Suppose ideal gas equation follows `VP^(3) = constant`. Initial temperature and volume of the gas are T and V respectively. If gas expand to `27V` temperature will become

Options

A. `T`
B. `9T`
C. `27T`
D. `T//9`

Explanation

To solve the problem, we start with the given relationship for the ideal gas that follows the equation \( VP^3 = \text{constant} \). We need to find the final temperature when the volume expands from \( V \) to \( 27V \).

### Step-by-Step Solution:

1. **Understand the Given Relationship**:
The equation \( VP^3 = \text{constant} \) implies that for an ideal gas, the product of pressure and volume raised to the power of three remains constant during the process.

2. **Initial Conditions**:
Let the initial conditions be:
- Initial volume, \( V_1 = V \)
- Initial temperature, \( T_1 = T \)
- Initial pressure, \( P_1 \)

3. **Final Conditions**:
The final conditions after expansion are:
- Final volume, \( V_2 = 27V \)
- Final temperature, \( T_2 \)
- Final pressure, \( P_2 \)

4. **Using the Ideal Gas Law**:
The ideal gas law states that:
\[ PV = nRT \]
From this, we can express pressure in terms of volume and temperature:
\[ P = \frac{nRT}{V} \]

5. **Relating Initial and Final States**:
From the relationship \( VP^3 = \text{constant} \), we can write:
\[ V_1 P_1^3 = V_2 P_2^3 \]

6. **Substituting the Values**:
Substitute \( V_1 = V \) and \( V_2 = 27V \):
\[ V P_1^3 = 27V P_2^3 \]
Dividing both sides by \( V \) (since \( V \neq 0 \)):
\[ P_1^3 = 27 P_2^3 \]

7. **Finding the Relationship Between Pressures**:
Taking the cube root of both sides:
\[ P_1 = 3 P_2 \]

8. **Using the Ideal Gas Law for Initial and Final States**:
For the initial state:
\[ P_1 = \frac{nRT_1}{V} \]
For the final state:
\[ P_2 = \frac{nRT_2}{27V} \]

9. **Substituting Pressures into the Equation**:
Substitute \( P_1 \) and \( P_2 \) into the relationship \( P_1 = 3 P_2 \):
\[ \frac{nRT_1}{V} = 3 \left(\frac{nRT_2}{27V}\right) \]

10. **Simplifying the Equation**:
Cancel \( nR \) and \( V \) from both sides:
\[ T_1 = 3 \left(\frac{T_2}{27}\right) \]
Simplifying gives:
\[ T_1 = \frac{T_2}{9} \]

11. **Finding Final Temperature**:
Rearranging gives:
\[ T_2 = 9 T_1 \]
Since \( T_1 = T \):
\[ T_2 = 9T \]

### Final Answer:
The final temperature \( T_2 \) when the gas expands to \( 27V \) is \( 9T \).

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