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, we can use the ideal gas law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas (constant in this case)
- \( R \) = universal gas constant
- \( T \) = temperature

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
Let the initial volume be \( V \), the initial temperature be \( T \), and the initial pressure be \( P \).

2. **Determine Final Conditions:**
- The volume is increased to \( 2V \) (doubled).
- The temperature is decreased to \( \frac{T}{2} \) (half of the initial temperature).

3. **Apply the Ideal Gas Law:**
According to the ideal gas law for the initial state:
\[ P V = n R T \]
For the final state, we have:
\[ P_2 (2V) = n R \left(\frac{T}{2}\right) \]

4. **Set Up the Equation:**
We can express the final pressure \( P_2 \) in terms of the initial pressure \( P \):
\[ P_2 (2V) = n R \left(\frac{T}{2}\right) \]

5. **Rearranging the Equation:**
From the initial condition, we know:
\[ n R = P V / T \]
Substitute this into the final state equation:
\[ P_2 (2V) = \left(\frac{P V}{T}\right) \left(\frac{T}{2}\right) \]

6. **Simplifying:**
\[ P_2 (2V) = \frac{P V}{2} \]
Divide both sides by \( 2V \):
\[ P_2 = \frac{P}{4} \]

7. **Conclusion:**
Thus, the final pressure \( P_2 \) becomes:
\[ P_2 = \frac{P}{4} \]

### Final Answer:
The pressure becomes \( \frac{1}{4} \) of its initial value.

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