A gas is compressed adiabatically till its pressure becomes 27 times its initial pressure. Calculate final temperature if initial temperature is`27^@ C` and value of `gamma` is 3/2.

Options

A. `300 K`
B. `600 K`
C. `900 K`
D. `1200 K`

Explanation

To solve the problem step by step, we will use the principles of adiabatic processes in thermodynamics.

### Step 1: Understand the Given Data
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final pressure, \( P_2 = 27 \times P_1 \)
- Value of \( \gamma = \frac{3}{2} \)

### Step 2: Use the Adiabatic Relation
In an adiabatic process, the relationship between pressure and temperature can be expressed as:
\[ \frac{P_1^{1 - \gamma}}{T_1^{\gamma}} = \frac{P_2^{1 - \gamma}}{T_2^{\gamma}} \]
Rearranging this gives:
\[ \left( \frac{P_1}{P_2} \right)^{1 - \gamma} = \left( \frac{T_2}{T_1} \right)^{\gamma} \]

### Step 3: Substitute the Known Values
Substituting \( P_2 = 27 P_1 \) and \( \gamma = \frac{3}{2} \):
\[ \left( \frac{P_1}{27 P_1} \right)^{1 - \frac{3}{2}} = \left( \frac{T_2}{300} \right)^{\frac{3}{2}} \]
This simplifies to:
\[ \left( \frac{1}{27} \right)^{- \frac{1}{2}} = \left( \frac{T_2}{300} \right)^{\frac{3}{2}} \]

### Step 4: Simplify the Equation
The left side becomes:
\[ \sqrt{27} = \left( \frac{T_2}{300} \right)^{\frac{3}{2}} \]
Since \( \sqrt{27} = 3\sqrt{3} \), we can write:
\[ 3\sqrt{3} = \left( \frac{T_2}{300} \right)^{\frac{3}{2}} \]

### Step 5: Raise Both Sides to the Power of \(\frac{2}{3}\)
To eliminate the exponent on the right side, we raise both sides to the power of \(\frac{2}{3}\):
\[ (3\sqrt{3})^{\frac{2}{3}} = \frac{T_2}{300} \]

### Step 6: Calculate \(T_2\)
Calculating \( (3\sqrt{3})^{\frac{2}{3}} \):
\[ 3^{\frac{2}{3}} \cdot (\sqrt{3})^{\frac{2}{3}} = 3^{\frac{2}{3}} \cdot 3^{\frac{1}{3}} = 3^{1} = 3 \]
Thus:
\[ \frac{T_2}{300} = 3 \implies T_2 = 3 \times 300 = 900 \, K \]

### Step 7: Final Answer
The final temperature \( T_2 \) is:
\[ T_2 = 900 \, K \]