A weather balloon filled with hydrogen at 1 atm and `27^(@)C` has volume equal to 1200 litres. On ascending, it reaches a place where temperture is `-23^(@)C` and pressure is 0.5 atm. The volume of the balloon is

Options

A. 24000 litres
B. 20000 litres
C. 10000 litres
D. 12000 litres

Answer

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:

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

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial volume, \( V_1 = 1200 \, \text{liters} \)
- Initial temperature, \( T_1 = 27^\circ C \)
- Final pressure, \( P_2 = 0.5 \, \text{atm} \)
- Final temperature, \( T_2 = -23^\circ C \)

2. **Convert temperatures from Celsius to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \( T_2 \):
\[ T_2 = -23 + 273 = 250 \, K \]

3. **Substitute the known values into the combined gas law:**
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substituting the values:
\[ \frac{1 \times 1200}{300} = \frac{0.5 \times V_2}{250} \]

4. **Cross-multiply to solve for \( V_2 \):**
\[ 1 \times 1200 \times 250 = 0.5 \times V_2 \times 300 \]
Simplifying gives:
\[ 300000 = 150 V_2 \]

5. **Solve for \( V_2 \):**
\[ V_2 = \frac{300000}{150} = 2000 \, \text{liters} \]

### Final Answer:
The volume of the balloon at the new conditions is **2000 liters**.