Question

The total number of degrees of freedom associated with \( 2 \, cm^3 \) of Nitrogen gas at normal temperature and pressure is:
[Given Avogadro number as 'N']

• A. \( \frac{N}{44800} \)
• B. \( \frac{N}{4480} \)
• C. \( \frac{N}{2240} \)
• D. \( \frac{N}{22400} \)

Hint:

At room temperature, diatomic gases have 5 degrees of freedomAlways use molar volume \( 22400 \, cm^3 \) at NTP to find number of molecules.

Answer 1

The Correct Option is C

Step 1: Recall degrees of freedom for nitrogen gas.
Nitrogen \( (N_2) \) is a diatomic gas.
At normal temperature, it has:
\[ f = 5 \quad \text{(3 translational + 2 rotational)} \]

Step 2: Find number of molecules in given volume.
At NTP, \( 1 \) mole of gas occupies \( 22400 \, cm^3 \).
Given volume = \( 2 \, cm^3 \)
So number of moles is:
\[ \frac{2}{22400} = \frac{1}{11200} \]

Step 3: Calculate number of molecules.
\[ \text{Number of molecules} = \frac{N}{11200} \]

Step 4: Calculate total degrees of freedom.
Total degrees of freedom = number of molecules \( \times f \)
\[ = \frac{N}{11200} \times 5 \]
\[ = \frac{5N}{11200} \]

Step 5: Simplify expression.
\[ = \frac{N}{2240} \]

Step 6: Conclusion.
Thus, total degrees of freedom is:
\[ \boxed{\frac{N}{2240}} \]