A balloon contains `1500 m^(3)` of helium at `27^(@)C` and 4 atmospheric pressue. The volume of helium at `-3^(@)C` themperature and 2 atmospheric pressure will be

Options

A. `1500m^(3)`
B. `1700m^(3)`
C. `1900m^(3)`
D. `2700m^(3)`

Explanation

To solve the problem, we will use the combined gas law, which relates the pressure, volume, and temperature of a gas. The formula is given by:

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

Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures,
- \( V_1 \) and \( V_2 \) are the initial and final volumes,
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

### Step-by-Step Solution:

1. **Convert the Initial Temperature to Kelvin:**
\[ T_1 = 27^\circ C = 27 + 273 = 300 \, K \]

2. **Convert the Final Temperature to Kelvin:**
\[ T_2 = -3^\circ C = -3 + 273 = 270 \, K \]

3. **Identify the Initial Conditions:**
- Initial Volume, \( V_1 = 1500 \, m^3 \)
- Initial Pressure, \( P_1 = 4 \, atm \)

4. **Identify the Final Conditions:**
- Final Pressure, \( P_2 = 2 \, atm \)
- Final Volume, \( V_2 = ? \)

5. **Use the Combined Gas Law:**
Rearranging the combined gas law to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

6. **Substitute the Known Values:**
\[ V_2 = \frac{(4 \, atm) \cdot (1500 \, m^3) \cdot (270 \, K)}{(2 \, atm) \cdot (300 \, K)} \]

7. **Calculate \( V_2 \):**
\[ V_2 = \frac{4 \cdot 1500 \cdot 270}{2 \cdot 300} \]
\[ V_2 = \frac{1620000}{600} = 2700 \, m^3 \]

### Final Answer:
The volume of helium at \(-3^\circ C\) and \(2 \, atm\) will be \(2700 \, m^3\).

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