Pressure from Molecular Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Pressure from Molecular Motion MCQs with step-by-step solutions (31 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Pressure from Molecular Motion · easy · theory
According to kinetic theory, the pressure exerted by an ideal gas of density $\rho$ whose molecules have rms speed $v_{rms}$ is:
A. $\dfrac{2}{3}\rho v_{rms}^2$
B. $\dfrac{1}{2}\rho v_{rms}^2$
C. $\rho v_{rms}^2$
D. $\dfrac{1}{3}\rho v_{rms}^2$ ✓ Correct
Solution: Averaging the momentum delivered to the walls over the three coordinate directions gives $P = \dfrac{1}{3}\rho v_{rms}^2$.
Q2 — Pressure from Molecular Motion · easy · theory
The rms speed of gas molecules in terms of the pressure $P$ and density $\rho$ of the gas is:
A. $\sqrt{\dfrac{3P}{\rho}}$ ✓ Correct
B. $\sqrt{\dfrac{P}{3\rho}}$
C. $\sqrt{\dfrac{2P}{\rho}}$
D. $\sqrt{\dfrac{3\rho}{P}}$
Solution: Rearranging $P = \dfrac{1}{3}\rho v_{rms}^2$ gives $v_{rms} = \sqrt{\dfrac{3P}{\rho}}$.
Q3 — Pressure from Molecular Motion · medium · numerical
A gas has pressure $10^5\text{ Pa}$ and density $1.3\text{ kg/m}^3$. The rms speed of its molecules is approximately:
A. $170\text{ m/s}$
B. $480\text{ m/s}$ ✓ Correct
C. $960\text{ m/s}$
D. $340\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3P}{\rho}} = \sqrt{\dfrac{3 \times 10^5}{1.3}} = \sqrt{2.31 \times 10^5} \approx 480\text{ m/s}$.
Q4 — Pressure from Molecular Motion · medium · theory
The pressure of an ideal gas is related to the translational kinetic energy $E$ per unit volume by:
A. $P = \dfrac{3}{2}E$
B. $P = \dfrac{1}{3}E$
C. $P = \dfrac{2}{3}E$ ✓ Correct
D. $P = 3E$
Solution: With $P = \dfrac{1}{3}\rho v_{rms}^2$ and $E = \dfrac{1}{2}\rho v_{rms}^2$, dividing gives $P = \dfrac{2}{3}E$.
Q5 — Pressure from Molecular Motion · easy · numerical
If the rms speed of the molecules of a gas is doubled while its density is kept constant, the pressure becomes:
A. Unchanged
B. $2$ times
C. $8$ times
D. $4$ times ✓ Correct
Solution: From $P = \dfrac{1}{3}\rho v_{rms}^2$, pressure varies as the square of the rms speed, so doubling it multiplies the pressure by $4$.
Q6 — Pressure from Molecular Motion · easy · numerical
If the density of a gas is doubled while the rms speed of its molecules is unchanged, its pressure:
A. Halves
B. Doubles ✓ Correct
C. Becomes four times
D. Remains unchanged
Solution: $P = \dfrac{1}{3}\rho v_{rms}^2$ is directly proportional to density at fixed molecular speed.
Q7 — Pressure from Molecular Motion · easy · theory
The pressure exerted by a gas on the walls of its container arises from:
A. The mutual attraction between the gas molecules
B. The weight of the gas molecules
C. The rate of change of momentum of molecules colliding with the walls ✓ Correct
D. The elastic deformation of the container walls
Solution: Each molecule rebounding elastically from a wall reverses its normal momentum. The total rate of momentum transfer per unit area is exactly the pressure.
Q8 — Pressure from Molecular Motion · easy · theory
A molecule of mass $m$ moving with speed $v$ strikes a wall normally and rebounds elastically. The magnitude of the change in its momentum is:
A. $2mv$ ✓ Correct
B. $mv$
C. Zero
D. $\dfrac{mv}{2}$
Solution: The momentum changes from $+mv$ to $-mv$, so the change has magnitude $2mv$. This factor of two is what enters the kinetic-theory derivation of pressure.
Q9 — Pressure from Molecular Motion · medium · theory
In terms of the number density $n$ of molecules, the pressure of an ideal gas at absolute temperature $T$ is:
A. $P = \dfrac{n k_B}{T}$
B. $P = n R T$
C. $P = n k_B T$ ✓ Correct
D. $P = \dfrac{n T}{k_B}$
Solution: Writing $PV = N k_B T$ and dividing by $V$ with $n = \dfrac{N}{V}$ gives $P = n k_B T$.
Q10 — Pressure from Molecular Motion · medium · theory
The total translational kinetic energy of $n$ moles of an ideal gas at absolute temperature $T$ is:
A. $\dfrac{3}{2}nRT$ ✓ Correct
B. $\dfrac{1}{2}nRT$
C. $\dfrac{5}{2}nRT$
D. $3nRT$
Solution: Each mole contains $N_A$ molecules, each carrying $\dfrac{3}{2}k_B T$, so the total is $n N_A \times \dfrac{3}{2}k_B T = \dfrac{3}{2}nRT$.
Q11 — Pressure from Molecular Motion · medium · numerical
The total translational kinetic energy of $2\text{ moles}$ of an ideal gas at $300\text{ K}$ is ($R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$):
A. $4988\text{ J}$
B. $12472\text{ J}$
C. $7483\text{ J}$ ✓ Correct
D. $2494\text{ J}$
Solution: $E = \dfrac{3}{2}nRT = \dfrac{3}{2}(2)(8.314)(300) = 3 \times 8.314 \times 300 \approx 7483\text{ J}$.
Q12 — Pressure from Molecular Motion · medium · theory
The pressure exerted by an ideal gas is independent of:
A. The shape of the containing vessel ✓ Correct
B. The number density of molecules
C. The temperature of the gas
D. The mass of each molecule
Solution: Pressure depends on the number density, molecular mass and mean square speed, but not on the geometry of the container — the molecular bombardment is isotropic.
Q13 — Pressure from Molecular Motion · easy · numerical
For an ideal gas kept at constant volume, doubling the absolute temperature causes the pressure to:
A. Become four times
B. Remain unchanged
C. Double ✓ Correct
D. Halve
Solution: At constant $V$ and $n$, $PV = nRT$ reduces to $P \propto T$, so the pressure doubles.
Q14 — Pressure from Molecular Motion · medium · theory
The relation $PV = \dfrac{1}{3}M v_{rms}^2$, where $M$ is the total mass of the enclosed gas, follows directly from:
A. Stefan's law of radiation
B. Newton's law of cooling
C. Charles' law alone
D. The kinetic theory expression for pressure ✓ Correct
Solution: Multiplying $P = \dfrac{1}{3}\rho v_{rms}^2$ by $V$ and using $\rho V = M$ gives the stated result.
Q15 — Pressure from Molecular Motion · medium · numerical
Two vessels of equal volume contain the same gas at the same temperature, but the pressure in the first is twice that in the second. The ratio of the number of molecules in them is:
A. $4 : 1$
B. $2 : 1$ ✓ Correct
C. $1 : 1$
D. $1 : 2$
Solution: From $P = n k_B T$ with $T$ and $V$ common, $P \propto N$. So twice the pressure means twice as many molecules.
Q16 — Pressure from Molecular Motion · medium · theory
The translational kinetic energy per unit volume of an ideal gas at pressure $P$ is:
A. $\dfrac{2}{3}P$
B. $\dfrac{3}{2}P$ ✓ Correct
C. $\dfrac{1}{2}P$
D. $3P$
Solution: Inverting $P = \dfrac{2}{3}E$ gives the energy density $E = \dfrac{3}{2}P$.
Q17 — Pressure from Molecular Motion · easy · theory
In deriving the expression for gas pressure, the collisions of molecules with the container walls are assumed to be:
A. Perfectly elastic, so that no kinetic energy is lost ✓ Correct
B. Rare enough to be neglected entirely
C. Accompanied by a steady loss of energy to the wall
D. Perfectly inelastic, so that molecules stick to the wall
Solution: If collisions were inelastic the gas would continuously lose energy to the walls and could never remain in thermal equilibrium at constant temperature.
Q18 — Pressure from Molecular Motion · medium · numerical
A gas of density $1.2\text{ kg/m}^3$ has molecules with an rms speed of $500\text{ m/s}$. Its pressure is:
A. $3 \times 10^5\text{ Pa}$
B. $2 \times 10^5\text{ Pa}$
C. $6 \times 10^2\text{ Pa}$
D. $10^5\text{ Pa}$ ✓ Correct
Solution: $P = \dfrac{1}{3}\rho v_{rms}^2 = \dfrac{1}{3}(1.2)(2.5 \times 10^5) = 10^5\text{ Pa}$.
Q19 — Pressure from Molecular Motion · medium · numerical
A gas at pressure $2 \times 10^5\text{ Pa}$ has density $1.5\text{ kg/m}^3$. The rms speed of its molecules is:
A. $632\text{ m/s}$ ✓ Correct
B. $1095\text{ m/s}$
C. $447\text{ m/s}$
D. $365\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3P}{\rho}} = \sqrt{\dfrac{6 \times 10^5}{1.5}} = \sqrt{4 \times 10^5} \approx 632\text{ m/s}$.
Q20 — Pressure from Molecular Motion · medium · numerical
The translational kinetic energy per unit volume of a gas at a pressure of $10^5\text{ Pa}$ is:
A. $6.7 \times 10^4\text{ J/m}^3$
B. $3 \times 10^5\text{ J/m}^3$
C. $1.5 \times 10^5\text{ J/m}^3$ ✓ Correct
D. $10^5\text{ J/m}^3$
Solution: Since $P = \dfrac{2}{3}E$, the energy density is $E = \dfrac{3}{2}P = 1.5 \times 10^5\text{ J/m}^3$.
Q21 — Pressure from Molecular Motion · hard · numerical
The number of molecules per unit volume of an ideal gas at $10^5\text{ Pa}$ and $300\text{ K}$ is approximately ($k_B = 1.38 \times 10^{-23}$):
A. $6.0 \times 10^{23}\text{ m}^{-3}$
B. $2.4 \times 10^{23}\text{ m}^{-3}$
C. $2.4 \times 10^{25}\text{ m}^{-3}$ ✓ Correct
D. $4.1 \times 10^{-21}\text{ m}^{-3}$
Solution: $n = \dfrac{P}{k_BT} = \dfrac{10^5}{1.38 \times 10^{-23} \times 300} = \dfrac{10^5}{4.14 \times 10^{-21}} \approx 2.4 \times 10^{25}\text{ m}^{-3}$.
Q22 — Pressure from Molecular Motion · medium · numerical
A gas of density $2\text{ kg/m}^3$ has molecules of rms speed $300\text{ m/s}$. Its pressure is:
A. $9 \times 10^4\text{ Pa}$
B. $2 \times 10^4\text{ Pa}$
C. $6 \times 10^4\text{ Pa}$ ✓ Correct
D. $1.8 \times 10^5\text{ Pa}$
Solution: $P = \dfrac{1}{3}\rho v_{rms}^2 = \dfrac{1}{3}(2)(9 \times 10^4) = 6 \times 10^4\text{ Pa}$.
Q23 — Pressure from Molecular Motion · medium · numerical
A gas at $1.5 \times 10^5\text{ Pa}$ has a density of $2\text{ kg/m}^3$. The rms speed of its molecules is approximately:
A. $474\text{ m/s}$ ✓ Correct
B. $237\text{ m/s}$
C. $822\text{ m/s}$
D. $274\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3P}{\rho}} = \sqrt{\dfrac{4.5 \times 10^5}{2}} = \sqrt{2.25 \times 10^5} \approx 474\text{ m/s}$.
Q24 — Pressure from Molecular Motion · medium · numerical
The translational kinetic energy per unit volume of a gas at $2 \times 10^5\text{ Pa}$ is:
A. $1.33 \times 10^5\text{ J/m}^3$
B. $3 \times 10^5\text{ J/m}^3$ ✓ Correct
C. $4 \times 10^5\text{ J/m}^3$
D. $2 \times 10^5\text{ J/m}^3$
Solution: Since $P = \dfrac{2}{3}E$, the energy density is $E = \dfrac{3}{2}P = 3 \times 10^5\text{ J/m}^3$.
Q25 — Pressure from Molecular Motion · medium · numerical
A molecule of mass $5 \times 10^{-26}\text{ kg}$ strikes a wall normally at $400\text{ m/s}$ and rebounds elastically. The magnitude of its momentum change is:
A. Zero
B. $8 \times 10^{-23}\text{ kg}\cdot\text{m/s}$
C. $4 \times 10^{-23}\text{ kg}\cdot\text{m/s}$ ✓ Correct
D. $2 \times 10^{-23}\text{ kg}\cdot\text{m/s}$
Solution: The momentum reverses, so the change is $2mv = 2 \times 5 \times 10^{-26} \times 400 = 4 \times 10^{-23}\text{ kg}\cdot\text{m/s}$.
Q26 — Pressure from Molecular Motion · hard · numerical
The number of molecules per cubic metre of an ideal gas at $2 \times 10^5\text{ Pa}$ and $400\text{ K}$ is approximately ($k_B = 1.38 \times 10^{-23}$):
A. $1.8 \times 10^{25}$
B. $3.6 \times 10^{23}$
C. $7.2 \times 10^{25}$
D. $3.6 \times 10^{25}$ ✓ Correct
Solution: $n = \dfrac{P}{k_BT} = \dfrac{2 \times 10^5}{1.38 \times 10^{-23} \times 400} = \dfrac{2 \times 10^5}{5.52 \times 10^{-21}} \approx 3.6 \times 10^{25}\text{ m}^{-3}$.
Q27 — Pressure from Molecular Motion · medium · numerical
The total translational kinetic energy of $5\text{ moles}$ of an ideal gas at $300\text{ K}$ is ($R = 8.314$):
A. $1.25 \times 10^4\text{ J}$
B. $3.74 \times 10^4\text{ J}$
C. $6.23 \times 10^3\text{ J}$
D. $1.87 \times 10^4\text{ J}$ ✓ Correct
Solution: $E = \dfrac{3}{2}nRT = 1.5 \times 5 \times 8.314 \times 300 \approx 1.871 \times 10^4\text{ J}$.
Q28 — Pressure from Molecular Motion · easy · numerical
If the rms speed of gas molecules is tripled at constant density, the pressure becomes:
A. $9$ times ✓ Correct
B. $27$ times
C. $\dfrac{1}{3}$ times
D. $3$ times
Solution: $P = \dfrac{1}{3}\rho v_{rms}^2 \propto v_{rms}^2$, so tripling the speed multiplies the pressure by $9$.
Q29 — Pressure from Molecular Motion · easy · numerical
If the density of a gas is tripled while the rms speed of its molecules is unchanged, its pressure becomes:
A. $3$ times ✓ Correct
B. Unchanged
C. $\dfrac{1}{3}$ times
D. $9$ times
Solution: $P \propto \rho$ at fixed molecular speed, so the pressure is tripled.
Q30 — Pressure from Molecular Motion · hard · numerical
A vessel of volume $0.02\text{ m}^3$ contains $0.1\text{ kg}$ of a gas whose molecules have an rms speed of $400\text{ m/s}$. The product $PV$ for the gas is:
A. $5333\text{ J}$ ✓ Correct
B. $800\text{ J}$
C. $16000\text{ J}$
D. $1778\text{ J}$
Solution: $PV = \dfrac{1}{3}Mv_{rms}^2 = \dfrac{1}{3}(0.1)(1.6 \times 10^5) \approx 5333\text{ J}$.