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

Options

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

Answer

To solve the problem, we will use the principles of an adiabatic process for an ideal gas. Here are the steps:

### Step 1: Identify the initial conditions
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)

### Step 2: Determine the final pressure
- Final pressure, \( P_2 = 8 \times P_1 = 8 \times 1 \, \text{atm} = 8 \, \text{atm} \)

### Step 3: Use the adiabatic relation
For an adiabatic process, the relation between pressure and temperature can be expressed as:
\[ \frac{P_1 T_1^{\frac{\gamma}{\gamma - 1}}}{P_2 T_2^{\frac{\gamma}{\gamma - 1}}} = 1 \]
Where \( \gamma = \frac{3}{2} \).

### Step 4: Rearranging the equation
Rearranging the equation gives us:
\[ \frac{P_1}{P_2} = \left( \frac{T_2}{T_1} \right)^{\frac{\gamma}{\gamma - 1}} \]
Substituting the known values:
\[ \frac{1}{8} = \left( \frac{T_2}{300} \right)^{\frac{3/2}{(3/2) - 1}} \]
Calculating \( \frac{3/2}{(3/2) - 1} = \frac{3/2}{1/2} = 3 \):
\[ \frac{1}{8} = \left( \frac{T_2}{300} \right)^{3} \]

### Step 5: Solving for \( T_2 \)
Taking the cube root of both sides:
\[ \frac{T_2}{300} = \left( \frac{1}{8} \right)^{\frac{1}{3}} = \frac{1}{2} \]
Thus, we have:
\[ T_2 = 300 \times 2 = 600 \, \text{K} \]

### Step 6: Convert \( T_2 \) to Celsius
To convert the final temperature back to Celsius:
\[ T_2 = 600 - 273 = 327^\circ C \]

### Final Answer
The final temperature of the gas after adiabatic compression is:
\[ \boxed{327^\circ C} \]

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