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: Understanding the Question:
We are given a specific numeric ratio between the universal gas constant $R$ and the molar specific heat capacity at constant volume $C_v$ for an ideal gas. We need to identify the atomicity of the gas molecules based on this relation.

Step 2: Key Formula or Approach:
We use Mayer's formula relating specific heats of an ideal gas:
$$C_p - C_v = R$$ The ratio of specific heats is defined as $\gamma = \frac{C_p}{C_v}$. The standard values of $\gamma$ for different molecular structures are:
Monoatomic gas: $\gamma = \frac{5}{3} \approx 1.67$
Diatomic gas: $\gamma = \frac{7}{5} = 1.40$

Step 3: Detailed Explanation:
The given condition is $R = \frac{2}{3} C_v$. Let's substitute this value of $R$ directly into Mayer's relation:
$$C_p - C_v = \frac{2}{3} C_v$$ Add $C_v$ to both sides to find $C_p$:
$$C_p = C_v + \frac{2}{3} C_v = \frac{5}{3} C_v$$ Now, let's compute the specific heat ratio $\gamma$:
$$\gamma = \frac{C_p}{C_v} = \frac{\frac{5}{3} C_v}{C_v} = \frac{5}{3}$$ A specific heat ratio of $\gamma = \frac{5}{3}$ is the characteristic theoretical value for a monoatomic gas (which possesses exactly 3 translational degrees of freedom).

Step 4: Final Answer:
The gas consists of monoatomic molecules, which corresponds to option (C).