Temperature of an ideal gas is 300 K. The change in temperature of the gas when its volume changes from V to 2V in the process p = aV (Here, a is a positive constant) is

Options

A. 900 K
B. 1200 K
C. 600 K
D. 300 K

Answer

To solve the problem, we need to analyze the relationship between pressure, volume, and temperature of an ideal gas during the given process \( p = aV \), where \( a \) is a positive constant.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The initial temperature \( T_i = 300 \, K \).
- The initial volume \( V_i = V \).

2. **Use the Ideal Gas Law:**
- The ideal gas law is given by \( PV = nRT \).
- Rearranging this gives us \( T = \frac{PV}{nR} \).

3. **Express Initial Temperature:**
- For the initial state, we can write:
\[ T_i = \frac{P_i V_i}{nR} \]
- Substituting \( T_i = 300 \, K \):
\[ 300 = \frac{P_i V}{nR} \quad \text{(Equation 1)} \]

4. **Express Initial Pressure:**
- Given that \( P = aV \), the initial pressure \( P_i \) can be expressed as:
\[ P_i = aV \quad \text{(Equation 2)} \]

5. **Calculate Final Conditions:**
- The final volume \( V_f = 2V \).
- The final pressure \( P_f \) can be expressed as:
\[ P_f = aV_f = a(2V) = 2aV \quad \text{(Equation 3)} \]

6. **Express Final Temperature:**
- For the final state, we can write:
\[ T_f = \frac{P_f V_f}{nR} \]
- Substituting \( P_f \) and \( V_f \):
\[ T_f = \frac{(2aV)(2V)}{nR} = \frac{4aV^2}{nR} \]

7. **Relate Final Temperature to Initial Temperature:**
- From Equation 1, we know:
\[ T_i = \frac{aV^2}{nR} \]
- Therefore, we can express \( T_f \) in terms of \( T_i \):
\[ T_f = 4 \left(\frac{aV^2}{nR}\right) = 4T_i \]
- Substituting \( T_i = 300 \, K \):
\[ T_f = 4 \times 300 = 1200 \, K \]

8. **Calculate Change in Temperature:**
- The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T_f - T_i = 1200 - 300 = 900 \, K \]

### Final Answer:
The change in temperature of the gas when its volume changes from \( V \) to \( 2V \) is \( 900 \, K \).