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, we will use the ideal gas law and the combined gas law. Here’s a step-by-step solution:

### Step 1: Identify the given values
- 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 the units
1. Convert P1 from atm to mm Hg:
\[ P1 = 1 \text{ atm} = 760 \text{ mm Hg} \]

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

### Step 3: Use the combined gas law
The combined gas law is given by:
\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]
We need to find the final volume (V2).

### Step 4: Rearrange the equation to solve for V2
Rearranging the equation gives:
\[ V2 = \frac{P1 \cdot V1 \cdot T2}{P2 \cdot T1} \]

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

### Step 6: Calculate V2
Calculating the above expression:
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 results
- Initial volume (V1) = 200 L
- Final volume (V2) = 360 L

Since V2 > V1, the balloon expands.

### 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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