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 states that for a given amount of gas, the ratio of pressure (P) times volume (V) to temperature (T) is constant. This can be expressed mathematically as:

\[ \frac{PV}{T} = \text{constant} \]

Given that the amount of gas (n) and the ideal gas constant (R) remain unchanged, we can set up the following relationship for two different states of the gas:

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

### Step 1: Identify the known values
- Initial conditions:
- \( P_1 = 1 \, \text{atm} \)
- \( V_1 = 500 \, \text{m}^3 \)
- \( T_1 = 27^{\circ}C = 27 + 273 = 300 \, \text{K} \)

- Final conditions:
- \( P_2 = 0.5 \, \text{atm} \)
- \( T_2 = -3^{\circ}C = -3 + 273 = 270 \, \text{K} \)

### Step 2: Rearrange the equation to solve for \( V_2 \)
From the equation:

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

We can rearrange it to find \( V_2 \):

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

### Step 3: Substitute the known values into the equation
Now, substituting the known values into the equation:

\[ V_2 = \frac{(1 \, \text{atm}) \times (500 \, \text{m}^3) \times (270 \, \text{K})}{(0.5 \, \text{atm}) \times (300 \, \text{K})} \]

### Step 4: Calculate \( V_2 \)
Calculating the numerator and denominator:

- Numerator:
\[ 1 \times 500 \times 270 = 135000 \]

- Denominator:
\[ 0.5 \times 300 = 150 \]

Now, substituting these values:

\[ V_2 = \frac{135000}{150} = 900 \, \text{m}^3 \]

### Final Answer
The volume of helium gas contained in the balloon at -3°C and 0.5 atm pressure will be **900 m³**.

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