Suppose ideal gas equation follows `VP^(3) = contant`. 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

Answer

To solve the problem, we start with the given relationship for an ideal gas, which states that \( V P^3 = \text{constant} \). We need to find the new temperature \( T' \) when the gas expands from volume \( V \) to \( 27V \).

### Step-by-step Solution:

1. **Understand the relationship**:
We know that \( V P^3 = \text{constant} \). This means that as the volume \( V \) changes, the pressure \( P \) will also change in such a way that the product \( V P^3 \) remains constant.

2. **Express pressure in terms of temperature and volume**:
From the ideal gas law, we have:
\[ P = \frac{nRT}{V} \]
Substituting this expression for \( P \) into our relationship gives:
\[ V \left(\frac{nRT}{V}\right)^3 = \text{constant} \]
Simplifying this, we get:
\[ V \cdot \frac{n^3 R^3 T^3}{V^3} = \text{constant} \]
This simplifies to:
\[ \frac{n^3 R^3 T^3}{V^2} = \text{constant} \]

3. **Establish the relationship between temperature and volume**:
From the above equation, we can deduce that:
\[ T^3 \propto V^2 \]
This means that the cube of the temperature is proportional to the square of the volume.

4. **Set up the equation for initial and final states**:
Let the initial temperature be \( T \) and the initial volume be \( V \). When the volume expands to \( 27V \), we denote the new temperature as \( T' \). Thus, we can write:
\[ \frac{T^3}{(T')^3} = \frac{V^2}{(27V)^2} \]
Simplifying the right side:
\[ (27V)^2 = 729V^2 \]
Therefore, we have:
\[ \frac{T^3}{(T')^3} = \frac{V^2}{729V^2} \]
This simplifies to:
\[ \frac{T^3}{(T')^3} = \frac{1}{729} \]

5. **Cross-multiply to find the relationship**:
Rearranging gives:
\[ T^3 = \frac{(T')^3}{729} \]

6. **Express \( T' \) in terms of \( T \)**:
Taking the cube root of both sides, we find:
\[ T' = T \cdot \sqrt[3]{729} \]
Since \( \sqrt[3]{729} = 9 \), we have:
\[ T' = 9T \]

### Final Answer:
Thus, the new temperature \( T' \) when the gas expands to \( 27V \) is \( 9T \).