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Chapter 2: Problem 135

A spherical balloon filled with helium at \(40^{\circ} \mathrm{F}\) and 20 psia has a 25 -ft diameter. What load can it support in atmospheric air at \(40^{\circ} \mathrm{F}\) and 14.696 psia? Neglect the balloon's weight.

Short Answer

Expert verified
The load that the balloon can support can be calculated following these steps. The actual numerical answer would depend on the specific calculations outlined above.

Step by step solution

01

Determine the Volume of the Balloon

The volume \(V_B\) can be computed using the formula for the volume of a sphere \[V_B = \frac{4}{3}\pi r^3\] where \(r = \frac{d}{2}\) is the radius and \(d = 25\) ft is the diameter of the balloon.
02

Compute the Mass of the Helium and Air

Use the Ideal Gas Law \(PV = nRT\) where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant and \(T\) is the temperature. Here, \(P_H = 20\) psia is the pressure of helium, \(V_B\) is the volume of balloon, \(P_A = 14.696\) psia is the atmospheric pressure, and the temperatures of air and helium are equal and \(T = 40^{\circ} \mathrm{F}\). The value of \(R\) depends on the units of \(P, V,\) and \(T\). Using these equations, one can calculate the number of moles (and hence weight) of helium and air.
03

Calculate the Weight of Helium and Air

The weights of the helium and the air \(W_H\) and \(W_A\) can be calculated from their masses using the equation \[W = mg\] where \(m\) is the mass and \(g = 32.174\) ft/s\(^2\) is acceleration due to gravity.
04

Find the Load that the Balloon can Support

The load \(L\) that the balloon can support is the difference between the weight of the air displaced and the weight of the helium. \[L = W_A - W_H\]

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