A monoatomic gas at a pressure p, having a volume 2V and then adiabatically to a volume 16 V. The final pressure of the gas is (take `gamma = (5)/(3)` )

Options

A. 64p
B. 32 p
C. `(p)/(32)`
D. 16 p

Answer

To solve the problem, we will use the principles of adiabatic processes for an ideal gas. The key relationship we will use is that for an adiabatic process, the product of pressure and volume raised to the power of gamma (γ) remains constant.

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- Initial Pressure: \( P_1 = P \)
- Initial Volume: \( V_1 = 2V \)

2. **Identify the Final Conditions**:
- Final Volume: \( V_2 = 16V \)
- Final Pressure: \( P_2 \) (This is what we need to find)

3. **Use the Adiabatic Condition**:
The relationship for an adiabatic process is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( \gamma = \frac{5}{3} \).

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

5. **Substituting the Known Values**:
Substitute \( P_1 = P \), \( V_1 = 2V \), and \( V_2 = 16V \):
\[ P_2 = P \left( \frac{2V}{16V} \right)^\gamma \]

6. **Simplifying the Volume Ratio**:
The volume ratio simplifies as follows:
\[ \frac{2V}{16V} = \frac{2}{16} = \frac{1}{8} \]
Thus, we can rewrite \( P_2 \):
\[ P_2 = P \left( \frac{1}{8} \right)^\gamma \]

7. **Substituting the Value of Gamma**:
Now, substitute \( \gamma = \frac{5}{3} \):
\[ P_2 = P \left( \frac{1}{8} \right)^{\frac{5}{3}} \]

8. **Calculating \( \left( \frac{1}{8} \right)^{\frac{5}{3}} \)**:
We can express \( \frac{1}{8} \) as \( 2^{-3} \):
\[ P_2 = P \left( 2^{-3} \right)^{\frac{5}{3}} = P \cdot 2^{-5} = \frac{P}{32} \]

9. **Final Result**:
Therefore, the final pressure of the gas is:
\[ P_2 = \frac{P}{32} \]