Question

Find the number of degrees of freedom of molecules in a gas. Whose molar heat capacity at constant pressure ${C_{\text{P}}} = 29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - 1}}$.

Answer 1

In this question use the concept of the adiabatic index, that is it is the ratio of the specific heat at constant pressure and the specific heat at constant volume. First, calculate the specific heat at constant volume and then calculate the adiabatic index.

Complete step by step answer:
As we know that the molar heat capacity of any substance is the amount of energy that is in the form of heat added to the substance to one mole of the substance because of this the temperature of the substance increases.

We know that the constant pressure means that the heat transfer to the system does work but it changes the internal energy of the system. We are given molar heat capacity at constant pressure ${C_{\text{P}}} = 29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}\,{{\text{K}}^{ - 1}}$. First we calculate the value of ${C_{\text{v}}}$, use the ideal gas equation formula ${\text{R}}$ is the gas constant and the value of ${\text{R}}$ is $8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}$.
${C_{\text{p}}} - {C_{\text{v}}} = {\text{R}}$
$\Rightarrow {C_{\text{v}}} = {C_{\text{p}}} - {\text{R}}$
Now we substitute the given in the above equation,
${C_{\text{v}}} = 29 - 8.314$
$\Rightarrow {C_{\text{v}}} = 20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$

Now, use the formula to calculate the adiabatic index $\gamma$.
$\gamma = \dfrac{{{C_{\text{P}}}}}{{{C_{\text{V}}}}}$
Now we substitute the values in the above equation,
$\gamma = \dfrac{{29\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}{{20.686\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}}}$
By simplification we get,
$\Rightarrow \gamma = 1.4$

Now, we calculate the degrees of freedom of the molecules $\left( {\text{f}} \right)$ in a gas,
$\sqrt {\dfrac{3}{\gamma }} = \sqrt {\dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}}$

After squaring both sides, we get

$\Rightarrow \dfrac{3}{\gamma } = \dfrac{{3{\text{f}}}}{{{\text{f}} + 2}}$

By simplification we get,
$\Rightarrow {\text{f}} + 2 = \gamma {\text{f}}$

Now, Substitute the value of $\gamma$ in the above equation,
${\text{f}} + 2 = 1.4{\text{f}}$
$\therefore {\text{f}} = 5$
Therefore, the number of degrees of freedom of molecules in a gas is $5$.

Note:
Do not confuse the universal gas constant and the characteristic gas constant. As we know that the value of universal gas constant is $8.314\;{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - {\text{1}}}}{{\text{K}}^{ - {\text{1}}}}$ and it is same for each gas, but characteristic gas constant depends on the molecular mass of the gas.