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, we will use the Ideal Gas Law, which states that \( PV = nRT \). Since the number of moles of gas in the balloon remains constant as it rises, we can use the relationship between the initial and final states of the gas.

### Step-by-step Solution:

1. **Identify the Initial Conditions:**
- Initial Volume, \( V_1 = 30 \, m^3 \)
- Initial Pressure, \( P_1 = 76 \, cm \, Hg \)
- Initial Temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)

2. **Identify the Final Conditions:**
- Final Pressure, \( P_2 = 7.6 \, cm \, Hg \)
- Final Temperature, \( T_2 = -54^\circ C = -54 + 273 = 219 \, K \)

3. **Apply the Ideal Gas Law:**
Since the number of moles remains constant, we can use the equation:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

4. **Rearranging the Equation:**
We can rearrange the equation to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

5. **Substituting the Known Values:**
Substitute \( P_1 = 76 \, cm \, Hg \), \( V_1 = 30 \, m^3 \), \( T_2 = 219 \, K \), \( P_2 = 7.6 \, cm \, Hg \), and \( T_1 = 300 \, K \):
\[ V_2 = \frac{76 \times 30 \times 219}{7.6 \times 300} \]

6. **Calculating \( V_2 \):**
- First, calculate the numerator:
\[ 76 \times 30 \times 219 = 499560 \]
- Then, calculate the denominator:
\[ 7.6 \times 300 = 2280 \]
- Now, divide the numerator by the denominator:
\[ V_2 = \frac{499560}{2280} \approx 219 \, m^3 \]

7. **Calculate the Increase in Volume:**
The increase in volume is given by:
\[ \Delta V = V_2 - V_1 = 219 \, m^3 - 30 \, m^3 = 189 \, m^3 \]

### Final Answer:
The increase in volume of the gas when the balloon rises is \( 189 \, m^3 \).

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