An ideal monoatomic gas at 300K expands adiabatically to twice its volume. What is the final temperature?

Explanation

To solve the problem of finding the final temperature of an ideal monoatomic gas that expands adiabatically to twice its volume, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial Temperature, \( T_1 = 300 \, K \)
- Initial Volume, \( V_1 = V \)
- Final Volume, \( V_2 = 2V \)
- For a monoatomic gas, \( \gamma = \frac{5}{3} \)

2. **Use the Adiabatic Process Equation:**
The relationship for an adiabatic process is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

3. **Rearranging the Equation:**
We can rearrange this equation to solve for the final temperature \( T_2 \):
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

4. **Substituting Known Values:**
Substitute \( V_1 = V \) and \( V_2 = 2V \):
\[ T_2 = T_1 \left( \frac{V}{2V} \right)^{\gamma - 1} = T_1 \left( \frac{1}{2} \right)^{\gamma - 1} \]

5. **Calculating \( \gamma - 1 \):**
Since \( \gamma = \frac{5}{3} \):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]

6. **Final Temperature Calculation:**
Now, substituting \( T_1 = 300 \, K \) and \( \gamma - 1 = \frac{2}{3} \):
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

7. **Calculating \( \left( \frac{1}{2} \right)^{\frac{2}{3}} \):**
\[ \left( \frac{1}{2} \right)^{\frac{2}{3}} = \frac{1}{2^{\frac{2}{3}}} \]
This can be calculated as approximately \( 0.39685 \).

8. **Final Calculation:**
\[ T_2 \approx 300 \times 0.39685 \approx 119.06 \, K \]

9. **Conclusion:**
The final temperature \( T_2 \) is approximately \( 189 \, K \).