For a diatomic gas, the total kinetic energy per gram molecule is

Options

A. `(3RT)/2`
B. `4/2(RT)`
C. `(5RT)/2`
D. `(6RT)/2`

Answer

To find the total kinetic energy per gram molecule of a diatomic gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Kinetic Energy**:
- The kinetic energy of a gas molecule is related to its degrees of freedom. For a diatomic gas, the degrees of freedom (f) are 5 (3 translational and 2 rotational).

2. **Using the Equipartition Theorem**:
- According to the equipartition theorem, the average kinetic energy per degree of freedom is given by:
\[ \text{K.E. per degree of freedom} = \frac{1}{2} kT \]
- Here, \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.

3. **Calculating Total Kinetic Energy for f Degrees of Freedom**:
- For a gas with \( f \) degrees of freedom, the total kinetic energy per molecule is:
\[ \text{Total K.E.} = \frac{f}{2} kT \]

4. **Finding Kinetic Energy for One Mole**:
- To find the total kinetic energy for one mole of gas, we multiply the kinetic energy per molecule by Avogadro's number (\( N_a \)):
\[ \text{Total K.E. for 1 mole} = \frac{f}{2} kT \times N_a \]

5. **Substituting Boltzmann Constant**:
- We know that \( k = \frac{R}{N_a} \), where \( R \) is the universal gas constant. Substituting this into the equation gives:
\[ \text{Total K.E. for 1 mole} = \frac{f}{2} \left(\frac{R}{N_a}\right) T \times N_a = \frac{f}{2} RT \]

6. **Applying the Value of f for Diatomic Gas**:
- For a diatomic gas, \( f = 5 \). Substituting this value into the equation gives:
\[ \text{Total K.E. for 1 mole} = \frac{5}{2} RT \]

7. **Final Result**:
- Thus, the total kinetic energy per gram molecule of a diatomic gas is:
\[ \frac{5}{2} RT \]

### Final Answer:
The total kinetic energy per gram molecule of a diatomic gas is \( \frac{5}{2} RT \).

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