Two litres of gas are maintained at `25^(@)C` and two atmospheric pressure. If the pressure is double and absolute temperature is halved, the gas will now occupy

Options

A. 2.0 litre
B. 4.0 litre
C. 0.5 litre
D. 1.0 litre

Explanation

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

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \(P_1\) = initial pressure
- \(V_1\) = initial volume
- \(T_1\) = initial temperature (in Kelvin)
- \(P_2\) = final pressure
- \(V_2\) = final volume
- \(T_2\) = final temperature (in Kelvin)

### Step 1: Convert the initial temperature from Celsius to Kelvin
The initial temperature \(T_1\) is given as \(25^\circ C\). To convert this to Kelvin:

\[ T_1 = 25 + 273 = 298 \, K \]

### Step 2: Identify the initial conditions
From the problem:
- \(P_1 = 2 \, \text{atm}\)
- \(V_1 = 2 \, \text{liters}\)
- \(T_1 = 298 \, K\)

### Step 3: Determine the final conditions
According to the problem:
- The pressure is doubled:
\[ P_2 = 2 \times 2 = 4 \, \text{atm} \]
- The absolute temperature is halved:
\[ T_2 = \frac{298}{2} = 149 \, K \]

### Step 4: Substitute the values into the ideal gas law
Now we can substitute the known values into the ideal gas law equation:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Substituting the values we have:

\[ \frac{2 \, \text{atm} \times 2 \, \text{liters}}{298 \, K} = \frac{4 \, \text{atm} \times V_2}{149 \, K} \]

### Step 5: Simplify and solve for \(V_2\)
Cross-multiplying gives:

\[ 2 \times 2 \times 149 = 4 \times V_2 \times 298 \]

This simplifies to:

\[ 596 = 1192 V_2 \]

Now, solving for \(V_2\):

\[ V_2 = \frac{596}{1192} = \frac{1}{2} \, \text{liters} \]

### Final Answer
The gas will now occupy \(0.5\) liters or \(\frac{1}{2}\) liters.

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