Question

For an ideal gas, $R = \frac{2}{3} C_v$. This suggests that the gas consists of molecules, which are ($R = $ universal gas constant)

• A. polyatomic
• B. diatomic
• C. monoatomic
• D. a mixture of diatomic and polyatomic molecules

Hint:

You can also find this directly using degrees of freedom ($f$) where $C_v = \frac{f}{2}R$. Rewriting the problem's condition gives $\frac{C_v}{R} = \frac{3}{2}$. Equating $\frac{f}{2} = \frac{3}{2}$ instantly shows that $f = 3$, which uniquely characterizes a monoatomic gas.

Answer 1

The Correct Option is C

Step 1: Read the given relation.
For an ideal gas we are told $R = \dfrac{2}{3} C_v$, and must decide whether the gas is monoatomic, diatomic, or polyatomic.
Step 2: Use Mayer's relation.
For an ideal gas, $C_p - C_v = R$, linking the two molar specific heats to the gas constant.
Step 3: Substitute the given $R$.
$C_p - C_v = \dfrac{2}{3} C_v$, so $C_p = C_v + \dfrac{2}{3} C_v = \dfrac{5}{3} C_v$.
Step 4: Compute the ratio $\gamma$.
$\gamma = \dfrac{C_p}{C_v} = \dfrac{\tfrac{5}{3} C_v}{C_v} = \dfrac{5}{3} \approx 1.67$.
Step 5: Match $\gamma$ to the molecular type.
Standard values: monoatomic $\gamma = \dfrac{5}{3}$, diatomic $\gamma = \dfrac{7}{5}$, polyatomic $\approx \dfrac{4}{3}$. Our $\gamma = \dfrac{5}{3}$ is the monoatomic value.
Step 6: Conclude.
A monoatomic gas has only $3$ translational degrees of freedom, giving exactly $\gamma = 5/3$. So the gas is monoatomic, option (3).
\[ \boxed{\gamma = \dfrac{5}{3}\ \Rightarrow\ \text{monoatomic}} \]