Properties of Gases MCQ Quiz in मल्याळम - Objective Question with Answer for Properties of Gases - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 15, 2025

നേടുക Properties of Gases ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Properties of Gases MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Properties of Gases MCQ Objective Questions

Top Properties of Gases MCQ Objective Questions

Properties of Gases Question 1:

Two vapour power cycles having efficiencies of 0.5 and 0.6 are coupled in series then the efficiency of coupled cycle will be ____

  1. 0.6
  2. 0.8
  3. 0.3
  4. 0.7

Answer (Detailed Solution Below)

Option 2 : 0.8

Properties of Gases Question 1 Detailed Solution

For two coupled cycle in series

\(\begin{array}{l} \eta = 1 - \left( {1 - {\eta _1}} \right)\left( {1 - {\eta _2}} \right)\\ = 1 - \left[ {1 - {\eta _1} - {\eta _2} + {\eta _1} \cdot {\eta _2}} \right]\\ = {\eta _1} + {\eta _2} - {\eta _1} \cdot {\eta _2} \end{array}\)

= 0.5 + 0.6 – 0.5 × 0.6 = 0.80

Properties of Gases Question 2:

Dalton's law states that the total pressure of the mixture of gases is equal to -

  1. Sum of the partial pressures of all multiplied by the average atomic weight
  2. Sum of the partial pressures of all the gases
  3. Average of the partial pressures of all the gases
  4. Product of the partial pressures of all the gases

Answer (Detailed Solution Below)

Option 2 : Sum of the partial pressures of all the gases

Properties of Gases Question 2 Detailed Solution

Explanation:

Dalton’s law of partial pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

Suppose there are n1 moles of gas A1, n2 moles of gas A2, n3 moles of gas A3,… and up to nk moles of gas Ak, are there in a homogeneous mixture of inert ideal gases at a temperature T, a pressure P, and a volume V.

Since there is no chemical reaction, the mixture is in a state of equilibrium with the equation of state

pV = (n1 + n2 + n+ …+ nk) R̅ .T Where, R̅ is universal gas constant.

and R̅ = 8.3143 kJ / kg-mol-K

\(p = \frac{{{n_1}\bar R\ T}}{V} + \;\frac{{{n_2}\bar R\;T}}{V} + \frac{{{n_3}\bar R\;T}}{V} + \ldots + \frac{{{n_k}\bar R\;T}}{V}\)

As \({p_1} = \;\frac{{{n_1}\bar R\;T}}{V},\;{p_2} = \frac{{{n_2}\bar R\;T}}{V},\;{p_3} = \frac{{{n_3}\bar R\;T}}{V}, \ldots {p_k} = \frac{{{n_k}\bar R\;T}}{V}\)

Thus, p = p1 + p2 + p3 +…+ pk

Properties of Gases Question 3:

The general law of expansion or compression is PVn = C, The process is said to be hyperbolic, if n is equal to

  1. ∞ 
  2. 1
  3. 0
  4. Y

Answer (Detailed Solution Below)

Option 2 : 1

Properties of Gases Question 3 Detailed Solution

Explanation:

Polytropic Process is represented by:

PVn = C

where P is pressure, V is volume, and n is the polytropic index.

Equation value of n  Type of process Type of curve
P = C n = 0

Constant Pressure Process

(Isobaric Process)

Straight horizontal line
PV = C n = 1 Constant Temperature Process (Isothermal process) Rectangular Hyperbolic Curve
PVγ = C n = γ Adiabatic Process Polynomial Curve
PV = C n = ∞

Constant Volume Process

(Isochoric process)

Straight vertical line

 RRB JE ME 46 11Q TE CH 1 Hindi Diag(Shashi) images Q6

  • The isothermal process is governed by Boyle’s law.
  • The temperature remains constant in this process. Pressure and volume are inversely proportional to each other. Hence, when P increases V decreases, and when V increases P decreases.
  • Thus the curve developed is a rectangular hyperbola. So, the graph for the Isothermal process shows a hyperbolic curve.

T = C

i.e. PV = C (from ideal gas equation)
where P = Pressure, T = Temperature, V = Volume

F1 S.C Madhu 08.06.20 D6

Properties of Gases Question 4:

The ratio of specific heat at constant pressure to the specific heat at constant volume for argon and helium is-

  1. 1.11
  2. 1.3
  3. 1.4
  4. 1.667

Answer (Detailed Solution Below)

Option 4 : 1.667

Properties of Gases Question 4 Detailed Solution

Concept:

  • The molar specific heat capacity of a gas at constant volume is defined as the amount of heat required to raise the temperature of 1 mol of the gas by 1 °C at the constant volume.

\({C_v} = {\left( {\frac{{\Delta Q}}{{n\Delta T}}} \right)_{constant\;volume}}\)

  • The molar specific heat of a gas at constant pressure is defined as the amount of heat required to raise the temperature of 1 mol of the gas by 1 °C at the constant pressure.

\({C_p} = {\left( {\frac{{\Delta Q}}{{n\Delta T}}} \right)_{constant\;pressure}}\)

  • The relation between the ratio of Cp and Cv with a degree of freedom is given by

\(\gamma = \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{f}\)

Where f = degree of freedom

EXPLANATION:

  • The relation between the ratio of Cp and Cv with a degree of freedom is given by

\(\Rightarrow \gamma = \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{f}\)

Monoatomic gas has 3 degrees of freedom

\(\Rightarrow \gamma = 1 + \frac{2}{3} = \frac{3 +2}{3} = \frac{5}{3} =1.67\)

Properties of Gases Question 5:

Critical pressure of a liquid is the pressure

  1. above which liquid will remain liquid
  2. above which liquid will become gas
  3. above which liquid becomes vapour
  4. above which liquid becomes solid

Answer (Detailed Solution Below)

Option 3 : above which liquid becomes vapour

Properties of Gases Question 5 Detailed Solution

Explanation:

Critical Pressure/Temperature: 

F1 S.S Madhu 23.11.19 D3

  • It is defined as the pressure/temperature at which liquid flashes into vapour.
  • And there does not exist any distinction between the liquid and gaseous phase.
  • Degree of freedom at critical pressure is zero.
  • At a critical point, dryness fraction is undefined.
  • Latent heat becomes zero at critical pressure.
  • For water, Pcr = 221.2 bar, Tcr = 374.15°C

Properties of Gases Question 6:

Steam which is formed in contact with a water is known as:

  1. Saturated steam
  2. Supersaturated steam
  3. Dry saturated steam
  4. Superheated steam

Answer (Detailed Solution Below)

Option 1 : Saturated steam

Properties of Gases Question 6 Detailed Solution

Explanation:

F1 S.S Madhu 13.01.20 D5

The steam may be in any one of the following three conditions:

  • Saturated steam which may be either wet or dry
  • Superheated steam
  • Supersaturated steam


​​​Saturated steam:

  • Steam which is formed in contact with water is known as saturated steam.
  • If heat is supplied to the saturated vapour and to the liquid with which it is in contact, more liquid evaporates but the temperature remains the same. Similarly, if heat is removed, more of the vapour condenses but the temperature will remain the same.
    • If saturated steam contains liquid particles, then it is known as wet steam.
    • If saturated steam does not contain any liquid particles, then it is known as dry saturated steam.


Superheated Steam:

If the temperature of the steam is greater than the boiling point temperature corresponding to the pressure of steam generation, then it is called superheated steam.

Properties of Gases Question 7:

Air with initial condition of p1, v1 expands to final condition of p1/2, 3v1. The process is

  1. Hyperbolic
  2. Adiabatic
  3. Polytropic with n > 1
  4. Polytropic with n < 1

Answer (Detailed Solution Below)

Option 4 : Polytropic with n < 1

Properties of Gases Question 7 Detailed Solution

Concept:

The initial condition is P1, V1

And the final condition is \({P_2} = \frac{{{P_1}}}{2}\), V2 = 3V1

Hyperbolic/Isothermal Process:

T = C i.e. PV = C 

Here: \(P_2V_2=\frac{P_1}{2}(3V_1)\ne P_1V_1\)

Polytropic Process:

PVn = C

\({P_1}V_1^n = {P_2}V_2^n\)

\( \Rightarrow {P_1}V_1^n = \left( {\frac{{{P_1}}}{2}} \right){\left( {3{V_1}} \right)^n }\)

\( \Rightarrow V_1^n = \frac{{{3^n }}}{2}V_1^n \)

⇒ 3n = 2

⇒ For n < 1 ⇒ n = 0.63

Adiabatic Process:

In adiabatic process, γ = 1.4

Properties of Gases Question 8:

The ratio of specific heat at constant pressure and specific heat at constant volume for a diatomic gas is

  1. 5/3
  2. 2/3
  3. 7/5
  4. 5/7

Answer (Detailed Solution Below)

Option 3 : 7/5

Properties of Gases Question 8 Detailed Solution

Concept:

  • The molar specific heat capacity of a gas at constant volume is defined as the amount of heat required to raise the temperature of 1 mol of the gas by 1 °C at the constant volume.

\({C_v} = {\left( {\frac{{\Delta Q}}{{n\Delta T}}} \right)_{constant\;volume}}\)

  • The molar specific heat of a gas at constant pressure is defined as the amount of heat required to raise the temperature of 1 mol of the gas by 1 °C at the constant pressure.

\({C_p} = {\left( {\frac{{\Delta Q}}{{n\Delta T}}} \right)_{constant\;pressure}}\)

  • The relation between the ratio of Cp and Cv with a degree of freedom is given by

\(\gamma = \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{f}\)

Where f = degree of freedom

For mono atomic gas f = 3

For Diatomic gas f = 5

Calculation:

Given:

Gas is diatomic i.e. f = 5

We know that,

\(\gamma = \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{f}\)

\(\gamma = \frac{{{C_p}}}{{{C_v}}} = 1 + \frac{2}{5}\)

\(\gamma = \frac{{{C_p}}}{{{C_v}}} = \frac{7}{5}\)

Properties of Gases Question 9:

The difference between two specific heats, \({C_p} - {C_v} = \frac{R}{J}\). This relation is valid for

  1. Any gas
  2. Perfect gases
  3. Real gases
  4. Pure gases

Answer (Detailed Solution Below)

Option 2 : Perfect gases

Properties of Gases Question 9 Detailed Solution

Concept:

A perfect gas is also an Ideal gas, which follows the Ideal gas equation of states i.e. PV = mRT all temperature.

where, P = pressure of gas, V = volume occupied, m = mass of a gas, R = universal gas constant. 

The universal gas constant (R) is the difference between specific heat constants for constant pressure (Cp) and constant volume (Cv)

i.e.R = Cp - Cv

26 June 1

A real gas behaves as an Ideal gas at low pressure and very high temperature. Air is a perfect gas.

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 behavior 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

\(\frac{P}{T} = Const\)

Boyle's Law

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

\(P\propto \frac{1}{V}\)

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

Charles' Law

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

\(V\propto T \ or, \ \frac{V}{T} = Const\)

The Combined gas law or General Gas Equation is obtained by combining Boyle's Law, Charles's law, and Gay-Lussac's Law. It shows the relationship between the pressure, volume, and temperature for a fixed mass (quantity) of gas:

\(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

\({C_P} - {C_V} = \frac{R}{J}\) is valid for a perfect gas only.

Properties of Gases Question 10:

Adiabatic index of a triatomic gas is 

  1. 5/3
  2. 7/5
  3. 4/3
  4. 6/5

Answer (Detailed Solution Below)

Option 3 : 4/3

Properties of Gases Question 10 Detailed Solution

CONCEPT:

Adiabatic index:

  • The adiabatic index is also known as the heat capacity ratio and is defined as the ratio of heat capacity at constant pressure CP to heat capacity at constant volume CV.

​It is given as,

\(\Rightarrow γ=\frac{C_P}{C_V}\)

Gas Monoatomic Diatomic Polyatomic
γ 5/3 7/5 4/3


EXPLANATION:

Gas Monoatomic Diatomic Polyatomic
γ 5/3 7/5 4/3

 

  • From the above table, it is clear that the adiabatic index for the monoatomic gas is equal to 5/3. Hence, option 3 is correct.
Get Free Access Now
Hot Links: teen patti apk teen patti club teen patti joy 51 bonus teen patti gold online mpl teen patti