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. `450 K`
B. `375 K`
C. `225 K`
D. `405 K`

Explanation

To solve the problem of finding the rise in temperature of an ideal gas that is compressed adiabatically, we can follow these steps:

### Step 1: Understand the Given Information
- Initial temperature, \( T_1 = 27^\circ C \)
- Volume after compression, \( V_2 = \frac{8}{27} V_1 \)
- Adiabatic process constant, \( \gamma = \frac{5}{3} \)

### Step 2: Convert Initial Temperature to Kelvin
To convert the initial temperature from Celsius to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Use the Adiabatic Relation
For an adiabatic process, we can use the relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Rearranging this gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
- \( V_2 = \frac{8}{27} V_1 \)
- Therefore, \( \frac{V_1}{V_2} = \frac{V_1}{\frac{8}{27} V_1} = \frac{27}{8} \)

Now substituting into the equation for \( T_2 \):
\[ T_2 = 300 \left( \frac{27}{8} \right)^{\frac{5}{3} - 1} \]
Calculating \( \gamma - 1 \):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
Thus:
\[ T_2 = 300 \left( \frac{27}{8} \right)^{\frac{2}{3}} \]

### Step 5: Calculate \( \left( \frac{27}{8} \right)^{\frac{2}{3}} \)
Calculating \( \left( \frac{27}{8} \right)^{\frac{2}{3}} \):
\[ \left( \frac{27}{8} \right)^{\frac{2}{3}} = \frac{27^{\frac{2}{3}}}{8^{\frac{2}{3}}} = \frac{9}{4} \]
Thus:
\[ T_2 = 300 \times \frac{9}{4} = 675 \, K \]

### Step 6: Calculate the Rise in Temperature
The rise in temperature \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = 675 \, K - 300 \, K = 375 \, K \]

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

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