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, we will use the relationship for an adiabatic process, which states that:

\[ T_1 \cdot V_1^{\gamma - 1} = T_2 \cdot V_2^{\gamma - 1} \]

Where:
- \( T_1 \) = initial temperature
- \( V_1 \) = initial volume
- \( T_2 \) = final temperature
- \( V_2 \) = final volume
- \( \gamma \) = ratio of specific heats (given as 1.5)

### Step-by-Step Solution:

**Step 1: Identify the initial conditions.**
- Given that the gas is at Normal Temperature and Pressure (NTP), the initial temperature \( T_1 \) is 0°C, which is equivalent to 273 K.
- The initial volume \( V_1 \) is denoted as \( V \).

**Step 2: Determine the final volume.**
- The gas is compressed to \( \frac{1}{9} \) of its original volume, so:
\[ V_2 = \frac{V_1}{9} = \frac{V}{9} \]

**Step 3: Use the adiabatic process equation.**
- Substitute the known values into the adiabatic equation:
\[ T_1 \cdot V_1^{\gamma - 1} = T_2 \cdot V_2^{\gamma - 1} \]
This can be rewritten as:
\[ 273 \cdot V^{1.5 - 1} = T_2 \cdot \left(\frac{V}{9}\right)^{1.5 - 1} \]
Simplifying the equation gives:
\[ 273 \cdot V^{0.5} = T_2 \cdot \left(\frac{V}{9}\right)^{0.5} \]

**Step 4: Simplify the equation.**
- Rearranging gives:
\[ T_2 = 273 \cdot \frac{V^{0.5}}{\left(\frac{V}{9}\right)^{0.5}} \]
This can be simplified to:
\[ T_2 = 273 \cdot \frac{V^{0.5}}{V^{0.5} \cdot \left(\frac{1}{9}\right)^{0.5}} = 273 \cdot 9^{0.5} \]

**Step 5: Calculate \( 9^{0.5} \).**
- Since \( 9^{0.5} = 3 \), we have:
\[ T_2 = 273 \cdot 3 = 819 \text{ K} \]

**Step 6: Convert the final temperature to Celsius.**
- To convert from Kelvin to Celsius:
\[ T_2 = 819 \text{ K} - 273 = 546 \text{ °C} \]

### Final Answer:
The final temperature of the mixture is **546 °C**.