Temperature of an ideal gas is 300 K . The final temperature of the gas when its volume changes from `V " to " 2V` in the process `p=alphaV` (here `alpha` is a positive constant) is

Options

A. 900 K
B. 1200 K
C. 300 K
D. 500 K

Answer

To solve the problem, we need to find the final temperature of an ideal gas when its volume changes from \( V \) to \( 2V \) under the process defined by \( P = \alpha V \), where \( \alpha \) is a positive constant. The initial temperature of the gas is given as \( 300 \, K \).

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
Rearranging this gives:
\[ T = \frac{PV}{nR} \]

2. **Calculate Initial Temperature**:
The initial temperature \( T_i \) can be expressed as:
\[ T_i = \frac{P_i V_i}{nR} \]
Given that \( T_i = 300 \, K \), we have:
\[ 300 = \frac{P_i V_i}{nR} \tag{1} \]

3. **Express Final Temperature**:
The final temperature \( T_f \) can be expressed similarly:
\[ T_f = \frac{P_f V_f}{nR} \]

4. **Relate Initial and Final Pressures**:
From the process \( P = \alpha V \), we can express the final pressure \( P_f \) in terms of the initial pressure \( P_i \):
\[ P_f = \alpha V_f \]
Since \( V_f = 2V \) and \( V_i = V \), we can substitute:
\[ P_f = \alpha (2V) = 2\alpha V \]
Now, we can find the ratio of final pressure to initial pressure:
\[ \frac{P_f}{P_i} = \frac{2\alpha V}{\alpha V} = 2 \tag{2} \]

5. **Substituting into Final Temperature**:
Now substituting \( P_f \) and \( V_f \) into the equation for \( T_f \):
\[ T_f = \frac{P_f V_f}{nR} = \frac{(2P_i)(2V)}{nR} = \frac{4P_i V}{nR} \]

6. **Using Initial Temperature**:
From equation (1), we know:
\[ \frac{P_i V}{nR} = 300 \]
Therefore:
\[ T_f = 4 \times 300 = 1200 \, K \]

### Final Answer:
The final temperature \( T_f \) of the gas is:
\[ \boxed{1200 \, K} \]