A mixture of gases at NTP for which `gamma = 1.5` si suddenly compressed to `1/9th` of its original volume. The final temperature of mixture is

Options

A. `300^@C`
B. `546^@C`
C. `420^@C`
D. `872^@C`

Answer

To solve the problem of finding the final temperature of a gas mixture that is suddenly compressed to \( \frac{1}{9} \) of its original volume, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The gas is at Normal Temperature and Pressure (NTP).
- At NTP, the temperature \( T_1 \) is \( 0^\circ C \) which converts to Kelvin as:
\[ T_1 = 0 + 273 = 273 \, \text{K} \]

2. **Determine the Volume Ratio:**
- The gas is compressed to \( \frac{1}{9} \) of its original volume. Therefore, we can express the volumes as:
\[ V_2 = \frac{1}{9} V_1 \]
- This gives us the volume ratio:
\[ \frac{V_1}{V_2} = 9 \]

3. **Use the Adiabatic Process Relation:**
- For an adiabatic process, we know that:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
- Here, \( \gamma = 1.5 \).

4. **Substitute the Known Values:**
- Rearranging the equation to find \( T_2 \):
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
- Substitute \( T_1 = 273 \, \text{K} \), \( \frac{V_1}{V_2} = 9 \), and \( \gamma - 1 = 0.5 \):
\[ T_2 = 273 \left( 9 \right)^{0.5} \]

5. **Calculate \( (9)^{0.5} \):**
- The square root of 9 is:
\[ (9)^{0.5} = 3 \]

6. **Final Calculation for \( T_2 \):**
- Substitute back into the equation:
\[ T_2 = 273 \times 3 = 819 \, \text{K} \]

7. **Convert Kelvin to Celsius:**
- To convert back to Celsius:
\[ T_2 = 819 - 273 = 546 \, \text{°C} \]

### Final Answer:
The final temperature of the gas mixture after compression is \( 546 \, \text{°C} \).

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