Dry air at one atmospheric pressure is suddenly compressed so that its volume becomes one-fourth. Its pressure will become –`(gamma= 1.5)`

Options

A. 4 atm
B. 8 atm
C. 16 atm
D. 32 atm

Answer

To solve the problem of determining the pressure of dry air after it is suddenly compressed to one-fourth of its original volume, we will use the principles of an adiabatic process. Here are the steps to arrive at the solution:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, the relationship between pressure (P) and volume (V) is given by the equation:
\[ P V^\gamma = \text{constant} \]
where \(\gamma\) (gamma) is the heat capacity ratio. In this case, \(\gamma = 1.5\).

### Step 2: Define Initial and Final Conditions
Let:
- Initial pressure, \(P_1 = 1 \text{ atm}\)
- Initial volume, \(V_1 = V\)
- Final volume, \(V_2 = \frac{V}{4}\)
- Final pressure, \(P_2\) (which we need to find)

### Step 3: Apply the Adiabatic Condition
Using the adiabatic condition, we can write:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 1 \text{ atm} \cdot V^\gamma = P_2 \cdot \left(\frac{V}{4}\right)^\gamma \]

### Step 5: Simplify the Equation
Rearranging the equation gives:
\[ P_2 = \frac{1 \text{ atm} \cdot V^\gamma}{\left(\frac{V}{4}\right)^\gamma} \]

### Step 6: Simplify Further
This can be simplified as follows:
\[ P_2 = \frac{1 \text{ atm} \cdot V^\gamma}{\frac{V^\gamma}{4^\gamma}} \]
\[ P_2 = 1 \text{ atm} \cdot 4^\gamma \]

### Step 7: Calculate \(4^\gamma\)
Now, substituting \(\gamma = 1.5\):
\[ P_2 = 1 \text{ atm} \cdot 4^{1.5} \]
\[ P_2 = 1 \text{ atm} \cdot (4^{3/2}) \]
\[ P_2 = 1 \text{ atm} \cdot (2^3) \]
\[ P_2 = 1 \text{ atm} \cdot 8 \]

### Step 8: Final Result
Thus, the final pressure \(P_2\) is:
\[ P_2 = 8 \text{ atm} \]

### Conclusion
The pressure of the dry air after it is compressed to one-fourth of its volume is 8 atm.

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