An ideal gas at pressure of 1 atmosphere and temperature of `27^(@)C` is compressed adiabatically until its pressure becomes 8 times the initial pressure, then the final temperature is `(gamma=3//2)`

Options

A. `627^(@)C`
B. `527^(@)C`
C. `427^(@)C`
D. `327^(@)C`

Explanation

To find the final temperature of an ideal gas that is compressed adiabatically, we can use the adiabatic process equation for an ideal gas:

\[ \frac{P_1}{T_1^{\gamma}} = \frac{P_2}{T_2^{\gamma}} \]

Where:
- \( P_1 \) = initial pressure
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( T_2 \) = final temperature (in Kelvin)
- \( \gamma \) = heat capacity ratio (given as \( \frac{3}{2} \))

### Step 1: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as \( 27^\circ C \). To convert this to Kelvin:

\[ T_1 = 27 + 273 = 300 \, K \]

### Step 2: Determine the initial and final pressures
The initial pressure \( P_1 \) is given as \( 1 \, atm \). The final pressure \( P_2 \) is 8 times the initial pressure:

\[ P_2 = 8 \times P_1 = 8 \times 1 \, atm = 8 \, atm \]

### Step 3: Substitute the known values into the adiabatic equation
Now we can substitute the values into the adiabatic equation:

\[ \frac{1 \, atm}{(300 \, K)^{\frac{3}{2}}} = \frac{8 \, atm}{T_2^{\frac{3}{2}}} \]

### Step 4: Rearrange the equation to solve for \( T_2 \)
Cross-multiplying gives:

\[ 1 \, atm \cdot T_2^{\frac{3}{2}} = 8 \, atm \cdot (300 \, K)^{\frac{3}{2}} \]

Now, divide both sides by \( 1 \, atm \):

\[ T_2^{\frac{3}{2}} = 8 \cdot (300)^{\frac{3}{2}} \]

### Step 5: Calculate \( (300)^{\frac{3}{2}} \)
Calculating \( (300)^{\frac{3}{2}} \):

\[ (300)^{\frac{3}{2}} = 300 \cdot \sqrt{300} \approx 300 \cdot 17.32 \approx 5196 \]

### Step 6: Substitute back to find \( T_2 \)
Now substituting back:

\[ T_2^{\frac{3}{2}} = 8 \cdot 5196 \approx 41568 \]

To find \( T_2 \), we take the \( \frac{2}{3} \) power:

\[ T_2 = (41568)^{\frac{2}{3}} \]

Calculating \( (41568)^{\frac{2}{3}} \):

\[ T_2 \approx 335.5 \, K \]

### Final Answer
The final temperature \( T_2 \) is approximately \( 335.5 \, K \).