A balloon partially filled with helium has a volume of `30 m^3`, at the earth's surface, where pressure is `76 cm` of (Hg) and temperature is `27^@ C` What will be the increase in volume of gas if balloon rises to a height, where pressure is `7.6 cm` of `Hg` and temperature is `-54^@ C` ?

Explanation

To solve the problem of the helium balloon rising to a higher altitude, we will use the ideal gas law, which states that for a given amount of gas, the relationship between pressure (P), volume (V), and temperature (T) can be expressed as:

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

Where:
- \( P_1, V_1, T_1 \) are the initial pressure, volume, and temperature.
- \( P_2, V_2, T_2 \) are the final pressure, volume, and temperature.

### Step 1: Convert the temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin using the formula:

\[ T(K) = T(°C) + 273 \]

- Initial temperature \( T_1 = 27°C = 27 + 273 = 300 K \)
- Final temperature \( T_2 = -54°C = -54 + 273 = 219 K \)

### Step 2: Identify the known values
From the problem, we have:
- Initial pressure \( P_1 = 76 \, \text{cm of Hg} \)
- Initial volume \( V_1 = 30 \, \text{m}^3 \)
- Initial temperature \( T_1 = 300 \, K \)
- Final pressure \( P_2 = 7.6 \, \text{cm of Hg} \)
- Final temperature \( T_2 = 219 \, K \)
- Final volume \( V_2 = ? \)

### Step 3: Set up the equation
Using the ideal gas law, we can set up the equation:

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

Substituting the known values into the equation:

\[ \frac{76 \, \text{cm of Hg} \times 30 \, \text{m}^3}{300 \, K} = \frac{7.6 \, \text{cm of Hg} \times V_2}{219 \, K} \]

### Step 4: Solve for \( V_2 \)
Cross-multiply to solve for \( V_2 \):

\[ 76 \times 30 \times 219 = 7.6 \times V_2 \times 300 \]

Calculating the left side:

\[ 76 \times 30 \times 219 = 499560 \]

Now, we can rearrange the equation to solve for \( V_2 \):

\[ V_2 = \frac{499560}{7.6 \times 300} \]

Calculating the denominator:

\[ 7.6 \times 300 = 2280 \]

Now, substituting back:

\[ V_2 = \frac{499560}{2280} \approx 219.0 \, \text{m}^3 \]

### Step 5: Calculate the change in volume
The change in volume \( \Delta V \) is given by:

\[ \Delta V = V_2 - V_1 \]

Substituting the values:

\[ \Delta V = 219.0 \, \text{m}^3 - 30.0 \, \text{m}^3 = 189.0 \, \text{m}^3 \]

### Final Answer
The increase in volume of the gas when the balloon rises is:

\[ \Delta V = 189.0 \, \text{m}^3 \]

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