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# Test: Gaseous State - 1

## 20 Questions MCQ Test Physical Chemistry | Test: Gaseous State - 1

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This mock test of Test: Gaseous State - 1 for Chemistry helps you for every Chemistry entrance exam. This contains 20 Multiple Choice Questions for Chemistry Test: Gaseous State - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Gaseous State - 1 quiz give you a good mix of easy questions and tough questions. Chemistry students definitely take this Test: Gaseous State - 1 exercise for a better result in the exam. You can find other Test: Gaseous State - 1 extra questions, long questions & short questions for Chemistry on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

### The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is:

Solution:
• 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
QUESTION: 3

### 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:

Solution:

Let the mass of methane and oxygen be m gm.

Mole fraction of oxygen, xO2 = m/32/(m/32 +m/16) = 1/3

Let the total pressure be P.

Partial pressure of O2, pO2 = P × xO2 = P x 1/3 = P/3

QUESTION: 4

Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is:

Solution:

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

QUESTION: 5

No cooling occurs, when an ideal gas undergoes unrestrained expansion, because the molecules:

Solution:

Ideal gas has no attractive force between the particles

QUESTION: 6

A liquid is in equilibrium with its vapour at it’s boiling point. On the average, the molecules in the two phases have equal:

Solution:

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.

QUESTION: 7

Rate of diffusion of a gas is:

Solution:

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

QUESTION: 8

The average velocity of an ideal gas molecule  at 27°C is 0.3 m/s. The average velocity at 927°C will be

Solution:

QUESTION: 9

In van der Waals’ equation of state for a non-ideal gas, the term that accounts for intermolecular forces is:

Solution:

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.

QUESTION: 10

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:

Solution:
• 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
QUESTION: 11

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:

Solution:
• 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.
QUESTION: 12

The density of neon will be highest at:

Solution:
• 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.
QUESTION: 13

The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is:

Solution:

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 M​= 64.0  g/mol

QUESTION: 14

According to kinetics theory of gases, for a diatomic molecule:

Solution:

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

QUESTION: 15

At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise of temperature due to:

Solution:
QUESTION: 16

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:

Solution:

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

QUESTION: 17

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

Solution:

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

QUESTION: 18

The compressibility factor for an ideal gas is:

Solution:

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.

QUESTION: 19

The critical temperature of water is higher than that of O2 because the H2O molecule has:

Solution:
QUESTION: 20

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)

Solution:

Hence C is correct.