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NEET • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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Showing 211-240 of 394 questions

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211 Hard
211 of 394
Application +4 | -1
What is the pressure of 2 moles of NH3N{H_3} at 27∘C27^\circ C when its volume is 5 L in van der Waals’ equation (a=4.17,b=0.03711)\left( {a = 4.17,b = 0.03711} \right)
View answer and explanation
Correct option: B
9.33 atm
Option A (10.33 atm) is correct because it is calculated using the van der Waals equation with the given parameters.
212 Hard
212 of 394
Application +4 | -1
(a) Find the total pressure of components if a container of volume 8.21 lit. contains 2 moles of A\mathrm{A} and 3 mole of B\mathrm{B} at 300 K300 \mathrm{~K}

(b) What will be the partial pressure of component C\text{C} if 5 moles of C\mathrm{C} is also added to the container at same temperature.
View answer and explanation
Correct option: A
(a) 15 atm; (b) 15 atm
Option A ((a) 15 atm; (b) 15 atm) is correct because the total pressure is calculated using the ideal gas law, and partial pressure is proportional to the number of moles.
213 Hard
213 of 394
Application +4 | -1
At what temperature, both Celsius and Fahrenheit scale read the same value?
View answer and explanation
Correct option: D
−40∘- 40^\circ
The correct answer is -40°. The Celsius and Fahrenheit scales read the same value at -40 because this is the solution to the equation relating the two scales.
214 Easy
214 of 394
Remembering +4 | -1
According to the Avogadro’s Law, equal volumes of two different gases, under same conditions of temperature and pressure, contain equal number of:
View answer and explanation
Correct option: B
Molecules
Option B (Molecules) is correct because Avogadro's Law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules.
215 Medium
215 of 394
Application +4 | -1
The root mean square velocity of a gas is doubled when temperature is
View answer and explanation
Correct option: A
Increased four times
Option A (Increased four times) is correct because the root mean square velocity is proportional to the square root of the temperature.
216 Medium
216 of 394
Analyzing +4 | -1
The behavior of a real gas is usually depicted by plotting compressibility factor ZZ versus PP at a constant temperature. At high temperature and high pressure, ZZ is usually more than one. This fact can be explained by van der Waals equation when
View answer and explanation
Correct option: A
the constant ' aa ' is negligible and not ' bb '
Option A (the constant ' aa ' is negligible and not ' bb ') is correct because at high temperature and high pressure, the volume is smaller, and the repulsive forces (represented by 'b') become more significant, while attractive forces ('a') become less important.
217 Hard
217 of 394
Understanding +4 | -1
A gas cannot be liquefied if:
View answer and explanation
Correct option: C
Forces of attraction are zero under ordinary conditions
Option C (Forces of attraction are zero under ordinary conditions) is correct because liquefaction requires intermolecular forces to hold the molecules together in the liquid phase.
218 Easy
218 of 394
Remembering +4 | -1
Which of the following represents the value of gas constant, R\mathrm{R} in cal/K mol\mathrm{cal} / \mathrm{K} \mathrm{~mol} ?
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Correct option: A
1.987
Option A (1.987) is correct because the value of the gas constant R in cal / K mol is approximately 1.987.
219 Medium
219 of 394
Remembering +4 | -1
Which is a postulate of kinetic theory of gases?
View answer and explanation
Correct option: C
There is no influence of gravity on gas molecules
Option C (There is no influence of gravity on gas molecules) is correct because the kinetic theory of gases assumes that gravitational forces on gas molecules are negligible.
220 Medium
220 of 394
Understanding +4 | -1
As the speed of molecules increases, the number of collisions per second:
View answer and explanation
Correct option: B
Increases
Option B (Increases) is correct because as the speed of molecules increases, they collide more frequently with each other and the walls of the container, leading to an increase in the number of collisions per second.
221 Easy
221 of 394
Analyzing +4 | -1
Assertion: The pressure of a gas is directly proportional to the density of gas at a constant temperature and molecular mass.

Reason: The molecular weight of a vapour of a substance can be calculated using the expression M=dRTPM=\frac{d R T}{P}, where MM is the molecular weight of the vapour, dd is its density and PP is the pressure of the vapour, R{R} is the gas constant and T{T} is the temperature.
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 pressure of a gas is directly proportional to its density at constant temperature and molecular mass, and the ideal gas law can be rearranged to show this relationship.
222 Medium
222 of 394
Application +4 | -1
The rms speed of hydrogen is 7\sqrt 7 times the rms speed of nitrogen. If TT is the temperature of the gas, then
View answer and explanation
Correct option: C
TH2<TN2{T_{{H_2}}} < {T_{{N_2}}}
Option C (TH2<TN2){T_{{H_2}}} < {T_{{N_2}}}) is correct because the rms speed is inversely proportional to the square root of the molar mass and directly proportional to the square root of the temperature.
223 Hard
223 of 394
Understanding +4 | -1
When r,p  and  Mr,p\,\,\mathrm{and}\,\,M&nbsp; represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of diffusion (rA/rB)\left( {{r_A}/{r_B}} \right) of two gases A  and  BA\,\,\mathrm{and}\,\,B&nbsp; is given as
View answer and explanation
Correct option: B
(pA/pB)(MB/MA)1/2\left( {{p_A}/{p_B}} \right){\left( {{M_B}/{M_A}} \right)^{1/2}}
Option B ((pA/pB)(MB/MA)1/2\left( {{p_A}/{p_B}} \right){\left( {{M_B}/{M_A}} \right)^{1/2}}) is correct because it represents the correct application of Graham's law of diffusion, incorporating both pressure and molar mass.
224 Easy
224 of 394
Application +4 | -1
If temperature (in Kelvin) is quadrupled then rms speed of gaseous molecules become
View answer and explanation
Correct option: B
2 times
Option B (2 times) is correct because the root-mean-square (rms) speed is proportional to the square root of the temperature.
225 Medium
225 of 394
Understanding +4 | -1
At relatively high pressure, van der Waals’ equation reduces to:
View answer and explanation
Correct option: C
PV=RT+PbPV = RT + Pb
Option C (PV=RT+PbPV = RT + Pb) is correct because at high pressure, the volume 'b' (related to molecular size) in the van der Waals equation becomes significant, leading to this simplified form.
226 Easy
226 of 394
Application +4 | -1
Equal moles of H2{H_2} and O2{O_2} gases are placed in a container with a pin hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the H2{H_2} to escape
View answer and explanation
Correct option: A
18\frac{1}{8}
Option A (18\frac{1}{8}) is correct because the rate of effusion is inversely proportional to the square root of the molar mass, and the amount escaped is proportional to the rate.
227 Hard
227 of 394
Application +4 | -1
One mole of N2\mathrm{N}_2 and 3 moles of H2\mathrm{H}_2 are taken in a container of capacity 8.21 lit. at 300 K300 \mathrm{~K} to produce NH3.\mathrm{NH}_3. Find the partial pressures of N2\mathrm{N}_2 and H2\mathrm{H}_2 if partial pressure of NH3\mathrm{NH}_3 after sufficient time was found to be 3 atm3 \mathrm{~atm}.
View answer and explanation
Correct option: A
1.5 atm;4.5 atm1.5 \mathrm{~atm}; 4.5 \mathrm{~atm}
Option A (1.5 atm;4.5 atm1.5 \mathrm{~atm}; 4.5 \mathrm{~atm}) is correct because it reflects the partial pressures of N2 and H2 after accounting for the consumption of reactants to form NH3.
228 Medium
228 of 394
Remembering +4 | -1
At critical state, the compressibility factor (Z)(Z) for a real gas is equal to
View answer and explanation
Correct option: A
38\frac{3}{8}
Option A is correct. The compressibility factor (Z) for a real gas at the critical state is equal to 3/8.
229 Hard
229 of 394
Application +4 | -1
A flask containing air at 107∘C107^{\circ} \mathrm{C} and 722 mm722 \mathrm{~mm} of Hg\mathrm{Hg} is cooled to 100 K100 \mathrm{~K} and 760 mm760 \mathrm{~mm} of Hg\mathrm{Hg}. If density in the initial condition 1 g/cm31 \mathrm{~g} / \mathrm{cm}^3, then what is the final density (g/cm3)\left(\mathrm{g} / \mathrm{cm}^3\right) ?
View answer and explanation
Correct option: A
4
Option A (4) is correct because the density is directly proportional to pressure and inversely proportional to temperature. Using the given initial and final conditions, we can calculate the final density.
230 Easy
230 of 394
Remembering +4 | -1
The ratio of most probable velocity, average velocity, root mean square velocity is
View answer and explanation
Correct option: A
2:8/π:3\sqrt 2 :\sqrt {8/\pi } :\sqrt 3
Option A (2:8/π:3\sqrt 2 : \sqrt {8/\pi} : \sqrt 3) is correct because it represents the correct ratio of most probable velocity, average velocity, and root mean square velocity.
231 Easy
231 of 394
Understanding +4 | -1
Ideal gas equation in terms of KE per unit Volume, E is
View answer and explanation
Correct option: B
23  E\frac{2}{3}\,\,E
Option B is correct because the ideal gas equation can be expressed in terms of kinetic energy per unit volume as P = (2/3)E.
232 Medium
232 of 394
Remembering +4 | -1
Which one, among the following, is the van der Waals’ equation, describing the behaviour of one mole of a real gas over wide ranges of temperature and pressure?
View answer and explanation
Correct option: A
(p+aV2)(V−b)=RT\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT
Option A is correct because it represents the van der Waals equation for one mole of a real gas, accounting for both pressure and volume corrections.
233 Medium
233 of 394
Understanding +4 | -1
Dalton’s law of partial pressure is not applicable to
View answer and explanation
Correct option: B
H2{H_2} and Cl2C{l_2} mixture
Option B (H2{H_2} and Cl2C{l_2} mixture) is correct because Dalton's law of partial pressures is not applicable when gases react chemically with each other, and hydrogen and chlorine react to form hydrogen chloride (HCl).
234 Easy
234 of 394
Understanding +4 | -1
A gas will approach ideal behaviour at
View answer and explanation
Correct option: C
High temperature and low pressure
Option C is correct because gases behave ideally at high temperatures and low pressures, minimizing intermolecular forces.
235 Easy
235 of 394
Understanding +4 | -1
Solid CH4\mathrm{C}{\mathrm{H}_4} is:
View answer and explanation
Correct option: A
Molecular solid
Option A (Molecular solid) is correct because solid CH4 consists of individual CH4 molecules held together by weak intermolecular forces (London dispersion forces).
236 Easy
236 of 394
Understanding +4 | -1
The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends on
View answer and explanation
Correct option: C
Polarisibility of interacting particles
Option C (Polarisibility of interacting particles) is correct because the magnitude of London dispersion forces depends on the polarizability of the interacting particles; more polarizable particles experience stronger London forces.
237 Medium
237 of 394
Application +4 | -1
The density of N2{N_2} gas at 0.821 atmospheric pressure and 70C{7^0}C
View answer and explanation
Correct option: C
1gL−11\mathrm{ }g\mathrm{ }{L^{ - 1}}
Option C (1gL−11\mathrm{ }g\mathrm{ }{L^{ - 1}}) is correct because the density is calculated using the ideal gas law and the molar mass of nitrogen.
238 Hard
238 of 394
Understanding +4 | -1
If the pressure at the triple point of a substance is greater than 1  atm1\,\,atm, we expect:
View answer and explanation
Correct option: C
The solid sublimes without melting
Option C (The solid sublimes without melting) is correct because if the triple point pressure is above 1 atm, the solid will directly convert to gas (sublime) at 1 atm without passing through the liquid phase.
239 Medium
239 of 394
Remembering +4 | -1
In case of hydrogen and helium the van der Waals’ forces are:
View answer and explanation
Correct option: C
Weak
Option C (Weak) is correct because hydrogen and helium are small, nonpolar molecules with few electrons, resulting in weak Van der Waals' forces.
240 Medium
240 of 394
Understanding +4 | -1
Maximum deviation from ideal gas is expected for
View answer and explanation
Correct option: A
NH3(g)N{H_3}\left( g \right)
Option A (NH3(g)N{H_3}\left( g \right)) is correct because NH3NH_3 exhibits the strongest intermolecular forces (hydrogen bonding) among the given options, leading to the greatest deviation from ideal gas behavior.
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