Problem 61 - University Physics with Modern Physics

Question

A balloon whose volume is 750 \(\mathrm{m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of 1.90 \(\mathrm{m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{Pa},\) how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at \(15.0^{\circ} \mathrm{C}\) ? The molar mass of hydro\(\operatorname{gen}\left(\mathrm{H}_{2}\right)\) is 2.02 \(\mathrm{g} / \mathrm{mol}\) . The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}\) . See Chapter 14 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 \(\mathrm{g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) Calculate moles of gas for both balloon and cylinders, then divide. (b) Use buoyant force minus gas weight for hydrogen. (c) Use the same for helium. Solving gives specific numbers.

Step 1: Calculate moles of hydrogen needed for the balloon

Using the ideal gas law \(PV = nRT\), find the number of moles \(n\) of hydrogen needed to fill the balloon. \(P = 1.01 \times 10^5 \text{ Pa}\), \(V = 750 \text{ m}^3\), and \(R = 8.314 \text{ J/mol K}\). Since T is consistent, it cancels out:\[ n = \frac{PV}{RT} = \frac{1.01 \times 10^5 \times 750}{R T}\].

Step 2: Calculate moles of hydrogen per cylinder

Use the ideal gas law for one cylinder. \(P = (1.01 \times 10^5 + 1.20 \times 10^6) \text{ Pa}\), \(V = 1.90 \text{ m}^3\). Use the equation:\[ n = \frac{PV}{RT} = \frac{(1.01 \times 10^5 + 1.20 \times 10^6) \times 1.90}{R T}\].

Step 3: Calculate the number of cylinders required

Divide the number of moles in the balloon by the number of moles per cylinder:\[ \text{Number of cylinders} = \frac{\text{moles in balloon}}{\text{moles per cylinder}} \].

Step 4: Calculate buoyant force for hydrogen-filled balloon

Buoyant force = Weight of air displaced by the balloon = Volume x Density of air x g.\[ F_b = 750 \times 1.23 \times 9.81 \text{ N} \]

Step 5: Calculate weight of hydrogen in balloon

Weight of hydrogen = moles of H2 x molar mass H2 x g:\[ \text{Weight of hydrogen} = \text{moles of H2} \times 2.02 \times 10^{-3} \times 9.81 \text{ N} \].

Step 6: Calculate net weight supported by hydrogen balloon

Net weight = Buoyant force - Weight of hydrogen. Subtract the weight of hydrogen from the buoyant force to find the additional load the balloon can lift.

Step 7: Calculate weight of helium in balloon

Repeat the weight calculation using the molar mass of helium (4.00 g/mol). Use the same equation as Step 5 but with the helium molar mass.

Step 8: Calculate net weight supported by helium-filled balloon

Net weight = Buoyant force - Weight of helium. Subtract the weight of helium from the buoyant force to find the additional load the balloon can lift.

Key Concepts

Buoyant Force

The buoyant force is the upward force exerted on an object submerged in a fluid. This force is a result of pressure differences at different depths in the fluid. When a balloon is filled with a gas, it displaces a volume of air equal to the volume of the balloon. The weight of this displaced air is what provides the buoyant force. To calculate it, multiply the volume of the balloon by the density of the air and the acceleration due to gravity \(g\), thus \( F_b = \text{Volume} \times \text{Density of air} \times g \).
For example, in our problem, a balloon with a volume of 750 m³ in an environment where the air density is 1.23 kg/m³ yields a buoyant force \( F_b = 750 \times 1.23 \times 9.81 = 9046.35 \text{ N} \). This is the force lifting the balloon and effectively supporting additional weight in the air.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It plays a crucial role in converting between the amount of substance (moles) and mass. The molar mass allows us to determine the weight of a gas within a balloon by multiplying the number of moles of the gas by its molar mass.
For hydrogen, the molar mass is 2.02 g/mol, while for helium, it is 4.00 g/mol. To find the weight of hydrogen gas in a balloon, calculate the number of moles using the ideal gas law and then multiply by 2.02 g/mol.
This understanding helps in computing how much weight the balloon gas itself contributes when solving problems involving buoyancy and lifting.

Gauge Pressure

Gauge pressure is the pressure of a system compared to the atmospheric pressure. It is often used when measuring pressures in contained environments like gas cylinders. Gauge pressure does not include atmospheric pressure, while absolute pressure does. Therefore, to find the true pressure exerted by a gas, add atmospheric pressure to the gauge pressure.
In the problem scenario, the hydrogen tank gauge pressure is given as \(1.20 \times 10^6 \) Pa, and atmospheric pressure is \(1.01 \times 10^5 \) Pa. Therefore, the absolute pressure used in computations is the sum of these two values.
This distinction is vital to correctly applying the ideal gas law when calculating the number of moles of gas in a cylinder. Understanding how to handle gauge pressure ensures precision in such thermodynamic equations.

Density of Air

Density of air is a measurement of air mass per unit volume and varies with temperature and pressure. At a specified condition (15°C in this problem), air density is given as 1.23 kg/m³. This value is integral in calculating the buoyant force since it determines how much air is displaced by the balloon.
The concept of density is straightforward: as it increases, so does the buoyant force experienced by an object in air. When we speak about the density of air in physics problems, it often refers to standard temperature and pressure conditions, although specifics can vary.

• Density increase can enhance lift potential.
• Always relate density to local altitude and weather conditions for precision.

Having a solid grasp on air density allows you to accurately predict how much an object will be buoyed when suspended in the air.