The temperature of a definite amount of an ideal monoatomic gas becomes four times in a reversible process for which heat exchange is zero. Which of the following is correct relation between the final and initial parameteres of gas?

Options

A. `V_(f)=8V_(i)`
B. `P_(f)=32P_(i)`
C. `V_(f)=16V_(i)`
D. `P_(f)=(1)/(16)P_(i)`

Answer

To solve the problem, we need to analyze the behavior of an ideal monoatomic gas undergoing a reversible process where the temperature increases to four times its initial value, and there is no heat exchange (adiabatic process).

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions**:
- Let the initial temperature be \( T_1 \).
- The final temperature is given as \( T_2 = 4T_1 \).
- We denote the initial volume as \( V_1 \) and the final volume as \( V_2 \).
- The initial pressure is \( P_1 \) and the final pressure is \( P_2 \).

2. **Use the Adiabatic Condition**:
- For an ideal gas undergoing an adiabatic process, we can use the relation:
\[ PV^{\gamma} = \text{constant} \]
- For a monoatomic gas, \( \gamma = \frac{5}{3} \).

3. **Apply the Adiabatic Relation**:
- From the adiabatic condition, we have:
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]
- Rearranging gives:
\[ \frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma} \]

4. **Use the Ideal Gas Law**:
- The ideal gas law states that:
\[ PV = nRT \]
- For two states of the gas, we can write:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
- Rearranging gives:
\[ \frac{P_2}{P_1} = \frac{T_2 V_1}{T_1 V_2} \]

5. **Substitute Known Values**:
- Substitute \( T_2 = 4T_1 \) into the equation:
\[ \frac{P_2}{P_1} = \frac{4T_1 V_1}{T_1 V_2} = \frac{4 V_1}{V_2} \]

6. **Combine the Two Relations**:
- Now we have two expressions for \( \frac{P_2}{P_1} \):
1. From the adiabatic relation: \( \frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\frac{5}{3}} \)
2. From the ideal gas law: \( \frac{P_2}{P_1} = \frac{4 V_1}{V_2} \)

7. **Set the Two Expressions Equal**:
- Equating the two expressions:
\[ \left( \frac{V_1}{V_2} \right)^{\frac{5}{3}} = \frac{4 V_1}{V_2} \]

8. **Solve for Volume Ratio**:
- Let \( x = \frac{V_1}{V_2} \):
\[ x^{\frac{5}{3}} = 4x \implies x^{\frac{5}{3} - 1} = 4 \implies x^{\frac{2}{3}} = 4 \implies x = 4^{\frac{3}{2}} = 8 \]
- Thus, \( \frac{V_1}{V_2} = 8 \) or \( V_1 = 8V_2 \).

9. **Find Pressure Relation**:
- Substitute \( V_1 = 8V_2 \) back into the pressure relation:
\[ \frac{P_2}{P_1} = \frac{4 \cdot 8V_2}{V_2} = 32 \implies P_2 = 32 P_1 \]

### Final Relations:
- The correct relations are:
- \( V_1 = 8V_2 \)
- \( P_2 = 32P_1 \)