A mass of dry air at NTP is compressed to (1)/(20)th of its original volume suddenly . If `gamma=1.4` , the final pressure would be

Options

A. 20 atm
B. 66.28 atm
C. 30 atm
D. 150 atm

Explanation

To find the final pressure of a mass of dry air that is compressed to \( \frac{1}{20} \) of its original volume, we can use the adiabatic process equation for an ideal gas, which states that:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Where:
- \( P_1 \) is the initial pressure,
- \( V_1 \) is the initial volume,
- \( P_2 \) is the final pressure,
- \( V_2 \) is the final volume,
- \( \gamma \) is the heat capacity ratio (given as 1.4).

### Step 1: Identify the initial conditions
At Normal Temperature and Pressure (NTP), the standard conditions are:
- \( P_1 = 1 \, \text{atm} \)
- \( V_1 = V \) (original volume)

### Step 2: Determine the final volume
The final volume \( V_2 \) is given as:
\[ V_2 = \frac{1}{20} V_1 = \frac{1}{20} V \]

### Step 3: Substitute values into the adiabatic equation
Using the adiabatic process equation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Substituting the known values:
\[ 1 \, \text{atm} \cdot V^\gamma = P_2 \left(\frac{1}{20} V\right)^\gamma \]

### Step 4: Simplify the equation
Rearranging the equation to solve for \( P_2 \):
\[ P_2 = \frac{1 \, \text{atm} \cdot V^\gamma}{\left(\frac{1}{20} V\right)^\gamma} \]

This simplifies to:
\[ P_2 = 1 \, \text{atm} \cdot \frac{V^\gamma}{\left(\frac{1}{20^\gamma} V^\gamma\right)} = 1 \, \text{atm} \cdot 20^\gamma \]

### Step 5: Calculate \( P_2 \)
Now substituting \( \gamma = 1.4 \):
\[ P_2 = 1 \, \text{atm} \cdot 20^{1.4} \]

Calculating \( 20^{1.4} \):
\[ 20^{1.4} \approx 66.2890 \]

Thus, the final pressure \( P_2 \) is approximately:
\[ P_2 \approx 66.2890 \, \text{atm} \]

### Final Answer
The final pressure after compression is approximately \( 66.2890 \, \text{atm} \).

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