Question

A spherical inextensible balloon filled with hydrogen is at rest in air (Molar mass 30 gm/mol). The volume of balloon is V = 1m³. The temperature (300 K) and pressure (100kPa) of hydrogen can be assumed to be the same as that of the surrounding air. Assume the temperature to be constant. (A) Mass of the balloon without hydrogen is nearly 1.12 kg (B) Assuming barometric formula, the balloon is in stable equilibrium for displacement in vertical direction. (C) Assuming barometric formula, the balloon is in unstable equilibrium for displacement in vertical direction. (D) The buoyant force is 15 times the weight of hydrogen filled in the balloon (in magnitude)

• A. Mass of the balloon without hydrogen is nearly 1.12 kg
• B. Assuming barometric formula, the balloon is in stable equilibrium for displacement in vertical direction.
• C. Assuming barometric formula, the balloon is in unstable equilibrium for displacement in vertical direction.
• D. The buoyant force is 15 times the weight of hydrogen filled in the balloon (in magnitude)

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The number of moles of hydrogen is calculated using the ideal gas law: $n = \frac{PV}{RT} = \frac{(100 \times 10^3 \text{ Pa})(1 \text{ m}^3)}{(8.314 \text{ J/(mol·K)})(300 \text{ K})} \approx 40.09 \text{ mol}$. The mass of hydrogen is $m_{H_2} = n \times M_{H_2} = 40.09 \text{ mol} \times 0.002 \text{ kg/mol} \approx 0.0802 \text{ kg}$. The mass of displaced air is $m_{air} = n \times M_{air} = 40.09 \text{ mol} \times 0.03 \text{ kg/mol} \approx 1.203 \text{ kg}$. The mass of the balloon is $m_{balloon} = m_{air} - m_{H_2} = 1.203 \text{ kg} - 0.0802 \text{ kg} \approx 1.123 \text{ kg}$. (Option A is correct). The buoyant force is $F_B = m_{air} g$ and the weight of hydrogen is $W_{H_2} = m_{H_2} g$. The ratio $\frac{F_B}{W_{H_2}} = \frac{m_{air}}{m_{H_2}} \approx \frac{1.203}{0.0802} \approx 15.0$. (Option D is correct). For vertical equilibrium, the buoyant force $F_B(h) = \rho_{air}(h) V g$. The atmospheric density decreases with height ($\rho_{air}(h) = \rho_0 e^{-M_{air}gh/RT}$). A small upward displacement increases height, decreasing density and thus buoyant force, resulting in a downward restoring force. A small downward displacement decreases height, increasing density and buoyant force, resulting in an upward restoring force. This indicates stable equilibrium. (Option B is correct, C is incorrect).