An ideal gas at `27^(@)C` is compressed adiabatically to `8//27` of its original volume. If `gamma = 5//3`, then the rise in temperature is

Options

A. `475K`
B. `402K
C. `275K`
D. `375K`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Initial temperature \( T_1 = 27^\circ C \)
- Convert \( T_1 \) to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The volume is compressed to \( \frac{8}{27} \) of its original volume, so:
\[ V_2 = \frac{8}{27} V_1 \]
- The value of \( \gamma = \frac{5}{3} \).

### Step 2: Use the adiabatic process relation
For an adiabatic process, the relation between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Rearranging this gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

### Step 3: Substitute the values into the equation
Substituting \( V_2 = \frac{8}{27} V_1 \):
\[ T_2 = T_1 \left( \frac{V_1}{\frac{8}{27} V_1} \right)^{\gamma - 1} \]
This simplifies to:
\[ T_2 = T_1 \left( \frac{27}{8} \right)^{\gamma - 1} \]

### Step 4: Calculate \( \gamma - 1 \)
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]

### Step 5: Substitute \( T_1 \) and \( \gamma - 1 \) into the equation
Now substituting \( T_1 = 300 \, K \):
\[ T_2 = 300 \left( \frac{27}{8} \right)^{\frac{2}{3}} \]

### Step 6: Calculate \( \left( \frac{27}{8} \right)^{\frac{2}{3}} \)
Calculating \( \left( \frac{27}{8} \right)^{\frac{2}{3}} \):
\[ \frac{27}{8} = \left( \frac{3^3}{2^3} \right) = \left( \frac{3}{2} \right)^3 \]
Thus,
\[ \left( \frac{27}{8} \right)^{\frac{2}{3}} = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \]

### Step 7: Substitute back to find \( T_2 \)
Now substituting back:
\[ T_2 = 300 \times \frac{9}{4} = 675 \, K \]

### Step 8: Calculate the rise in temperature \( \Delta T \)
The rise in temperature is given by:
\[ \Delta T = T_2 - T_1 = 675 \, K - 300 \, K = 375 \, K \]

### Final Answer
The rise in temperature is \( \Delta T = 375 \, K \).

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