A balloon is filled at `27^(@)C` and 1 atm pressure by `500 m^(3)` He. At-`3^(@)C` and 0.5 atm pressures, the volume of He-gas contained in balloon will be

Options

A. 700 `m^(3)`
B. 900 `m^(3)`
C. 1000 `m^(3)`
D. 500 `m^(3)`

Explanation

To solve the problem, we will use the Ideal Gas Law, which can be expressed in the form of a proportional relationship for two different states of a gas:

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

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 temperatures from Celsius to Kelvin
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = -3^\circ C = -3 + 273 = 270 \, K \)

### Step 2: Identify the pressures and volumes
- Initial pressure \( P_1 = 1 \, atm \)
- Initial volume \( V_1 = 500 \, m^3 \)
- Final pressure \( P_2 = 0.5 \, atm \)

### Step 3: Rearrange the Ideal Gas Law to solve for \( V_2 \)
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 4: Substitute the known values into the equation
\[ V_2 = \frac{(1 \, atm)(500 \, m^3)(270 \, K)}{(0.5 \, atm)(300 \, K)} \]

### Step 5: Calculate \( V_2 \)
\[ V_2 = \frac{135000}{150} = 900 \, m^3 \]

Thus, the volume of helium gas contained in the balloon at -3°C and 0.5 atm pressure will be **900 m³**.