, and the ideal gas law assume a fixed amount of gas. Adding air to the balloon changes the amount of gas, so none of these laws are violated.
137 Easy
137 of 394
Understanding +4 | -1
Dalton’s law of partial pressure will not apply to which of the following mixture
View answer and explanation
Correct option: B
H2+Cl2{H_2} + C{l_2}
Option B (H2+Cl2{H_2} + C{l_2}) is correct because hydrogen and chlorine react to form HCl, thus Dalton's law is not applicable.
138 Medium
138 of 394
Application +4 | -1
At what temperature will the volume of a gas at 0 ∘C0\,^\circ C double itself, pressure remaining constant?
View answer and explanation
Correct option: D
546  K546\,\,\mathrm{K}
Option D is correct because, according to Charles's Law, volume is directly proportional to temperature in Kelvin. Doubling the volume requires doubling the absolute temperature.
139 Medium
139 of 394
Understanding +4 | -1
The total kinetic energy associated with n moles of monoatomic gas is
View answer and explanation
Correct option: B
32 nRT\frac{3}{2}\,nRT
Option B (32 nRT\frac{3}{2}\,nRT) is correct because the total kinetic energy of n moles of a monoatomic gas is given by this formula.
140 Easy
140 of 394
Understanding +4 | -1
At ordinary temperature and pressure, among halogens, the chlorine is a gas, bromine is a liquid and iodine is a solid. This is because:
View answer and explanation
Correct option: B
Intermolecular forces among molecules of chlorine are the weakest and those in iodine are the strongest
Option B is correct because the strength of intermolecular forces increases down the halogen group, leading to chlorine being a gas, bromine a liquid, and iodine a solid.
141 Easy
141 of 394
Analyzing +4 | -1
Which pair of the gases diffuse through a small hole with the same rate of diffusion at same ‘P’ at ‘T’
View answer and explanation
Correct option: D
NO,C2H6NO,\mathrm{ }{C_2}{H_6}
Option D (NO,C2H6NO,\mathrm{ }{C_2}{H_6}) is correct because NO and C2H6 have nearly the same molar mass, leading to similar diffusion rates.
142 Medium
142 of 394
Remembering +4 | -1
Compressibility factor for a gas under critical conditions is
View answer and explanation
Correct option: C
38\frac{3}{8}
Option C is correct. The compressibility factor under critical conditions is Z = 3/8.
143 Medium
143 of 394
Analyzing +4 | -1
For one mole of an ideal gas, increasing the temperature from 10∘C10^\circ C to 20  ∘C20\,\,^\circ C
View answer and explanation
Correct option: D
Increases both the average kinetic energy and rms velocity, but not significantly
Option D is correct because the increase in temperature from 10°C to 20°C is not significant enough to double the average kinetic energy or increase the rms velocity by a factor of 2\sqrt{2} or 2.
144 Medium
144 of 394
Understanding +4 | -1
Slope between pV  and  ppV\,\,\mathrm{and}\,\,p at constant temperature is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because, according to the ideal gas law, pVpV is constant at a constant temperature, so the slope of the pVpV vs. pp graph is zero.
145 Hard
145 of 394
Application +4 | -1
Calculate the total pressure in a 10.0 L cylinder which contains 0.4 g helium, 1.6 g oxygen and 1.4 g nitrogen at 27 ∘C27\,^\circ C.
View answer and explanation
Correct option: A
0.492  atm  0.492\mathrm{ }\,\,atm\;
Option D (Vms\mathrm{V}_{\mathrm{ms}}) is correct because the root-mean-square velocity depends on the ratio of temperature to molar mass, which is the same for both gases.
146 Medium
146 of 394
Application +4 | -1
The average speed of an ideal gas molecule at 270C{27^0}C is 0.3msec⁡−10.3\mathrm{m}{\sec ^{ - 1}}. The average speed at 9270C\mathrm{9}{27^0}C will be msec⁡−1\mathrm{m}{\sec ^{ - 1}}
View answer and explanation
Correct option: A
0.60.6
Option A (0.6) is correct because the average speed of gas molecules is proportional to the square root of the absolute temperature.
147 Medium
147 of 394
Application +4 | -1
O2{O_2} is collected over water at 20 ∘C20\,^\circ C. The pressure inside shown by the gas is 740 mm of Hg. What is the pressure due to O2{\mathrm{O}_2} alone if vapour pressure of H2O{\mathrm{H}_2}\mathrm{O} is 18 mm at 20  ∘C20\,\,^\circ C ?
View answer and explanation
Correct option: A
722  mm722\mathrm{ }\,\,mm
Option A (722  mm722\mathrm{ }\,\,mm) is correct because the partial pressure of oxygen is the total pressure minus the vapor pressure of water.
148 Medium
148 of 394
Understanding +4 | -1
Which gas can be most readily liquefied?
View answer and explanation
Correct option: C
SO2\mathrm{S}{\mathrm{O}_2}
Option C (SO2\mathrm{S}{\mathrm{O}_2}) is correct because SO2SO_2 has the highest critical temperature among the given gases, making it the easiest to liquefy.
149 Easy
149 of 394
Remembering +4 | -1
Value of universal gas constant (RR) depends upon
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the value of the universal gas constant (R) is independent of the number of moles, volume, or temperature of the gas.
150 Medium
150 of 394
Remembering +4 | -1
The concept of critical temperature for a gas was given by:
View answer and explanation
Correct option: A
Andrew
Option A is correct because Thomas Andrews discovered the concept of critical temperature while studying the behavior of carbon dioxide.
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