When the temperature of certain sample of a gas is chaged from `30^(@)C` to 606 K and its pressure is reduced of gas changes from V to `V^2` . The value of V is

Options

A. `2 dm^3`
B. `4 dm^3`
C. ` 8 dm^3`
D. Unprdictable

Answer

To solve the problem, we will use the ideal gas law and the relationship between volume and temperature. Here’s a step-by-step solution:

### Step 1: Convert the initial temperature from Celsius to Kelvin
The initial temperature \( T_1 \) is given as \( 30^\circ C \). To convert this to Kelvin:
\[ T_1 = 30 + 273 = 303 \, K \]

### Step 2: Identify the final temperature
The final temperature \( T_2 \) is given as \( 606 \, K \).

### Step 3: Write the relationship between volumes and temperatures
According to Charles's Law, for a given amount of gas at constant pressure, the volume is directly proportional to the temperature in Kelvin:
\[ \frac{V_1}{V_2} = \frac{T_1}{T_2} \]
Where:
- \( V_1 \) is the initial volume \( V \)
- \( V_2 \) is the final volume \( V^2 \)

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ \frac{V}{V^2} = \frac{303}{606} \]

### Step 5: Simplify the equation
The right side simplifies to:
\[ \frac{303}{606} = \frac{1}{2} \]
Thus, we have:
\[ \frac{V}{V^2} = \frac{1}{2} \]

### Step 6: Solve for \( V \)
Rearranging the equation gives:
\[ \frac{1}{V} = \frac{1}{2} \]
Taking the reciprocal of both sides:
\[ V = 2 \]

### Conclusion
The value of \( V \) is \( 2 \, \text{dm}^3 \).

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