A gas is suddenly compressed to `1/4th` of its original volume. Caculate the rise in temperature when original temperature is `27^(@)C. gamma= 1.5`.

Options

A. a.400 K
B. b.500 K
C. c.600 K
D. d.700 K

Explanation

To solve the problem of calculating the rise in temperature when a gas is suddenly compressed to 1/4th of its original volume, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Original temperature, \( T_1 = 27^\circ C = 300 \, K \) (Convert Celsius to Kelvin by adding 273).
- The gas is compressed to \( \frac{1}{4} \) of its original volume, so if the original volume is \( V_1 \), the final volume \( V_2 = \frac{V_1}{4} \).
- The value of \( \gamma = 1.5 \).

2. **Understand the Adiabatic Process:**
- Since the gas is compressed suddenly, we can assume the process is adiabatic. For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

3. **Rearranging the Equation:**
- We need to find \( T_2 \). Rearranging the equation gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

4. **Substituting the Values:**
- Substitute \( V_2 = \frac{V_1}{4} \) into the equation:
\[ T_2 = T_1 \left( \frac{V_1}{\frac{V_1}{4}} \right)^{\gamma - 1} = T_1 \left( 4 \right)^{\gamma - 1} \]
- Now substitute the values:
\[ T_2 = 300 \left( 4 \right)^{1.5 - 1} = 300 \left( 4 \right)^{0.5} \]

5. **Calculating \( 4^{0.5} \):**
- \( 4^{0.5} = 2 \), hence:
\[ T_2 = 300 \times 2 = 600 \, K \]

6. **Calculating the Rise in Temperature:**
- The rise in temperature \( \Delta T \) can be calculated as:
\[ \Delta T = T_2 - T_1 = 600 \, K - 300 \, K = 300 \, K \]

### Final Answer:
The rise in temperature when the gas is suddenly compressed to 1/4th of its original volume is \( 300 \, K \).

---