A monoatomic gas at pressure `P_(1)` and volume `V_(1)` is compressed adiabatically to `1/8th` of its original volume. What is the final pressure of gas.

Options

A. `64 P_(1)`
B. `P_(1)`
C. `16P_(1)`
D. `32P_(1)`

Explanation

To solve the problem of finding the final pressure of a monoatomic gas that is compressed adiabatically to 1/8th of its original volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given values**:
- Initial pressure: \( P_1 \)
- Initial volume: \( V_1 \)
- Final volume: \( V_2 = \frac{1}{8} V_1 \)

2. **Identify the relationship for adiabatic processes**:
For an adiabatic process involving an ideal gas, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( \gamma \) (gamma) is the heat capacity ratio. For a monoatomic gas, \( \gamma = \frac{5}{3} \).

3. **Substituting the known values**:
We can rearrange the equation to solve for \( P_2 \):
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

4. **Calculate \( \frac{V_1}{V_2} \)**:
Since \( V_2 = \frac{1}{8} V_1 \):
\[ \frac{V_1}{V_2} = \frac{V_1}{\frac{1}{8} V_1} = 8 \]

5. **Substituting into the equation for \( P_2 \)**:
Now substituting \( \frac{V_1}{V_2} = 8 \) into the equation:
\[ P_2 = P_1 \times 8^\gamma \]

6. **Substituting the value of \( \gamma \)**:
For a monoatomic gas, \( \gamma = \frac{5}{3} \):
\[ P_2 = P_1 \times 8^{\frac{5}{3}} \]

7. **Calculating \( 8^{\frac{5}{3}} \)**:
We can simplify \( 8^{\frac{5}{3}} \):
\[ 8 = 2^3 \implies 8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^{5} = 32 \]

8. **Final expression for \( P_2 \)**:
Therefore, we have:
\[ P_2 = 32 P_1 \]

### Final Answer:
The final pressure of the gas after adiabatic compression to 1/8th of its original volume is:
\[ P_2 = 32 P_1 \]