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

Explanation

To find the final temperature \( T_f \) of the gases in the two boxes after they are put in thermal contact, we can use the principle of conservation of energy. The heat lost by the hotter gas (helium) will be equal to the heat gained by the cooler gas (nitrogen).

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Box A (Nitrogen):
- Number of moles, \( n_A = 1 \)
- Initial temperature, \( T_A = T_0 \)
- Molar heat capacity at constant volume, \( C_{vA} = \frac{5}{2}R \) (since nitrogen is a diatomic gas)
- Box B (Helium):
- Number of moles, \( n_B = 1 \)
- Initial temperature, \( T_B = \frac{7}{3}T_0 \)
- Molar heat capacity at constant volume, \( C_{vB} = \frac{3}{2}R \) (since helium is a monatomic gas)

2. **Set Up the Heat Transfer Equation**:
- Heat lost by helium:
\[ Q_{\text{lost}} = n_B C_{vB} (T_B - T_f) = 1 \cdot \frac{3}{2}R \left(\frac{7}{3}T_0 - T_f\right) \]
- Heat gained by nitrogen:
\[ Q_{\text{gained}} = n_A C_{vA} (T_f - T_A) = 1 \cdot \frac{5}{2}R (T_f - T_0) \]

3. **Apply Conservation of Energy**:
- According to the conservation of energy, the heat lost by helium equals the heat gained by nitrogen:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
- Substituting the expressions from above:
\[ \frac{3}{2}R \left(\frac{7}{3}T_0 - T_f\right) = \frac{5}{2}R (T_f - T_0) \]

4. **Cancel \( R \) and Simplify**:
- Cancel \( R \) from both sides:
\[ \frac{3}{2} \left(\frac{7}{3}T_0 - T_f\right) = \frac{5}{2} (T_f - T_0) \]
- Multiply through by 2 to eliminate the fractions:
\[ 3 \left(\frac{7}{3}T_0 - T_f\right) = 5 (T_f - T_0) \]
- Distributing gives:
\[ 7T_0 - 3T_f = 5T_f - 5T_0 \]

5. **Rearranging the Equation**:
- Combine like terms:
\[ 7T_0 + 5T_0 = 3T_f + 5T_f \]
\[ 12T_0 = 8T_f \]

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

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