Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature `T_0`, while Box contains one mole of helium at temperature `(7/3)T_0`. The boxes are then put into thermal contact with each other, and heat flows between them until the gasses reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gasses, `T_f` in terms of `T_0` is

Options

A. `T_(f)=7/3 T_(0)`
B. `T_(f) = 3/2 T_(0)`
C. `T_(f) = 5/2 T_(0)`
D. `T_(f) = 3/7 T_(0)`

Explanation

To solve the problem, we will apply the principle of conservation of energy, which states that the heat lost by the hotter gas (helium) will be equal to the heat gained by the cooler gas (nitrogen) until they reach a common final temperature \( T_f \).

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- Box A (Nitrogen):
- Number of moles \( n_A = 1 \)
- Initial temperature \( T_A = T_0 \)
- Box B (Helium):
- Number of moles \( n_B = 1 \)
- Initial temperature \( T_B = \frac{7}{3} T_0 \)

2. **Determine the Type of Gases**:
- Nitrogen (N₂) is a diatomic gas, so its heat capacity at constant volume \( C_{v,A} = \frac{5}{2} R \).
- Helium (He) is a monoatomic gas, so its heat capacity at constant volume \( C_{v,B} = \frac{3}{2} R \).

3. **Set Up Heat Transfer Equations**:
- Heat lost by helium:
\[ Q_B = n_B C_{v,B} (T_B - T_f) = 1 \cdot \frac{3}{2} R \left(\frac{7}{3} T_0 - T_f\right) \]
- Heat gained by nitrogen:
\[ Q_A = n_A C_{v,A} (T_f - T_A) = 1 \cdot \frac{5}{2} R (T_f - T_0) \]

4. **Apply Conservation of Energy**:
- Since the heat lost by helium equals the heat gained by nitrogen:
\[ Q_B = Q_A \]
- Substitute the expressions for \( Q_B \) and \( Q_A \):
\[ \frac{3}{2} R \left(\frac{7}{3} T_0 - T_f\right) = \frac{5}{2} R (T_f - T_0) \]

5. **Cancel Out Common Factors**:
- We can cancel \( \frac{R}{2} \) from both sides:
\[ 3 \left(\frac{7}{3} T_0 - T_f\right) = 5 (T_f - T_0) \]

6. **Simplify the Equation**:
- Distributing on both sides:
\[ 7 T_0 - 3 T_f = 5 T_f - 5 T_0 \]
- Rearranging gives:
\[ 7 T_0 + 5 T_0 = 3 T_f + 5 T_f \]
\[ 12 T_0 = 8 T_f \]

7. **Solve for Final Temperature \( T_f \)**:
- Dividing both sides by 8:
\[ T_f = \frac{12 T_0}{8} = \frac{3 T_0}{2} \]

### Final Answer:
The final temperature \( T_f \) in terms of \( T_0 \) is:
\[ T_f = \frac{3 T_0}{2} \]