When volume of system is increased two times and temperature is decreased half of its initial temperature, then pressure becomes ………….

Answer

To solve the problem step by step, we will use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin

### Step-by-Step Solution:

1. **Define Initial Conditions:**
- Let the initial volume be \( V_i \).
- Let the initial pressure be \( P_i \).
- Let the initial temperature be \( T_i \).

2. **Define Final Conditions:**
- The volume is increased to two times its initial volume:
\[ V_f = 2V_i \]
- The temperature is decreased to half of its initial temperature:
\[ T_f = \frac{T_i}{2} \]

3. **Apply Ideal Gas Law for Initial and Final States:**
- For the initial state:
\[ P_i V_i = nRT_i \quad \text{(1)} \]
- For the final state:
\[ P_f V_f = nRT_f \quad \text{(2)} \]

4. **Substitute Final Conditions into the Ideal Gas Law:**
- Substitute \( V_f \) and \( T_f \) into equation (2):
\[ P_f (2V_i) = nR\left(\frac{T_i}{2}\right) \]

5. **Simplify the Final State Equation:**
- Rearranging gives:
\[ P_f \cdot 2V_i = \frac{nRT_i}{2} \]
- Dividing both sides by 2:
\[ P_f V_i = \frac{nRT_i}{4} \quad \text{(3)} \]

6. **Substitute Equation (1) into Equation (3):**
- From equation (1), we know that \( nRT_i = P_i V_i \).
- Substitute this into equation (3):
\[ P_f V_i = \frac{P_i V_i}{4} \]

7. **Solve for Final Pressure \( P_f \):**
- Divide both sides by \( V_i \) (assuming \( V_i \neq 0 \)):
\[ P_f = \frac{P_i}{4} \]

### Conclusion:
The final pressure \( P_f \) becomes one-fourth of the initial pressure \( P_i \):
\[ P_f = \frac{P_i}{4} \]