The quantity `(2U)/(fkT)` represents (where `U =` internal energy of gas)

Options

A. mass of the gas
B. kinetic energy of the gas
C. number of moles of the gas
D. number of molecules in the gas

Explanation

To solve the problem, we need to analyze the expression \( \frac{2U}{fkT} \) where \( U \) is the internal energy of the gas, \( f \) is the degrees of freedom, \( k \) is the Boltzmann constant, and \( T \) is the temperature.

### Step-by-Step Solution:

1. **Understand the Internal Energy of an Ideal Gas**:
The internal energy \( U \) of an ideal gas can be expressed as:
\[ U = \frac{nFRT}{2} \]
where \( n \) is the number of moles, \( F \) is the degrees of freedom, and \( R \) is the universal gas constant.

2. **Substituting \( R \) with \( k \)**:
We can also express the universal gas constant \( R \) in terms of the Boltzmann constant \( k \):
\[ R = N_A k \]
where \( N_A \) is Avogadro's number. Thus, we can rewrite the internal energy as:
\[ U = \frac{nF(N_A k) T}{2} \]

3. **Rearranging the Internal Energy Expression**:
Simplifying the expression for \( U \):
\[ U = \frac{nF N_A k T}{2} \]

4. **Finding \( \frac{2U}{fkT} \)**:
Now, we need to compute \( \frac{2U}{fkT} \):
\[ \frac{2U}{fkT} = \frac{2 \left(\frac{nF N_A k T}{2}\right)}{fkT} \]
The \( 2 \) in the numerator and denominator cancels out:
\[ = \frac{nF N_A k T}{fkT} \]

5. **Simplifying Further**:
The \( kT \) in the numerator and denominator also cancels out:
\[ = \frac{nF}{f} \]
Since \( F \) is the degrees of freedom, the expression simplifies to:
\[ = n \]

6. **Conclusion**:
Therefore, we find that:
\[ \frac{2U}{fkT} = n \]
This means that the quantity \( \frac{2U}{fkT} \) represents the number of moles of the gas.

### Final Answer:
The quantity \( \frac{2U}{fkT} \) represents the number of moles of the gas.