Properties of Gases MCQ Quiz - Objective Question with Answer for Properties of Gases - Download Free PDF

Last updated on Jun 4, 2025

Latest Properties of Gases MCQ Objective Questions

Properties of Gases Question 1:

In metallurgical and materials engineering systems, the Gibbs Phase Rule is given by : [pressure is maintained at one atmosphere]

[where F= Number of degrees of freedom, c = Number of components, p = Number of phases which can coexist at equilibrium]

  1. F = C +1 - P
  2. F = C +1 +P
  3. F = C - 1 + P
  4. F = C - 1 - P

Answer (Detailed Solution Below)

Option 1 : F = C +1 - P

Properties of Gases Question 1 Detailed Solution

Explanation:

Gibbs Phase Rule

Definition: The Gibbs Phase Rule is a fundamental principle in thermodynamics that defines the relationship between the number of phases, components, and degrees of freedom in a system at equilibrium. It is given by:

F = C + 1 - P

Where:

  • F = Number of degrees of freedom (the number of variables that can be independently changed without disturbing the equilibrium).
  • C = Number of components (chemically distinct species).
  • P = Number of phases (distinct physical states such as solid, liquid, or gas that coexist at equilibrium).

In metallurgical and materials engineering systems, this rule helps analyze and predict phase behavior under equilibrium conditions, especially when pressure is maintained constant at 1 atmosphere.

F = C + 1 - P

  • This equation is derived from the thermodynamic principles of phase equilibrium. The "1" in the formula accounts for the fixed pressure condition (1 atmosphere), reducing the number of variables required to describe the system. The degrees of freedom (F) represent the number of intensive variables (e.g., temperature, composition) that can be adjusted independently while maintaining equilibrium.

1. Understanding Components (C):

A component is a chemically distinct entity in a system. For example, in a binary alloy system consisting of copper (Cu) and nickel (Ni), there are two components.

In the context of metallurgical and materials engineering, components are often the elements or compounds that form the basis of the material. For example:

  • In a steel system, the components might be iron (Fe) and carbon (C).

  • In a ceramic system, the components could be alumina (Al₂O₃) and silica (SiO₂).

2. Understanding Phases (P):

A phase is a region of matter that is homogeneous in composition and physical properties. Common phases include solid, liquid, and gas. For example:

  • A single-phase system might consist of liquid water.

  • A two-phase system could include liquid water and ice (solid phase).

  • A three-phase system might consist of water in solid, liquid, and vapor forms.

3. Understanding Degrees of Freedom (F):

Degrees of freedom refer to the number of variables (e.g., temperature, pressure, composition) that can be independently changed while maintaining equilibrium. In a metallurgical system where pressure is fixed at 1 atmosphere, the degrees of freedom are reduced, simplifying the analysis.

Example: Consider a binary alloy system (C = 2) with two phases (P = 2). Using the Gibbs Phase Rule:

F = C + 1 - P

F = 2 + 1 - 2 = 1

This means that one variable (e.g., temperature or composition) can be independently adjusted while maintaining equilibrium between the two phases.

4. Application of Gibbs Phase Rule:

The Gibbs Phase Rule is particularly useful in metallurgical and materials engineering for analyzing phase diagrams, predicting phase stability, and understanding the behavior of multi-component systems. Examples include:

  • Determining the number of phases that can coexist in an alloy system.

  • Analyzing equilibrium conditions in ceramic or polymer systems.

  • Studying phase transformations, such as melting, solidification, or vaporization.

Properties of Gases Question 2:

Joule-Kelvin coefficient is given by [where T = absolute temperature, P = Pressure, s = Specific entropy, h = Specific enthalpy]

Answer (Detailed Solution Below)

Option 3 :

Properties of Gases Question 2 Detailed Solution

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

As a measure of the change in temperature which results from a drop in pressure across the construction.

  • For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
  • If μ is +ve, then the temperature will fall during throttling.
  • If μ is -ve, then the temperature will rise during throttling.

Properties of Gases Question 3:

A vessel of capacity 3 m3 contains 1 kg mole of Nitrogen at 90°C. Characteristic gas constant, R for nitrogen (molecular weight, M 28) is 296.9 J/ kg K. The pressure and specific volume of the gas will be:

  1. 13.26 bar, 0.178 m3/kg.
  2. 9.06 bar, 0.133 m3/kg.
  3. 8.12 bar, 0.167 m3/kg.
  4. 10.06 bar, 0.107 m3/kg.

Answer (Detailed Solution Below)

Option 4 : 10.06 bar, 0.107 m3/kg.

Properties of Gases Question 3 Detailed Solution

Concept:

We use the ideal gas law to determine the pressure and specific volume of nitrogen gas under given conditions.

Given:

  • Volume of vessel,
  • Amount of gas,
  • Temperature,
  • Characteristic gas constant for nitrogen,
  • Molecular weight of nitrogen,

Step 1: Calculate the universal gas constant (Ru)

The universal gas constant is related to the characteristic gas constant by:

Step 2: Determine the pressure using ideal gas law

The ideal gas law is given by:

Substituting the values:

Step 3: Calculate the specific volume

Specific volume () is the volume per unit mass:

Mass of nitrogen:

Thus:

 

Properties of Gases Question 4:

According to Dalton's law, the total pressure of the mixture of gases is equal to

  1. Greater of the partial pressures of all
  2. Average of the partial pressures of all
  3. Sum of the partial pressures of all
  4. Sum of the partial pressures of all divided by average molecular weight

Answer (Detailed Solution Below)

Option 3 : Sum of the partial pressures of all

Properties of Gases Question 4 Detailed Solution

Explanation:- 

  • Dalton's law of partial pressures:-   This law states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

At constant V, T, the total pressure of a gas is given by,

total = p1+p2+p3+p4+ ..........+ pn

Additional Information

  •  Raoult's Law: It states that for any solution the partial vapour pressure of the volatile component in the solution is directly proportional to its mole fraction.
  • Charles law:- It states that at constant pressure, the volume of fixed mass of gas is directly proportional to its absolute temperature.
  • Boyle's law:- At constant temperature, the volume of a gas is inversely proportional to its pressure.
  • Avogadro's law:-  It states that equal volume of all gases under the same conditions of temperature and pressure contains equal number of moles.

Properties of Gases Question 5:

An ideal gas at 27°C is heated at constant pressure till the volume becomes three times. The temperature of the gas will then be: 

  1. 81°C
  2. 627°C
  3. 900°C
  4. 927°C

Answer (Detailed Solution Below)

Option 2 : 627°C

Properties of Gases Question 5 Detailed Solution

Concept:

Ideal gas equation: PV = n R̅ T or PV = mRT

Where, P is absolute pressure in Pa, V is volume in m3, n is number of mole, R̅ is universal gas constant in kJ/kgmole-K and T is absolute temperature in K, m is mass in kg and R is universal gas constant in kJ/kg-K.

For constant pressure process P1 = P2

Calculation: 

Given, P1 = P2,T1 = 27°C = 300 K and V2 = 3V1

for constant pressure process 

⇒ T2 = 3 × 300 = 900 K 

⇒ T2 = 900 - 273 = 627°C

Top Properties of Gases MCQ Objective Questions

According to kinetic theory of gases, the absolute zero temperature is attained when

  1. Volume of the gas is zero
  2. Pressure of the gas is zero
  3. Kinetic energy of the molecules is zero 
  4. Specific heat of the gas is zero

Answer (Detailed Solution Below)

Option 3 : Kinetic energy of the molecules is zero 

Properties of Gases Question 6 Detailed Solution

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Explanation:
According to the kinetic theory of gases, the
average kinetic energy of translation per molecule of a gas is directly proportional to the absolute temperature of the gas.

The pressure exerted by an ideal gas is given by:

where C is the root mean square speed of molecules of the gas.

   {∵ PV = RT}

The square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecule.

Now, at C = 0, T = 0

  • Absolute zero of temperature is defined as that temperature at which the root mean square velocity of the gas molecules reduced to zero.
  • It means molecular motion ceases at absolute zero i.e. Kinetic energy of the molecules becomes zero at absolute zero temperature.

Zeroth law of thermodynamics is not valid for the following:

  1. 50 ml of water at 25°C is mixed with 150 ml of water at 25°C
  2. 500 ml milk at 15°C is mixed with 100 ml of water at 15°C
  3. 5 kg of wet steam at 100°C is mixed with 50 kg of dry and saturated steam.
  4. 10 ml of water at 20°C is mixed with 10 ml of sulphuric acid at 20°C

Answer (Detailed Solution Below)

Option 4 : 10 ml of water at 20°C is mixed with 10 ml of sulphuric acid at 20°C

Properties of Gases Question 7 Detailed Solution

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Explanation:

According to Zeroth Law, if system A is in thermal equilibrium with system C, and system B is thermal equilibrium with systems C, then system A is in thermal equilibrium with system B.

Now, two systems are said to be in (mutual) thermal equilibrium if, when they are placed in thermal contact (basically, contact that permits the exchange of energy between them), their state variables do not change.

In case of mixing of water and sulphuric acid, the enormous amount of heat is released as mixing is highly exothermic. So there is no more any thermal equilibrium. So Zeroth Law is not valid.

A perfect gas at 27°C is heated at constant pressure till its volume is double. The final temperature is

  1. 54°C
  2. 108°C
  3. 327°C
  4. 600°C

Answer (Detailed Solution Below)

Option 3 : 327°C

Properties of Gases Question 8 Detailed Solution

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Concept:

As the gas is heated at constant pressure, so by applying Charles’s law

Where V & T are volume and temperature respectively.

Calculation:

Given:

T1 = 27°C = 300 K, V1 = V, V2 = 2V

Therefore, 

T2 = 600 K = 600 - 273

= 327°C

When air expands from initial pressure P1 and volume V1 to final volume 5 V1 following the law PVn = C

  1. Greater the value of n, greater the work obtained
  2. Smaller the value of n, smaller the work obtained
  3. For n = 0, the work obtained is the greatest
  4. For n = 1.4, the work obtained is the greatest

Answer (Detailed Solution Below)

Option 3 : For n = 0, the work obtained is the greatest

Properties of Gases Question 9 Detailed Solution

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Concept:

Work obtained in the P – V diagram is the area under the curve.

As clearly seen that Larger the value of n, smaller is the area and so smaller is work obtained. In other words

So, W is maximum, when n = 0 i.e. constant pressure process

W is always zero for n = ∞ i.e. constant volume process

Generalized compressibility chart is drawn between:

  1. Compressibility factor(Z) on x-axis and reduced pressure(Pr) on y-axis
  2. Compressibility factor(Z) on y-axis and reduced temperature (Tr) on x-axis
  3. Reduced pressure(Pr) on y-axis and reduced temperature (Tr) on x-axis
  4. Compressibility factor(Z) on y-axis and reduced pressure(Pr) on x-axis

Answer (Detailed Solution Below)

Option 4 : Compressibility factor(Z) on y-axis and reduced pressure(Pr) on x-axis

Properties of Gases Question 10 Detailed Solution

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Explanation:

The compressibility factor is used to quantify the deviation of real behaviour from the ideal gas behaviour.

The compressibility factor Z is defined as the ratio of the actual volume to the volume predicted by the ideal gas law at a given temperature and pressure.

Z = (Actual volume) / (volume predicted by the ideal gas law)

For an ideal gas, Z = 1 at all temperatures and pressures.

Gasses behave differently at a given temperature and pressure but they behave very much the same at temperatures and pressure normalized to their critical temperatures and pressure.

  and

The experimentally determined Z values are plotted against PR and TR for several gasses. By curve fitting, all the data compressibility chart is obtained. In which Compressibility factor Z is on the y-axis and Reduced pressure PR is on the x-axis.

 

The observation made from the generalized compressibility chart is that at very low-pressure gasses behave as an ideal gas regardless of the temperature.

If the value of n = 0 in the equation PV= C, then the process is called

  1. Constant volume process
  2. Adiabatic process
  3. Constant pressure process
  4. Isothermal process

Answer (Detailed Solution Below)

Option 3 : Constant pressure process

Properties of Gases Question 11 Detailed Solution

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Explanation:

Polytropic Process is represented by

PVn = C

  • n = 0 ⇒ P = C ⇒ Constant Pressure Process (Isobaric Process)
  • n = 1 ⇒ PV = C ⇒ Constant Temperature Process (Isothermal process)
  • n = γ ⇒ PVγ = C ⇒ Adiabatic Process
  • n = ∞ ⇒ V = C ⇒ Constant Volume Process (Isochoric process)

Important point:

  • For n = 0, it is a constant pressure process and it is a horizontal line.
  • For n = ∞, it is a constant volume process and it is a vertical line.
  • So, as the value of n increases, the process line will come closer to the y-axis.

Which of the following property is used as thermometric property in constant – volume gas thermometer?

  1. Volume
  2. Pressure 
  3. Electric Resistance
  4. Voltage

Answer (Detailed Solution Below)

Option 2 : Pressure 

Properties of Gases Question 12 Detailed Solution

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Explanation:

  • In order to obtain a quantitative measure of temperature, a reference body is used, and a certain physical characteristic of this body which changes with temperature is selected.
  • The changes in the selected characteristics may be taken as an indication of the change in temperature.
  • The selected characteristic is called the thermometric property, and the reference body which is used in the determination of temperature is called a thermometer.
  • A constant-volume gas thermometer uses the thermometric property pressure for the measurement of temperature.
  • The volume of gas is set constant by raising or lowering a mercury reservoir while temperature changes.
  • The pressure is indicated by the level difference in mercury in the tube.

Various types of thermometers and their thermometric property are mentioned in the table below.

Sr. No

Thermometer

Thermometric property

Symbol

1.

Constant volume gas thermometer

Pressure

P

2.

Constant pressure gas thermometer

Volume

V

3.

Electrical resistance thermometer

Resistance

R

4.

Thermocouple

Thermal e.m.f.

ϵ

5.

Mercury in a gas thermometer

Length

L

The Vander Waals  equation of state is  where p is pressure, v is specific volume, T is temperature and R is characteristic gas constant. The SI unit of a is

  1. J/kg. K
  2. m3/kg
  3. m5/kgs2
  4. Pa/kg

Answer (Detailed Solution Below)

Option 3 : m5/kgs2

Properties of Gases Question 13 Detailed Solution

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Explanation:

Vander-Waals equation

  • The equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. Ideal gas equation fails to explain the behaviour of real gases. Therefore, the Van der Waals equation was devised and it helps us define the physical state of a real gas.
  • Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation is written as

  • The constant "a" provides a correction for the intermolecular forces.
  • Constant b adjusts for the volume occupied by the gas particles. It is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules.

here, p is pressure in N/m2, v is the specific volume in m3/kg, R is characteristic gas constant in J/kg-K,  is called the force of cohesion and b is called the co-volume.

() both terms should give the same unit since they are getting added

i.e. p =  in terms of unit.

The pressure p of an ideal gas and its mean kinetic energy E per unit volume are related by the relation

Answer (Detailed Solution Below)

Option 3 :

Properties of Gases Question 14 Detailed Solution

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Concept:

The pressure in a gas developed due to the collisions between the gaseous molecules.

The energy possessed by the molecule of the gas due to its motion is called kinetic energy of the gas molecules.

From the kinetic theory of gases, the pressure (P) exerted by an ideal gas is given by

where ​ = the density of ideal gas, C = its root mean square velocity.

We know,

So, for a unit volume of gas, mass= ρ × 1 = ρ 

Now, the mean kinetic energy of translation per unit volume of gas (KE) = 

Alternate Method:

From Kinetic Theory of Gases.

From the Ideal Gas Equation

PV = RT (for 1 mole)

Substituting the value of RT from the above equations in the equation of the kinetic theory of gases we get,

A perfect gas is heated at constant pressure. The final volume of the gas becomes 1.5 times the initial volume. If its initial temperature is 30°C, the final temperature will be:

  1. 45°C
  2. 20°C
  3. 181.5°C
  4. 330°C

Answer (Detailed Solution Below)

Option 3 : 181.5°C

Properties of Gases Question 15 Detailed Solution

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Concept:

Gases that obey the gas laws (Charles law, Boyles law and Universal Gas Law) are called ideal gases.

Boyle’s, Charles’, and Gay Lussac's Laws describe the basic behaviour of fluids with respect to volume, pressure, and temperature.

Gay Lussac’s Law

It states that at constant volume, the pressure of a fixed amount of a gas varies directly with temperature.

P ∝ T

Boyle's Law

For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.

PV = constant (If the temperature remains constant, the product of pressure and volume of a given mass of a gas is constant.)

Charle's Law

For a fixed mass of gas at constant pressure, the volume is directly proportional to the Kelvin temperature.

Calculation:

Given:

V2 = 1.5 V1 , T1 = 30° C = 303 K and P = Const.

from Charle's Law 

⇒ 

⇒ 

⇒ T2 = 1.5 × 303 = 454.5 K

⇒ T2 = 181.5° C

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