`1mm^3` of a gas is compressed at 1 atmospheric pressure and temperature `27^@C` to `627^@C`. What is the final pressure under adiabatic condition `(gamma` for the gas `=1.5`)

Options

A. `27xx10^5(N)/(m^2)`
B. `80xx10^5(N)/(m^2)`
C. `36xx10^5(N)/(m^2)`
D. `56xx10^5(N)/(m^2)`

Explanation

To solve the problem of finding the final pressure of the gas under adiabatic conditions, we can follow these steps:

### Step 1: Convert Temperatures to Kelvin
Given:
- Initial temperature, \( T_1 = 27^\circ C \)
- Final temperature, \( T_2 = 627^\circ C \)

Convert these temperatures to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
\[ T_2 = 627 + 273 = 900 \, K \]

### Step 2: Identify Given Values
We have:
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- \( \gamma = 1.5 \)

### Step 3: Use the Adiabatic Relation
For an adiabatic process, the relation between temperature and pressure is given by:
\[ \frac{T_1^\gamma}{P_1^{\gamma - 1}} = \frac{T_2^\gamma}{P_2^{\gamma - 1}} \]
Rearranging this gives:
\[ \frac{P_2}{P_1^{\gamma - 1}} = \frac{T_2}{T_1} \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ \frac{P_2}{1^{1.5 - 1}} = \frac{900}{300} \]
This simplifies to:
\[ P_2 = 1^{0.5} \cdot 3 = 3 \, \text{atm} \]

### Step 5: Convert Pressure to Newton per Meter Square
To convert the pressure from atm to Newton per meter square (N/m²), we use the conversion factor:
\[ 1 \, \text{atm} = 101325 \, \text{N/m}^2 \]
Thus,
\[ P_2 = 3 \, \text{atm} \times 101325 \, \text{N/m}^2/\text{atm} = 303975 \, \text{N/m}^2 \]

### Final Result
The final pressure \( P_2 \) under adiabatic conditions is:
\[ P_2 = 303975 \, \text{N/m}^2 \quad \text{(or approximately } 2.74 \times 10^5 \text{ N/m}^2\text{)} \]