Grade 11 : Physics - Kinetic Theory (Set 2)

Explanation:Using the ideal gas equation $PV = nRT$:
$$P = \frac{nRT}{V} = \frac{2 \times 8.31 \times 300}{0.0831} = \frac{4986}{0.0831} = 60,000\text{ Pa} = 6 \times 10^4\text{ Pa}$$
Thus, Option B is correct.

Explanation:The RMS speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$, quadrupling the absolute temperature ($T' = 4T$) gives:
$$v'_{rms} = \sqrt{4} \cdot v_{rms} = 2 v_{rms}$$
The RMS speed increases by a factor of 2.

Explanation:The average translational kinetic energy per molecule is $E_k = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is absolute temperature.
Therefore, translational kinetic energy is directly proportional to the absolute temperature $T$.

Explanation:A rigid diatomic molecule at room temperature has 3 translational degrees of freedom (movement along x, y, z axes) and 2 rotational degrees of freedom (rotation around two perpendicular axes). Total degrees of freedom = $3 + 2 = 5$.

Explanation:The law of equipartition of energy states that for any dynamical system in thermal equilibrium, the total energy is divided equally among all degrees of freedom, and the energy per degree of freedom per molecule is equal to $\frac{1}{2} k_B T$.

Explanation:For a monoatomic gas, degrees of freedom $f = 3$.
$C_v = \frac{f}{2}R = \frac{3}{2}R$, and $C_p = C_v + R = \frac{5}{2}R$.
$$\gamma = \frac{C_p}{C_v} = \frac{5/2}{3/2} = \frac{5}{3} \approx 1.67$$

Explanation:According to Dalton's Law of Partial Pressures, total pressure $P_{total} = P_1 + P_2$.
Since pressure is directly proportional to moles at fixed volume and temperature:
$$P_{\text{Argon}} = 3 \times P_{\text{Helium}} = 3 \times 20\text{ kPa} = 60\text{ kPa}$$
$$P_{total} = 20\text{ kPa} + 60\text{ kPa} = 80\text{ kPa}$$

Explanation:The mean free path equation is given by $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$, where $n$ is number density and $d$ is diameter of gas molecules.
Hence, $\lambda$ is inversely proportional to $d^2$.

Explanation:A monoatomic gas has $f = 3$ degrees of freedom per molecule.
Internal energy per mole is $\frac{3}{2} RT$. For $n$ moles, $U = n \cdot \frac{3}{2} RT = \frac{3}{2} nRT$.

Explanation:Kinetic Theory assumes that intermolecular forces of attraction or repulsion between ideal gas molecules are zero, except during actual collisions (which are perfectly elastic).

Explanation:$v_{rms} = \sqrt{\frac{3RT}{M}}$, so $v_{rms} \propto \frac{1}{\sqrt{M}}$ at constant temperature.
$$\frac{v_{H2}}{v_{O2}} = \sqrt{\frac{M_{O2}}{M_{H2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$
The ratio is 4 : 1.

Explanation:Pressure derived from kinetic theory is $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.
Since density $\rho = \frac{N m}{V}$, this simplifies directly to $P = \frac{1}{3} \rho v_{rms}^2$.

Explanation:The characteristic molecular speeds are:
$v_{mp} = \sqrt{2 RT / M} \approx 1.414 \sqrt{RT/M}$
$v_{avg} = \sqrt{\frac{8}{\pi} RT / M} \approx 1.595 \sqrt{RT/M}$
$v_{rms} = \sqrt{3 RT / M} \approx 1.732 \sqrt{RT/M}$
Therefore, $v_{mp} < v_{avg} < v_{rms}$.

Explanation:Mean free path in terms of pressure and temperature is $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$.
At constant temperature, $\lambda \propto \frac{1}{P}$. Doubling pressure reduces mean free path to half of its original value.

Explanation:Total degrees of freedom $f = 3 (\text{trans}) + 2 (\text{rot}) + 2 (\text{vib}) = 7$.
Molar heat capacity at constant volume $C_v = \frac{f}{2} R = \frac{7}{2} R$.

Explanation:The Boltzmann constant is $k_B = \frac{R}{N_A}$, representing the gas constant per individual molecule.

Explanation:The peak of the Maxwell-Boltzmann distribution curve represents the speed possessed by the maximum fraction of molecules, which is defined as the most probable speed ($v_{mp}$).

Explanation:Avogadro's Law states that under equal temperature and pressure, equal volumes of different ideal gases contain equal numbers of molecules (regardless of atomicity or mass).

Explanation:At $T = 0\text{ K}$, kinetic energy $E_k = \frac{3}{2} k_B T = 0$, meaning molecular speeds $v_{rms} = 0$. Since molecules stop colliding with container walls, pressure drops to zero.

Explanation:For an ideal gas, there are no intermolecular forces, so potential energy is zero. Internal energy consists solely of kinetic energy, which depends exclusively on absolute temperature ($U = \frac{f}{2} nRT$).