A mass of diatomic gas`(gamma=1.4)` at a pressure of 2 atomphere is compressed adiabitically so that its temperature rises from `27^(@)C` to `927^(@)C`. The pressure of the gas in the final state is

Options

A. 8 atm
B. 28 atm
C. 68.7 atm
D. 256 atm

Explanation

To solve the problem, we will use the adiabatic process equations for an ideal gas. The key relationship for an adiabatic process is given by:

\[ \frac{T_1}{T_2} = \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}} \]

Where:
- \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin,
- \(P_1\) and \(P_2\) are the initial and final pressures,
- \(\gamma\) is the heat capacity ratio (given as 1.4 for diatomic gas).

### Step 1: Convert Temperatures from Celsius to Kelvin
- Initial temperature \(T_1 = 27^\circ C = 27 + 273 = 300 \, K\)
- Final temperature \(T_2 = 927^\circ C = 927 + 273 = 1200 \, K\)

### Step 2: Identify Initial Pressure
- Initial pressure \(P_1 = 2 \, atm\)

### Step 3: Use the Adiabatic Relation
We rearrange the adiabatic relation to find \(P_2\):

\[ P_2 = P_1 \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma - 1}} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:

\[ P_2 = 2 \, atm \left(\frac{1200 \, K}{300 \, K}\right)^{\frac{1.4}{1.4 - 1}} \]

### Step 5: Simplify the Temperature Ratio
Calculate the temperature ratio:

\[ \frac{1200}{300} = 4 \]

### Step 6: Calculate the Exponent
Now calculate the exponent:

\[ \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = 3.5 \]

### Step 7: Substitute Back into the Pressure Equation
Now substituting back into the pressure equation:

\[ P_2 = 2 \, atm \cdot 4^{3.5} \]

### Step 8: Calculate \(4^{3.5}\)
Calculating \(4^{3.5}\):

\[ 4^{3.5} = (2^2)^{3.5} = 2^{7} = 128 \]

### Step 9: Final Calculation of Pressure
Now calculate \(P_2\):

\[ P_2 = 2 \cdot 128 = 256 \, atm \]

### Conclusion
The final pressure of the gas in the final state is:

\[ \boxed{256 \, atm} \]