How does volume of a given mass of gas change, when its pressure is reduced to half and absolute temperature is increased by four times ? 

Answer

To solve the problem of how the volume of a given mass of gas changes when its pressure is reduced to half and its absolute temperature is increased by four times, we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

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

### Step 1: Define Initial Conditions
Let’s assume the initial conditions of the gas are:
- Pressure = \( P \)
- Volume = \( V \)
- Temperature = \( T \)

### Step 2: Define Final Conditions
According to the problem:
- The pressure is reduced to half, so the final pressure \( P' = \frac{P}{2} \).
- The absolute temperature is increased by four times, so the final temperature \( T' = 5T \) (since \( T + 4T = 5T \)).

### Step 3: Apply the Ideal Gas Law for Initial and Final States
Using the Ideal Gas Law for the initial state:
\[ PV = nRT \] (Equation 1)

Using the Ideal Gas Law for the final state:
\[ P'V' = nRT' \] (Equation 2)

Substituting the values of \( P' \) and \( T' \) into Equation 2:
\[ \left(\frac{P}{2}\right)V' = nR(5T) \]

### Step 4: Rearranging the Final State Equation
Now, we can rearrange Equation 2 to solve for \( V' \):
\[ \frac{P}{2} V' = nR(5T) \]

Dividing both sides by \( \frac{P}{2} \):
\[ V' = \frac{nR(5T)}{\frac{P}{2}} \]

This simplifies to:
\[ V' = \frac{10nRT}{P} \]

### Step 5: Relate Final Volume to Initial Volume
From Equation 1, we know:
\[ PV = nRT \]

Thus, we can express \( nRT \) as:
\[ nRT = PV \]

Substituting this back into the equation for \( V' \):
\[ V' = \frac{10(PV)}{P} \]

This simplifies to:
\[ V' = 10V \]

### Conclusion
The volume of the gas increases to **10 times** its initial volume when the pressure is reduced to half and the absolute temperature is increased by four times.