A diatomic gas initially at `18^(@)` is compressed adiabatically to one- eighth of its original volume. The temperature after compression will b

Options

A. `18^(@)C`
B. `668.4 K`
C. `395.4^(@) C`
D. `144^(@)C`

Answer

To solve the problem of finding the final temperature of a diatomic gas after it is compressed adiabatically to one-eighth of its original volume, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial temperature \( T_i = 18^\circ C \)
- Convert to Kelvin:
\[ T_i = 18 + 273 = 291 \, K \]

2. **Determine the Volume Change:**
- The gas is compressed to one-eighth of its original volume:
\[ V_f = \frac{V_i}{8} \]

3. **Find the Value of \( \gamma \):**
- For a diatomic gas, the ratio of specific heats \( \gamma \) is given by:
\[ \gamma = \frac{C_p}{C_v} = \frac{7R/2}{5R/2} = \frac{7}{5} \]

4. **Use the Adiabatic Relation:**
- The adiabatic relation for an ideal gas is given by:
\[ T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1} \]
- Rearranging gives:
\[ T_f = T_i \left( \frac{V_i}{V_f} \right)^{\gamma - 1} \]

5. **Substituting Values:**
- Since \( V_f = \frac{V_i}{8} \):
\[ \frac{V_i}{V_f} = 8 \]
- Substitute into the equation:
\[ T_f = 291 \left( 8 \right)^{\frac{7}{5} - 1} \]
- Simplifying the exponent:
\[ \frac{7}{5} - 1 = \frac{2}{5} \]
- Therefore:
\[ T_f = 291 \times 8^{\frac{2}{5}} \]

6. **Calculate \( 8^{\frac{2}{5}} \):**
- Calculate \( 8^{\frac{2}{5}} \):
\[ 8^{\frac{2}{5}} = (2^3)^{\frac{2}{5}} = 2^{\frac{6}{5}} = 2^{1.2} \approx 2.297 \]

7. **Final Calculation:**
- Now calculate \( T_f \):
\[ T_f \approx 291 \times 2.297 \approx 668.4 \, K \]

### Final Answer:
The temperature after compression will be approximately \( 668.4 \, K \).