A balloon of volume 200 litre having ideal gas at 1 atm pressure and at `27^@C`, when rises to a height where atmospheric pressure is 380 mm Hg and temperature is `-3^@C` , balloon will

Options

A. Contract
B. Expand
C. No change in volume of balloon
D. Initially expand and then contract

Answer

To solve the problem of the balloon's behavior when it rises to a height with different pressure and temperature conditions, we can use the ideal gas law. Here’s a step-by-step solution:

### Step 1: Identify the Given Data
- Initial Volume (V1) = 200 liters
- Initial Pressure (P1) = 1 atm
- Initial Temperature (T1) = 27°C
- Final Pressure (P2) = 380 mm Hg
- Final Temperature (T2) = -3°C

### Step 2: Convert Units
1. Convert P1 from atm to mm Hg:
\[ P1 = 1 \text{ atm} \times 760 \text{ mm Hg/atm} = 760 \text{ mm Hg} \]

2. Convert temperatures from Celsius to Kelvin:
\[ T1 = 27°C + 273 = 300 \text{ K} \]
\[ T2 = -3°C + 273 = 270 \text{ K} \]

### Step 3: Use the Ideal Gas Law
The relationship between the initial and final states of the gas can be expressed using the ideal gas law:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]

### Step 4: Rearrange the Equation to Solve for V2
Rearranging the equation to find the final volume (V2):
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

### Step 5: Substitute the Values
Now substitute the known values into the equation:
\[ V2 = \frac{760 \text{ mm Hg} \cdot 200 \text{ L} \cdot 270 \text{ K}}{380 \text{ mm Hg} \cdot 300 \text{ K}} \]

### Step 6: Calculate V2
1. Calculate the numerator:
\[ 760 \cdot 200 \cdot 270 = 41040000 \]

2. Calculate the denominator:
\[ 380 \cdot 300 = 114000 \]

3. Now divide:
\[ V2 = \frac{41040000}{114000} \approx 360 \text{ L} \]

### Step 7: Analyze the Result
Since V2 (360 L) is greater than V1 (200 L), the volume of the balloon has increased.

### Conclusion
The balloon will **expand** when it rises to the height where the atmospheric pressure is 380 mm Hg and the temperature is -3°C.

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