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Chemistry Exam  >  Chemistry Test  >  Physical   >  Test: Gaseous State - 1 - Chemistry MCQ

Gaseous State - 1 - Chemistry Physical Free MCQ Test with solutions


MCQ Practice Test & Solutions: Test: Gaseous State - 1 (20 Questions)

You can prepare effectively for Chemistry Physical Chemistry with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Gaseous State - 1". These 20 questions have been designed by the experts with the latest curriculum of Chemistry 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 40 minutes
  • - Number of Questions: 20

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Test: Gaseous State - 1 - Question 1
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The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is:

Detailed Solution: Question 1

Test: Gaseous State - 1 - Question 2
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The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is:

Detailed Solution: Question 2

  • The temperature at which the real or non-ideal gas behaves as an ideal gas over a wide range of pressure is known as Boyle temperature.
  • Boyle temperature (TB) is related to the Vander Waal’s constant a, b as given below. At this temperature, the attractive and repulsive forces acting on the gas particles arrive at a balance for a real gas.
  • TB = a/Rb

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Test: Gaseous State - 1 - Question 3
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Equal weights of methane and oxygen and mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is:

Detailed Solution: Question 3

Equal masses of methane and oxygen are mixed in an empty container at 25oC. The fraction of the total pressure exerted by oxygen is 1/3. 

  • Let 32 g of oxygen are mixed with 32 g of methane.
  • The molar masses of methane and oxygen are 16 g/mol and 32 g/mole respectively.
  • The number of moles of methane = 32g/16g ​=2 moles.
  • The number of moles of oxygen = 32g/32g​ =1 mole.

  • Total number of moles =2+1=3. 

  • The mole fraction of oxygen  =1/1+2​= 1/3

  • The mole fraction is proportional to the fraction of the total pressure exerted.

  • Hence, the fraction of the total pressure exerted by oxygen is 1/3

Test: Gaseous State - 1 - Question 4
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Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is:

Detailed Solution: Question 4

Kinetic energy of gas molecule depends on temperature and does not depend upon the nature of gas.

K.E =(3/2)kT.

Thus hydrogen and helium have same K.E at same temperature

Test: Gaseous State - 1 - Question 5
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No cooling occurs, when an ideal gas undergoes unrestrained expansion, because the molecules:

Detailed Solution: Question 5

Ideal gas has no attractive force between the particles

Test: Gaseous State - 1 - Question 6
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A liquid is in equilibrium with its vapour at it’s boiling point. On the average, the molecules in the two phases have equal:

Detailed Solution: Question 6

When a liquid is in equilibrium with its vapout at its boiling point, the molecules in the two phases have equal kinetic energy.

At equilibrium, the rate of molecules going from the liquid to the gaseous phase equal to the rate of molecules going from the gaseous phase to the liquid surface and hence they have equal kinetic energy.

Test: Gaseous State - 1 - Question 7
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Rate of diffusion of a gas is:

Detailed Solution: Question 7

Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. 

Rate of Diffusion is proportional to 1/ √M

Test: Gaseous State - 1 - Question 8
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The average velocity of an ideal gas molecule  at 27°C is 0.3 m/s. The average velocity at 927°C will be

Detailed Solution: Question 8

Test: Gaseous State - 1 - Question 9
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In van der Waals’ equation of state for a non-ideal gas, the term that accounts for intermolecular forces is:

Detailed Solution: Question 9

In van der Waals’ equation of state:
(p + a/V2) (V - b) = RT (for 1 mole)

The first factor (p + a/V2) correct for intermolecular force while the second term (V - b) correct for molecular volume.

Test: Gaseous State - 1 - Question 10
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A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends the white ammonium chloride ring first formed will be:

Detailed Solution: Question 10

  • According to Graham's law of effusion:
    Rate of effusion = √(1/Molar mass)
    i.e. Gas of higher molar mass will effuse slowly, therefore, move a shorter distance.
  • NH3 molar mass is 17 and HCl molar mass is 36.5. So, HCl effuses slowly and the formation of NH4Cl white rings are formed near the HCl

Test: Gaseous State - 1 - Question 11
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The value of van der Waals’ constant ‘a’ for the gases O2, N2, NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 L2 atm mol–2 respectively. The gas which can most easily be liquefied is:

Detailed Solution: Question 11

  • The ease of liquification of a gas depends on their intermolecular force of attraction which in turn is measured in terms of van der Waals’ constant ‘a’.
  • Hence, the higher the value of ‘a’, the greater the intermolecular force of attraction, the easier the liquification.
  • In the present case, NH3 has the highest ‘a’, can most easily be liquefied.

Test: Gaseous State - 1 - Question 12
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The density of neon will be highest at:

Detailed Solution: Question 12

  • According to the ideal gas equation:
    P = dRT/M i.e. d = PM/RT
  • As per the above equation at high pressure and low-temperature, density is maximum.

Test: Gaseous State - 1 - Question 13
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The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is:

Detailed Solution: Question 13

The rate of diffusion of methane at a given temperature is twice that of gas X i.e. rCH4​ ​= 2rX​
∵ The rate of diffusion is inversely proportional to the square root of its molar mass.
Therefore,

Thus, molecular mass of gas X is MX ​= 64.0  g/mol

Test: Gaseous State - 1 - Question 14
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According to kinetics theory of gases, for a diatomic molecule:

Detailed Solution: Question 14

As we know: KE = (3/2) ​RT
⇒ KE ∝ T
i.e. mean transitional kinetic energy of the molecules is proportional to the absolute temperature.

Test: Gaseous State - 1 - Question 15
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At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise of temperature due to:

Detailed Solution: Question 15

Explanation:
Introduction:
When the volume of a gas is kept constant and the number of moles of the gas is fixed, the pressure of the gas increases with the rise in temperature. This is known as Gay-Lussac's law or the pressure-temperature law.
Reason:
The increase in pressure of the gas with a rise in temperature is due to the following factors:
1. Increase in average molecular speed:
- As the temperature increases, the average kinetic energy of the gas molecules also increases.
- The kinetic energy of gas molecules is directly proportional to the square of their speed.
- Therefore, an increase in temperature leads to an increase in the average molecular speed of the gas.
- This increase in molecular speed results in an increase in the frequency and force of molecular collisions with the walls of the container, which in turn increases the pressure of the gas.
2. Increase in rate of collisions amongst molecules:
- With an increase in temperature, the rate of collisions amongst gas molecules also increases.
- This is because the increased kinetic energy of the molecules leads to a higher probability of collisions.
- The increased rate of collisions leads to an increase in the pressure of the gas.
3. Increase in molecular attraction:
- The increase in temperature weakens the intermolecular forces of attraction between gas molecules.
- This weakening of molecular attraction allows the gas molecules to move more freely and collide with the container walls more frequently, resulting in an increase in pressure.
4. Decrease in mean free path:
- The mean free path is the average distance a gas molecule travels between collisions with other molecules.
- As the temperature increases, the molecules move faster and collide more frequently, resulting in a decrease in the mean free path.
- This decrease in mean free path leads to an increase in the frequency of collisions with the container walls, resulting in an increase in pressure.
Conclusion:
In conclusion, the pressure of a gas increases with the rise in temperature at constant volume and fixed number of moles of the gas due to an increase in average molecular speed, increase in the rate of collisions amongst molecules, increase in molecular attraction, and a decrease in the mean free path.

Test: Gaseous State - 1 - Question 16
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Equal weights of ethane and hydrogen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by hydrogen is:

Detailed Solution: Question 16


Partial pressure of hydrogen/total pressure = mole fraction of hydrogen =15:16

Test: Gaseous State - 1 - Question 17
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The ratio between the root mean square speed of H2 at 50 K and that of O2 at 800 K is:

Detailed Solution: Question 17

The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molecular weight.

The ratio between the root mean square velocity of H2​ at 50K and that of O2​ at 800K is

Test: Gaseous State - 1 - Question 18
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The compressibility factor for an ideal gas is:

Detailed Solution: Question 18

The deviation of ideal behaviour is introduced by compressibility factor Z.
i.e. Z = PV/nRT
for ideal gas PV = nRT, therefore, Z is 1.

Test: Gaseous State - 1 - Question 19
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The critical temperature of water is higher than that of O2 because the H2O molecule has:

Detailed Solution: Question 19

Explanation:
The critical temperature of a substance is the temperature at which a gas cannot be liquefied, regardless of the pressure applied. In the case of water (H2O) and oxygen (O2), the critical temperature of water is higher than that of O2. This can be explained by considering the molecular structure and properties of water.
Reasons:
1. Dipole moment: The water molecule (H2O) has a bent or V-shaped structure, with two hydrogen atoms bonded to an oxygen atom. This bent shape creates a permanent dipole moment in the molecule, where the oxygen atom carries a partial negative charge (δ-) and the hydrogen atoms carry partial positive charges (δ+). This dipole moment results in strong intermolecular forces of attraction between water molecules, known as hydrogen bonding. These hydrogen bonds require more energy to break, leading to a higher critical temperature.
2. Hydrogen bonding: The presence of hydrogen bonding in water molecules contributes to its higher critical temperature. Hydrogen bonds are stronger than the intermolecular forces in oxygen molecules, which only have London dispersion forces. These weaker forces in oxygen molecules require less energy to break, resulting in a lower critical temperature.
3. Electron density: Although the oxygen atom in water has more electrons compared to oxygen molecules, the electron density in water is more spread out due to the presence of hydrogen atoms. This reduced electron density in water molecules weakens the London dispersion forces, making them less significant in determining the critical temperature.
Therefore, the correct answer is D: Dipole moment. The dipole moment created by the V-shaped structure of water molecules and the resulting hydrogen bonding contribute to the higher critical temperature of water compared to oxygen.

Test: Gaseous State - 1 - Question 20
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According to Graham’s law, at a given temperature the ratio of the rates of diffusionof gases A and B is given by (where, p and M are pressure and molecular weights of gases A and B respectively)

Detailed Solution: Question 20

Hence C is correct.

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This test contains 20 MCQs on Gaseous State - 1 with detailed solutions for each question. It is part of the Physical Chemistry course on EduRev, designed to align with the current Chemistry syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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