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

CONCEPT:

Deviation from Ideal Gas Behavior

• In real gases, deviations from ideal gas behavior occur due to intermolecular forces and the volume occupied by gas molecules.
• The ideal gas law assumes no intermolecular interactions and that the volume of the gas molecules is negligible. However, real gases experience attractive and repulsive forces, leading to deviations from this behavior, especially at low pressures and high temperatures.
• The PV vs. P curve for a gas shows its deviation from the ideal gas law. Positive deviation occurs when the actual volume is greater than the ideal volume, while negative deviation occurs when the actual volume is smaller than the ideal volume.
• At low pressure, gases like hydrogen (H₂) and helium (He) show a positive deviation, meaning they are less compressible than ideal gases. Conversely, gases like CO₂, CH₄, and O₂ show negative deviations at low pressure due to intermolecular attractions.

EXPLANATION:

• The graph shows the behavior of various gases (H₂, He, CO₂, CH₄, O₂) at a particular temperature.
• • Carbon dioxide (CO₂), Methane (CH₄), and Oxygen (O₂) initially show a negative deviation at lower pressures and then transition to a positive deviation as pressure increases. This suggests that the attractive forces between molecules dominate at low pressures, leading to a negative deviation, but as the pressure increases, repulsive forces start to take over, leading to a positive deviation.
  Hydrogen (H₂) and Helium (He) show a positive deviation from the ideal gas law, meaning they are less compressible than an ideal gas. This indicates that the intermolecular forces are relatively weak, leading to a behavior that deviates positively from the ideal gas equation.
• Option 3 incorrectly states that H₂ and He show negative deviation, which is not supported by the graph. They actually show positive deviation. It also wrongly states that CO₂, CH₄, and O₂ show positive deviation, which is true only at higher pressures. Hence, option 3 is the incorrect statement.

Therefore, option C is the correct answer, which states that H₂ and He show positive deviation while CO₂, CH₄, and O₂ show negative deviation at low pressure.

At STP,

\(T = 273\ K\)

\(P = 1\ atm\)

\(M_{mix}=\cfrac {dRT}{P_{mix}}\).

Substitute values in the above expression:

\( M_{mix}=\cfrac {1.7\times 0.0821\times 273}{1}=1.7\times 22.4 \text{ gms}\)

The molecular weight of the mixture \(M_{mix}\) can also be calculated by using the expression,

\(M_{mix}=X_{CO}\times M_{CO}+(1-X_{CO}).M_{CO_2}\)

Here, \(X_{CO}\) and \(M_{CO}\) represents the mole fraction and molecular weight of CO and \((1-X_{CO})\) and \(M_{CO_2}\) represents the mole fraction and molecular weight of \(CO_2\).

Thus \(1.7\times 22.4=X_{CO}\times 28+(1-X_{CO})\times 44\)

\(X_{CO}=0.37\)

The mole fraction of carbon monoxide is \(0.37\).

Here, 1 represents \(SO_2\) and 2 represents \(CH_4\).

The rate of effusion is proportional to the ratio of pressure to the square root of molecular weight.

The rates of effusions of two gases are equal.

The number of moles are proportional to pressure.

Hence, \(\dfrac {p_1}{p_2}=\dfrac {\sqrt {M_1}}{\sqrt {M_2}}=\dfrac {n_1}{n_2}\)

Thus, \(\dfrac {n_{1}}{n_{2}}=\sqrt {\dfrac {M_{1}}{M_{2}}}=\sqrt {\dfrac {64}{16}}=2:1\)

Option B is correct.

The correct option is 3

Explanation: Boyle's Law states that for a given mass of an ideal gas at a constant temperature, the pressure P P" id="MathJax-Element-18-Frame" role="presentation" style="position: relative;" tabindex="0">P$P$ $P$ of the gas is inversely proportional to its volume V V" id="MathJax-Element-19-Frame" role="presentation" style="position: relative;" tabindex="0">V$V$ $V$ . Mathematically, it’s represented as: P∝1V P \propto \frac{1}{V}" id="MathJax-Element-20-Frame" role="presentation" style="position: relative;" tabindex="0">P∝1V$P \propto \frac{1}{V}$ $P \propto \frac{1}{V}$  or PV=constant PV = \text{constant}" id="MathJax-Element-21-Frame" role="presentation" style="position: relative;" tabindex="0">PV=constant$PV = \text{constant}$ $PV = \text{constant}$ .

• Graph (i): Shows a linear relationship between P P" id="MathJax-Element-22-Frame" role="presentation" style="position: relative;" tabindex="0">P$P$ $P$  (Pressure) and T T" id="MathJax-Element-23-Frame" role="presentation" style="position: relative;" tabindex="0">T$T$ $T$  (Temperature), which does not represent Boyle’s Law. Boyle’s Law applies when temperature is constant, so this option is incorrect.

• Graph (ii): Shows a linear relationship between P P" id="MathJax-Element-24-Frame" role="presentation" style="position: relative;" tabindex="0">P$P$ $P$  (Pressure) and V V" id="MathJax-Element-25-Frame" role="presentation" style="position: relative;" tabindex="0">V$V$ $V$  (Volume). According to Boyle’s Law, P P" id="MathJax-Element-26-Frame" role="presentation" style="position: relative;" tabindex="0">P$P$ $P$  and V V" id="MathJax-Element-27-Frame" role="presentation" style="position: relative;" tabindex="0">V$V$ $V$  are inversely proportional, meaning the graph should show a hyperbolic curve, not a straight line. Thus, this option is incorrect.

• Graph (iii): Shows a linear relationship between P P" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0">P$P$ $P$  (Pressure) and 1V \frac{1}{V}" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">1V$\frac{1}{V}$ $\frac{1}{V}$ (the inverse of Volume). This aligns with Boyle’s Law, as P∝1V P \propto \frac{1}{V}" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">P∝1V$P \propto \frac{1}{V}$ $P \propto \frac{1}{V}$ . Therefore, this option correctly represents Boyle’s Law.

• Graph (iv): Shows PV PV" id="MathJax-Element-31-Frame" role="presentation" style="position: relative;" tabindex="0">PV$PV$ $PV$  (the product of Pressure and Volume) as constant with respect to T T" id="MathJax-Element-32-Frame" role="presentation" style="position: relative;" tabindex="0">T$T$ $T$  (Temperature). Although PV PV" id="MathJax-Element-33-Frame" role="presentation" style="position: relative;" tabindex="0">PV$PV$ $PV$  remains constant in Boyle’s Law, this graph implies that PV PV" id="MathJax-Element-34-Frame" role="presentation" style="position: relative;" tabindex="0">PV$PV$ $PV$  is constant regardless of temperature, which is not accurate for Boyle's Law specifically, as Boyle’s Law only applies at a constant temperature. Hence, this option is incorrect.

Additional Information: 

Concept:

Average thermal energy of gas molecule depends upon the temperature.

Calculation:

For monoatomic gases, degree of freedom is 3.

Hence average thermal energy per molecule is

\(K{E_{avg}} = \frac{3}{2}{k_B}T\)

Correct answer: 4) 

Concept:

• Gas is a state of matter in which particles have maximum space between the particle as compared to solid and liquid states.
• An ideal gas is a gas that behaves ideally and obeys ideal gas laws.
• An ideal gas is a gas that conforms, in physical behavior, to a particular, idealized relation between pressure, volume, and temperature called the ideal gas law. This law is a generalization containing both Boyle's law and Charles's law as special cases and states that for a specified quantity of gas, the product of the volume, V, and pressure, P, is proportional to the absolute temperature T; i.e., in equation form, PV = kT, in which k is a constant.
• Such a relation for a substance is called its equation of state and is sufficient to describe its gross behaviour.

Explanation:

Main postulate of Kinetic molecular theory are:

• Gases consist of particles (molecules or atoms) that are in constant random motion.
• Gas particles are constantly colliding with each other and the walls of their container.
• These collisions are elastic; that is, there is no net loss of energy from the collisions.
• Gas particles are small and the total volume occupied by gas molecules is negligible relative to the total volume of their container.
• There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
• The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and all gases at the same temperature have the same average kinetic energy.
• Thus, all are correct for an ideal gas.

Conclusion:

• Hence, (i), (ii), and (iii) describe an ideal gas.

Correct answer: 2)

Concept:

• The kinetic molecular theory of gases should be understood to know about origins of ideal gas laws.
• According to the kinetic molecular theory, gaseous particles are in a state of constant random motion; individual particles move at different speeds, constantly colliding and changing directions.
• The most probable speed is the one which informs about the speed possessed by the maximum number of molecules of the gas.
• The most probable speed is the maximum value on Maxwell's distribution plot.
• The average speed is the sum of the speeds of all the molecules divided by the number of molecules.
• The root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas.

Explanation:

Given, root mean square (rms) speed of O₂​ is 500m/s at a constant temperature.

Root mean square speed in given by the following expression:

\(V_{rms}=\sqrt{\frac{3RT}{M}} \)

\(\Rightarrow V_{rms}\alpha \frac{1}{\sqrt{M}} \) For constant temperature

\(\frac{V_{H_{2}}}{V_{O_{2}}}=\sqrt{\frac{M_{O_{2}}}{M_{H_{2}}}} \;\)

V_O2 = 500 m/s

M_H2= 2g /mol and M_O2= 32 g/mol

\(\frac{V_{H_{2}}}{500}=\sqrt{\frac{32}{2}} \)

V_H2=4 x 500

VH2= 2000 m/s

The average kinetic energy is calculated as:

\(K_{av}=\frac{1}{2}mu^{2} \)

Here, u= 2000 m/s; m= H₂ = 2 g/mol= 0.002 kg/mol

\(K_{av}=\frac{1}{2}\times 0.002\times (2000)^{2} \)

\(K_{av}=4000 J/mol\)

\(K_{av}=4 kJ/mol\)

Conclusion:

Thus, the rms speed and the average kinetic energy of H2 at the same temperature is 2000 m/s and 4.0 kJ/mol respectively.