Question

A balloon carries a total load of 185 kg at normal pressure and temperature of 27°C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is -7°C. Assuming the volume constant ?

• A. 123.54 kg
• B. 214.15 kg
• C. 219.07 kg
• D. 181.46 kg

Hint:

For gas law problems involving ratios, you often don't need to convert pressures to Pascals if they are given in the same units (like cm of Hg), as the units will cancel out. However, always convert temperatures to the absolute scale (Kelvin).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The lifting capacity (load carrying ability) of a balloon depends on the buoyant force, which in turn depends on the density of the surrounding air. We are given the initial load at certain conditions and asked to find the new load at different atmospheric conditions, assuming the balloon's volume is constant.
Step 2: Key Formula or Approach:
The load (\(m\)) a balloon can carry is proportional to the buoyant force minus the weight of the gas inside. Assuming the weight of the gas inside is constant, the change in load capacity is primarily due to the change in the buoyant force. The buoyant force is \(F_B = \rho_{air} V g\). Since \(V\) and \(g\) are constant, the load \(m\) is proportional to the density of the outside air, \(\rho_{air}\).
So, \(m \propto \rho_{air}\).
From the ideal gas law, \(P = \rho \frac{RT}{M_{air}}\), we get \(\rho_{air} = \frac{P M_{air}}{RT}\). This means \(\rho_{air} \propto \frac{P}{T}\).
Therefore, the load \(m \propto \frac{P}{T}\).
This gives the relation: \(\frac{m_2}{m_1} = \frac{P_2/T_2}{P_1/T_1} = \frac{P_2}{P_1} \times \frac{T_1}{T_2}\).
Step 3: Detailed Explanation:
Let's list the initial and final conditions. Remember to convert temperatures to Kelvin.
Initial Conditions (1):
Load, \(m_1 = 185\) kg
Pressure, \(P_1 =\) normal pressure = 76 cm of Hg
Temperature, \(T_1 = 27^\circ\text{C} = 27 + 273 = 300\) K
Final Conditions (2):
Load, \(m_2 = ?\)
Pressure, \(P_2 = 45\) cm of Hg
Temperature, \(T_2 = -7^\circ\text{C} = -7 + 273 = 266\) K
Now, we use the proportionality relation: \[ m_2 = m_1 \times \left(\frac{P_2}{P_1}\right) \times \left(\frac{T_1}{T_2}\right) \] \[ m_2 = 185 \times \left(\frac{45}{76}\right) \times \left(\frac{300}{266}\right) \] \[ m_2 = 185 \times 0.5921 \times 1.1278 \] \[ m_2 \approx 123.54 \text{ kg} \] Step 4: Final Answer:
The new load the balloon will carry is approximately 123.54 kg.