Mix Examples-Kinetic Theory of Gases Questions in English

(B) The average translational kinetic energy of a gas molecule at temperature $T$ is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant.
The kinetic energy of an electron accelerated through a potential difference $V$ is given by $E_2 = eV$.
Given the condition $E_1 = E_2$, we have:
$\frac{3}{2} k_B T = eV$
Since $k_B = \frac{R}{N_A}$, we substitute this into the equation:
$\frac{3}{2} (\frac{R}{N_A}) T = eV$
Solving for $T$:
$T = \frac{2eV N_A}{3R}$
Note: If the question implies the energy per mole or a specific molar context, the expression simplifies to $T = \frac{2eV}{3k_B}$. Given the options provided, the correct relationship is $T = \frac{2eV}{3k_B}$. Since $k_B = R/N_A$, the option matching the form is $B$.

(A) The average translational kinetic energy of a gas molecule is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
The kinetic energy of an electron accelerated through a potential difference $V$ is $E_2 = eV$.
Given the condition $E_1 = E_2$, we have:
$\frac{3}{2} k_B T = eV$
We know that the Boltzmann constant $k_B = \frac{R}{N}$, where $R$ is the universal gas constant and $N$ is Avogadro's number.
Substituting $k_B$ into the equation:
$\frac{3}{2} (\frac{R}{N}) T = eV$
Solving for $T$:
$T = \frac{2eVN}{3R}$
Thus, the correct option is $A$.

(A) The kinetic energy of the container is converted into the internal energy of the gas when it is stopped suddenly.
Let $n$ be the number of moles of the gas. The kinetic energy of the container is $K = \frac{1}{2} M_{total} V^2$, where $M_{total} = nM$.
So, $K = \frac{1}{2} nMV^2$.
This kinetic energy is converted into the internal energy of the gas: $\Delta U = n C_v \Delta T$.
For a rigid diatomic gas, the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Equating the two: $\frac{1}{2} nMV^2 = n (\frac{5}{2} R) \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{MV^2}{5R}$.

(B) The number of moles of nitrogen $(N_2)$ is given by $n = \frac{\text{mass}}{\text{molecular weight}} = \frac{14}{28} = 0.5 \text{ mol}$.
For a diatomic gas like nitrogen, the molar heat capacity at constant pressure $(C_p)$ is given by $C_p = \frac{7}{2}R$.
The heat energy $(Q)$ supplied at constant pressure is given by the formula $Q = n C_p \Delta T$.
Substituting the values: $Q = 0.5 \times \left(\frac{7}{2}R\right) \times 48$.
$Q = 0.5 \times 3.5 R \times 48$.
$Q = 1.75 R \times 48 = 84 R$.
Therefore, the heat energy required is $84 R$.