Calculating Mass Needed to Ground Hot Air Balloon

[wislar]

Homework Statement

A good estimate for the volume of a particular hot air balloon is 2800 m3. Suppose the total load (passengers, fuel, balloon fabric, etc.) on a hot air balloon is 724 kg. In preparing to launch, the pilot heats the air inside to an average temperature of 210 oC, giving it a density of 0.95 kg/m3 (compared to a density of 1.2 kg/m3 for the air outside the balloon). [Note! This means that the total mass of the hot air balloon is 724 kg plus the mass of the hot air in the balloon!]

How much extra mass does the pilot need to keep in the basket in order to stay on the ground while all passengers are on board?

Homework Equations

The Attempt at a Solution

I'm really lost here. Any help would be great.

 

[SHISHKABOB]

well so you want the balloon to stay on the ground, in other words: zero velocity

this means that all the forces acting on the balloon need to sum up to be zero

there's only going to be two forces acting on the balloon. The force due to gravity acting on the total mass of the system and the buoyancy force due to the difference in the density of the air inside the balloon compared to outside

so obviously, as given in the problem, the 724 kg mass is not going to be enough to keep it down, so you need to add some other force to that to equal the buoyancy force

 

[wislar]

So...
-9.8(negative because pushing down)*724 + 9.8*___=0?

 

[SHISHKABOB]

well, no, you need to account for the buoyancy force

 

[Nickie]

I would approach this problem by using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

First, I would convert the temperature given in the problem from Celsius to Kelvin by adding 273.15, giving us 483.15 K.

Next, I would calculate the number of moles of hot air in the balloon using the ideal gas law. We know the volume (2800 m3), temperature (483.15 K), and density (0.95 kg/m3), so we can rearrange the ideal gas law to solve for n. This gives us n = (PV)/(RT) = ((0.95 kg/m3)(2800 m3))/(8.314 J/mol*K)(483.15 K) = 0.77 moles of hot air.

Now, we can use the molar mass of air (approximately 28 g/mol) to calculate the mass of hot air in the balloon. This gives us (0.77 moles)(28 g/mol) = 21.56 g of hot air.

Since the problem states that the total mass of the hot air balloon is 724 kg plus the mass of the hot air, we can subtract the mass of the hot air from the total mass to find the mass of the balloon fabric, fuel, and other items in the basket. This gives us 724 kg - 21.56 g = 723.97844 kg.

Finally, we can calculate the extra mass needed to keep the balloon grounded by subtracting the total load (724 kg) from the mass of the balloon fabric, fuel, and other items in the basket (723.97844 kg). This gives us an extra mass of 0.02156 kg, or approximately 22 grams.

So, in order to stay on the ground while all passengers are on board, the pilot would need to add an extra 22 grams of mass to the basket.