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 the Initial Conditions:**
- Initial temperature \( T_1 = 300 \, K \)
- Initial volume \( V_1 \)
- Final volume \( V_2 = 2V_1 \)

2. **Use the Adiabatic Condition:**
- For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Here, \( \gamma \) (gamma) for a monoatomic gas is \( \frac{5}{3} \).

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

4. **Substitute the Known Values:**
- Substitute \( T_1 = 300 \, K \), \( V_2 = 2V_1 \):
\[ T_2 = 300 \left( \frac{V_1}{2V_1} \right)^{\gamma - 1} \]
- This simplifies to:
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\gamma - 1} \]

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

6. **Substitute \( \gamma - 1 \) into the Equation:**
- Now substituting \( \gamma - 1 \):
\[ T_2 = 300 \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

7. **Calculate \( \left( \frac{1}{2} \right)^{\frac{2}{3}} \):**
- This can be calculated as:
\[ \left( \frac{1}{2} \right)^{\frac{2}{3}} = \frac{1}{\sqrt[3]{4}} \approx 0.7937 \]

8. **Final Calculation of \( T_2 \):**
- Now, substituting this value back into the equation for \( T_2 \):
\[ T_2 \approx 300 \times 0.7937 \approx 238.11 \, K \]

9. **Conclusion:**
- The final temperature \( T_2 \) after the adiabatic expansion is approximately \( 238.11 \, K \).

### Final Answer:
The final temperature of the gas after adiabatic expansion is approximately **238.11 K**.