An ideal monatomic gas at 300 K expands adiabatically to 8 times its volume . What is the final temperature ?

Options

A. 75 K
B. 300 K
C. 560 K
D. 340 K

Answer

To find the final temperature of an ideal monatomic gas that expands adiabatically to 8 times its volume, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial Temperature, \( T_1 = 300 \, \text{K} \)
- Initial Volume, \( V_1 = V \) (let's denote the initial volume as \( V \))
- Final Volume, \( V_2 = 8V \)

2. **Determine the Value of Gamma (\( \gamma \)):**
- For a monatomic ideal gas, \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \).

3. **Use the Adiabatic Condition:**
- The relationship for an adiabatic process is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Rearranging this gives us:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

4. **Substituting the Known Values:**
- Substitute \( T_1 = 300 \, \text{K} \), \( V_1 = V \), and \( V_2 = 8V \):
\[ T_2 = 300 \left( \frac{V}{8V} \right)^{\frac{5}{3} - 1} \]
- Simplifying the volume ratio:
\[ T_2 = 300 \left( \frac{1}{8} \right)^{\frac{2}{3}} \]

5. **Calculating \( \left( \frac{1}{8} \right)^{\frac{2}{3}} \):**
- Since \( 8 = 2^3 \), we have:
\[ \left( \frac{1}{8} \right)^{\frac{2}{3}} = \left( 2^{-3} \right)^{\frac{2}{3}} = 2^{-2} = \frac{1}{4} \]

6. **Final Calculation:**
- Substitute back into the equation for \( T_2 \):
\[ T_2 = 300 \times \frac{1}{4} = 75 \, \text{K} \]

### Final Answer:
The final temperature \( T_2 \) after the adiabatic expansion is \( 75 \, \text{K} \).