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 need to find the final pressure of a monoatomic gas after it undergoes an adiabatic expansion from a volume of \(2V\) to \(16V\). Given that the initial pressure is \(p\) and the adiabatic process is characterized by the adiabatic condition, we will use the relation between pressure and volume for adiabatic processes.

### 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, \(P_2 = ?\)

3. **Use the adiabatic condition:**
For a monoatomic gas, the relationship between pressure and volume during an adiabatic process is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \(\gamma = \frac{5}{3}\).

4. **Substitute the known values into the equation:**
\[ p \cdot (2V)^{\frac{5}{3}} = P_2 \cdot (16V)^{\frac{5}{3}} \]

5. **Simplify the equation:**
\[ p \cdot 2^{\frac{5}{3}} \cdot V^{\frac{5}{3}} = P_2 \cdot 16^{\frac{5}{3}} \cdot V^{\frac{5}{3}} \]
Since \(V^{\frac{5}{3}}\) appears on both sides, we can cancel it out:
\[ p \cdot 2^{\frac{5}{3}} = P_2 \cdot 16^{\frac{5}{3}} \]

6. **Express \(P_2\) in terms of \(p\):**
Rearranging gives:
\[ P_2 = p \cdot \frac{2^{\frac{5}{3}}}{16^{\frac{5}{3}}} \]

7. **Simplify the fraction:**
We know that \(16 = 2^4\), thus:
\[ P_2 = p \cdot \frac{2^{\frac{5}{3}}}{(2^4)^{\frac{5}{3}}} = p \cdot \frac{2^{\frac{5}{3}}}{2^{\frac{20}{3}}} = p \cdot 2^{\frac{5}{3} - \frac{20}{3}} = p \cdot 2^{-\frac{15}{3}} = p \cdot 2^{-5} \]

8. **Final expression for \(P_2\):**
\[ P_2 = \frac{p}{32} \]

### Conclusion:
The final pressure of the gas after adiabatic expansion to a volume of \(16V\) is:
\[ \boxed{\frac{p}{32}} \]