The Kinetic Theory of Gases MCQ Quiz - Objective Question with Answer for The Kinetic Theory of Gases - Download Free PDF

Correct option is: (2) 3 / 4

Using the First Law of Thermodynamics:

ΔQ = ΔU + PΔV

ΔQ is the same

ΔU is also the same

⇒ W_A = W_B

⇒ (PΔV)_A = (PΔV)_B

P is also the same

⇒ A_Ad_A = A_Bd_B

πr_A²d_A = πr_B²d_B

r_A / r_B = √(d_B / d_A) = √(9 / 16)

= 3 / 4

Correct option is: (2) 1.6 atm

Before partition

after partition removed

P₁V₁ + P₂V₂ = P(V₁ + V₂)

⇒ 1 × 2 + 2 × 3 = P × (2 + 3)

⇒ 8 / 5 = P

⇒ P = 1.6 atm

Correct option is: (3) 0.116 kg

Number of moles left

n = PV / RT = (12 × 1.01 × 10⁵ N/m² × 30 × 10⁻³ m³) / ((100/12) × 300)

n = (12 × 1.01 × 12) / 10 = 14.54 moles

Moles removed = 18.2 − 14.54

= 3.656 moles

Mass removed = 3.656 × 32 = 116.99 g = 0.116 kg

The Correct answer is Both Statement I and Statement II are true.

Key Points

Statement (I) : The mean free path of gas molecules is inversely proportional to square of molecular diameter.

• The mean free path (λ) is the average distance a molecule travels between collisions with other molecules.
• The formula for the mean free path is given by:

λ = \(\frac{\mathrm{RT}}{\sqrt{2} \pi \mathrm{d}^2 \mathrm{~N}_{\mathrm{A}} \mathrm{P}}\)

• According to this equation, the mean free path is indeed inversely proportional to the square of the molecular diameter 
• Thus Statement (I) is correct

Statement (II) : Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.

• The average kinetic energy of a gas molecule is given by the equation.

KE = \(\frac{\mathrm{f}}{2}\) nRT

• This shows that the average kinetic energy is directly proportional to the absolute temperature
• Thus Statement (II) is Correct. ​

Calculation:

Given that 

\( V_2=\dfrac{V_1}{8}, \gamma=\dfrac{4}{3} \)

According to the adiabatic compression

\( \dfrac{P_{2}}{P_{1}} = \left ( \dfrac{V_{1}}{V_{2}} \right )^{\gamma }P_{0}\left ( 8 \right )^{4/3}= 16 P_{0} \)

CONCEPT: 

• Gas Molecules are in Constant Motion.
• Root mean Square Speed: The root mean square speed of Gas average of roots of the speed of all Gas molecules present in the given volume.

It is given as 

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

R = Gas Constant

T = Absolute Temperature of Gas 

M = Molecular Weight of Gas

CALCULATION:

Given V_^rms = 300m/s. 

Let, initial Absolute Temperature is T and initial Molecular weight be M. So,

\(V_{rms} = \sqrt{\frac{3RT}{M}} = 300m/s\) .......  (1)

Now, the Absolute Temperature is halved, So the new absolute Temperature T' = T/2

Also, Molecular weight is doubled, So the new molecular weight M' = 2M

The New root mean square speed

 \(V'_{rms} = \sqrt{\frac{3RT'}{M'}}\).................. (2)

Putting the values of T' and M' in Equation (2) we get

\(V'_{rms} = \sqrt{\frac{3RT}{4M}}\)  

⇒ \(V'_{rms} = \frac{1}{2}\sqrt{\frac{3RT}{M}}\)............  (3)

Comparing Equation (3) and Equation (1)

\(V'_{rms} = \frac{V_{rms}}{2}\)

⇒ V'rms = 300 m/s / 2 = 150 m/s

So, 150 m/s is the Answer.

So, Option 2 is correct.

Additional Information

• The Average of all speeds of the Gas molecule is given as

\(V_{av} = \sqrt{\frac{8RT}{\pi M}}\)

• The Speed passed by the maximum fraction of the total number of molecules, or most probable speed is given as:

\((V_{mp} = \sqrt{\frac{2RT}{M}})\)

CONCEPT:

• Root Mean Square Speed is defined as the square root of the mean of squares of the speed of different molecules.
  • The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
  • The rms speed of any homogeneous gas sample is given by:

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

Where R = universal gas constant, T = temperature and M = Molecular mass

CALCULATION:

Given - Initial rms velocity (Vrms1) = V, initial temperature (T1) = 27° C = 300 K and final temperature (T2) = 327° C = 600 K

• As the sample is the same, therefore the molecular mass will be the same. Hence, 

⇒ Vrms ∝ \(\sqrt{T}\)

\(⇒ \frac{V_{rms1}}{V_{rms2}}=\sqrt{{\frac{T_1}{T_2}}}\)

\(⇒ \frac{V}{V_{rms2}}=\sqrt{{\frac{300}{600}}}= \frac{1}{\sqrt{{2}}}\)

⇒ V_rms2 = V\(\sqrt{2}\)

CONCEPT:

• According to kinetic energy theory, if we increase the temperature of a gas, it will increase the average kinetic energy of the molecule, which will increase the motion of the molecules.
  • This increased motion increases the outward pressure of the gas.

The average kinetic energy (KE) or energy (E) of translation per molecules of the gas is related to temperature by the relationship:

\(KE = \frac{3}{2}{k_B}T\)      (degree of freedom of a monoatomic gas = 3)

Where K E = kinetic energy, k_B = Boltzmann constant and T = temperature.

EXPLANATION:

The average energy of a molecule is given by:

KE = E = (3/2)kBT

So option 4 is correct.

CONCEPT:

RMS Velocity (Vrms):

• The effective velocity of gas particles in a gaseous sample is called Root-mean-square speed (RMS speed).
• RMS velocity is given by:

\({V_{rms}} = \sqrt{\frac{{3\;R\;T}}{M}} \)

Where R = Universal gas constant, T = temperature, and M = Molar mass of the gas.

CALCULATION:

Given - MO₂ = 32 and MH₂ = 2

• RMS velocity of 02 gas is given by:

\(\Rightarrow {V_{O_2}} = \sqrt{\frac{{3\;R\;T}}{M_{O_2}}} =\sqrt{\frac{{3\;R\;T}}{32}}\)    ------- (1)

• RMS velocity of H2 gas is given by:

\(\Rightarrow {V_{H_2}} = \sqrt{\frac{{3\;R\;T}}{M_{H_2}}} =\sqrt{\frac{{3\;R\;T}}{2}}\)  ------- (2)

On dividing equation 1 and 2, we get

\(\Rightarrow \frac{V_{O_2}}{V_{H_2}}= \frac{\sqrt{\frac{{3\;R\;T}}{32}}}{\sqrt{\frac{{3\;R\;T}}{2}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}\)

Different magnetic materials and their properties are shown in the Table:

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

\(P=\frac{1}{3}ρ \;C^2\)

where \(ρ\)​ is the density of ideal gas and C is its root mean square velocity.

We know, density = mass/volume which implies that mass = density × volume. So, for a unit volume of gas, mass= ρ × 1 = ρ 

Now, the mean kinetic energy of translation per unit volume of gas (E) = \(\frac{1}{2}ρ C^2\)

\(Therefore,\frac{P}{E}= \frac{(1/3)ρ C^2}{(1/2)ρ C^2}=\frac{2}{3}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;or, \;P=\frac{2}{3}E\)

EXPLANATION:

• The pressure(P) that is exerted by an ideal gas is equal to two-thirds of the mean kinetic energy of translation(E) per unit volume of the gas:

\(P=\frac{2}{3}E\)

So option 1 is correct.

EXTRA POINTS:

The pressure exerted by a gas:

• According to kinetic theory, the molecules of a gas are in a state of continuous random motion.
• They collide with one another and also with the walls of the vessel.
• Whenever a molecule collides with the wall, it returns with a changed momentum, and an equal momentum is transferred to the wall (conservation of momentum).
• According to Newton's second law of motion, the rate of transfer of momentum to the wall is equal to the force exerted on the wall.
• Since a large number of molecules collide with the wall, a steady force is exerted on the wall.
• The force exerted per unit area of the wall is the pressure of the gas.
• Hence a gas exerts pressure due to the continuous collisions of its molecules with the walls of the vessel.

CONCEPT:

• Basically Maxwell-Boltzmann distribution shows how the speed of molecules distributed for an ideal gas.
• It can be represented with the following graph

• If we draw Maxwell's curve between the number of particles and velocity, we can see that the curve first rises from zero to a maximum then again reaches zero.
• The total area under the graph represents the total no of molecules.

EXPLANATION:

• From the shape of the curve, it is clear that only a certain no of molecules achieves very high velocities and very low velocities.  Therefore option 2 is correct.
• The peak gives the no of molecules acquired high velocity and the flattened part gives the no of molecules that possess very low velocity.

CONCEPT:

• Kinetic energy: The energy possessed by a body due to its motion is called Kinetic energy.
  • Temperature is directly proportional to kinetic energy.
  • If the temperature increases, then the kinetic energy of the particles also increases.
  • Kinetic energy is usually measured in units of Joules(J)
  • Kelvin(K) is the SI unit of temperature.

EXPLANATION:

• The energy of a molecule in the solid is due to its randomness.
• If there is a change in the temperature of the solid, the randomness of the molecules will change.
• Once we increase the temperature, the randomness will increase, and hence the kinetic energy of the molecules increases. So option 3 is correct.

CONCEPT:

Boyle’s law: 

• For a given mass of an ideal gas at a constant temperature, the volume of a gas is inversely proportional to its pressure.
  i.e.\(\Rightarrow \frac{{{P_1}}}{{{P_2}}}\; = \;\frac{{{V_2}}}{{{V_1}}}\;\)
  or PV = constant
  ⇒ P1V1 = P2V2
• As temperature remains constant in the isothermal process, then Boyle's law holds for an ideal gas Isothermal changes.

EXPLANATION:

• From above it is clear that according to Boyle’s law for a fixed mass of gas at a constant temperature, the volume is inversely proportional to the pressure.
• That means that, for example, if you double the pressure, you will halve the volume.
• This can express this mathematically as

PV = constant

⇒ P1V1 = P2V2

\(\Rightarrow \frac{{{P_1}}}{{{P_2}}}\; = \;\frac{{{V_2}}}{{{V_1}}}\;\)

CONCEPT:

The motion of Molecules:

• As we know that all molecules are in constant motion.
• Molecules of a liquid have more freedom of movement than those in a solid whereas molecules in a gas have the greatest degree of motion.

EXPLANATION:

• According to kinetic energy theory, if we increase the temperature of a gas, it will increase the average kinetic energy of the molecule, which will increase the motion of the molecules.
• This increased motion increases the outward pressure of the gas.
• The average kinetic energy of translation per molecules of the gas is related to temperature by the relationship:

\(\Rightarrow KE = \frac{3}{2}{k_B}T\)

Where E = kinetic energy, kB = Boltzmann constant and T = temperature

• From the above equation, it is clear that the average kinetic energy of the gas molecule is directly proportional to the absolute temperature of the gas. 
• As the temperature of both gases is the same. Therefore, the average kinetic energy of O2 and H2 is the same. Hence, option 4 is correct.