Question

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is $\overset{\to }{\mathop{v}}\,$, m is its mass and ${{k}_{B}}$ is Boltzmann constant, then its temperature will be:

$A.\,\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{6{{k}_{B}}}$
$B.\,\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{5{{k}_{B}}}$
$C.\,\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{3{{k}_{B}}}$
$D.\,\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{7{{k}_{B}}}$

Answer 1

The formula that relates the motion of the gas molecules and the temperature should be used to solve this problem. The kinetic energy formula is equated with this formula to represent the equation in terms of the temperature of the molecules of the HCl molecules in the gaseous phase.
Formula used:
$E=\dfrac{3}{2}kT$

Complete answer:
The kinetic energy of the molecule is given by the formula as follows.
$E=\dfrac{3}{2}nRT$
Where T is the absolute temperature, k is Boltzmann constant, R is the molar gas constant and n is the number of molecules of the gas.
The average kinetic energy of the molecule is given by the formula as follows.

& \dfrac{E}{n}=\dfrac{3}{2}RT \\\ & \Rightarrow E =\dfrac{3}{2}RT \\\ \end{aligned}$$ Where T is the absolute temperature, k is Boltzmann constant and R is the molar gas constant. The kinetic energy of the molecules is given by the formula as follows. $$KE=\dfrac{1}{2}m{{v}^{2}}$$ Where m is the mass of the molecules and v is the velocity of the molecules. For the translational kinetic energy of the molecule, the above equation can be represented as follows. $$\dfrac{1}{2}m{{v}^{2}}=\dfrac{3}{2}RT$$ Rearrange the above terms to obtain the expression in terms of the temperature of the molecules. So, we get, $$T=\dfrac{m{{v}^{2}}}{3k}$$ Now represent the above equation in terms of the vector form. In the above equation, only the velocity “v” is the vector quantity. Thus, we get, $$T=\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{3k}$$ As, the equation for the temperature of the HCl molecules is obtained as $$\dfrac{m\overset{\to }{\mathop{{{v}^{2}}}}\,}{3k}$$. **So, the correct answer is “Option C”.** **Note:** The kinetic energy formula remains the same even in the case of the molecules of the gas. So, we have used this equation. Even, they can further extend the question, by asking for the value of the temperature by giving the values of the other parameters.