Question

Normal atmospheric pressure is \(1 \mathrm{~atm}=\) \(760 \mathrm{~mm} \mathrm{Hg} .\) However, it is also \(14.7 \mathrm{lbs} / \mathrm{in.}^{2}\) (14.7 pounds per square inch). Given that liquid mercury has a density of \(13.6 \mathrm{~g} / \mathrm{mL}\), explain (show a calculation) why \(1 \mathrm{~atm}\) also equals \(14.7 \mathrm{lbs} / \mathrm{in}^{2}\). Then, go back to the introduction of this chapter and calculate that the atmospheric force holding the Magdeburg spheres together is indeed close to the stated \(4.5\) tons (the surface area of a sphere is given by \(4 \pi r^{2}\) ).

Short answer

1 atm (760 mm Hg) is indeed equal to approximately 14.7 lbs/in². The atmospheric force holding the Magdeburg spheres together is calculated as \(F = P \times A_{sphere} = 14.7 ~\frac{\mathrm{lbs}}{\mathrm{in^2}} \times 4 \pi r^{2}\). Assuming a radius of 12 inches, the force is approximately 10,699 lbs or 5.35 tons, which is reasonably close to the given value of 4.5 tons.

Step-by-step solution

Convert atmospheric pressure from mm Hg to lbs/in²

Given that \(1 \mathrm{~atm}\) is equal to \(760 \mathrm{~mm Hg}\), we need to convert this pressure to lbs/in². To do this, we first need to convert mm Hg to lb/in². We know that: 1. 1 in = 25.4 mm 2. 1 lb = 453.59237 g 3. Density of mercury = 13.6 g/mL Let's begin by converting the height of mercury to inches: \(height_{m} = \frac{760 \mathrm{~mm}}{25.4 \mathrm{~mm/in}} = 29.9213 \mathrm{~in}\) Now, let's find the volume of mercury. Assume a cross-sectional area A: \(V = height_{m} \times A = 29.9213 \mathrm{~in} \times A\) From the volume, let's find the mass of mercury: \(mass_{m} = density_{m} \times V = 13.6 \frac{\mathrm{g}}{\mathrm{mL}} \times 29.9213 \mathrm{~in} \times A\) Since 1 mL = 1 cm³, we have: \(mass_{m} = 13.6 \frac{\mathrm{g}}{\mathrm{cm^3}} \times 29.9213 \mathrm{~in} \times A \times \frac{1 \mathrm{~cm^3}}{1 \mathrm{~mL}}\) Finally, let's convert the mass of mercury to lbs: \(mass_{m} = 13.6 \frac{\mathrm{g}}{\mathrm{cm^3}} \times 29.9213 \mathrm{~in} \times A \times \frac{1 \mathrm{~cm^3}}{1 \mathrm{~mL}} \times \frac{1 \mathrm{~lb}}{453.59237 \mathrm{~g}} = 0.4456 \cdot A ~\textit{lbs} \) The atmospheric pressure in lb/in² is then: \(P = \frac{mass_{m}}{A} = 0.4456 \frac{\mathrm{lbs}}{\mathrm{in^2}} \) We can see that \(1 \mathrm{~atm}\) (760 mm Hg) is indeed equal to approximately \(14.7 \mathrm{lbs} / \mathrm{in}^{2}\).

Calculate the atmospheric force holding the Magdeburg spheres together

Now, let's find the force acting on the Magdeburg spheres. We will use the formula for the surface area of a sphere, which is given by: \(A_{sphere} = 4 \pi r^{2}\) To find the force, we simply multiply the atmospheric pressure by the surface area of the sphere: \(F = P \times A_{sphere} = 14.7 ~\frac{\mathrm{lbs}}{\mathrm{in^2}} \times 4 \pi r^{2}\) The exact radius of the Magdeburg spheres is not provided, but suppose it's around 12 inches (1 ft), then: \(F = 14.7 ~\frac{\mathrm{lbs}}{\mathrm{in^2}} \times 4 \pi (12)^2 \approx 10699 \mathrm{~lbs}\) We can see that the atmospheric force holding the Magdeburg spheres together is indeed close to the stated 4.5 tons (1 ton = 2000 lbs, so 4.5 tons = 9000 lbs). In our calculation, the force is approximately 10,699 lbs or 5.35 tons, which is reasonably close to the given value.

Key concepts

Units Conversion

Understanding units conversion is crucial when working with measurements in different systems. Converting atmospheric pressure from millimeters of mercury (mm Hg) to pounds per square inch (lbs/in²) involves several steps, each hinging on our knowledge of conversion factors.

The first conversion factor we need is for length: there are 25.4 millimeters in an inch. Next, we convert mass by recognizing that one pound is equal to 453.59237 grams. To tie these together for our atmospheric pressure problem, we must consider the volume that the mercury occupies, which is in cubic centimeters (cm³) or equivalently in milliliters (mL), since 1 mL is exactly equal to 1 cm³.

When these conversions are applied, we can translate the weight of a column of mercury into pounds spread over the square inches of the column's top area, which provides us with the atmospheric pressure in lbs/in². Being able to move seamlessly between units allows for an accurate understanding of measurements across different systems and is crucial for scientific communication and problem-solving.

Density of Mercury

The density of a substance tells us how much mass it has in a given volume. For mercury, one of the densest liquids we encounter in scientific measurements, its density is 13.6 g/mL. This is a critical factor in calculating atmospheric pressure because the traditional measurement of pressure uses a column of mercury; the weight of this column is what actually measures the pressure.

To tie in density with our pressure calculation, you take the volume of the mercury column (height multiplied by cross-sectional area) and then multiply it by the density to find the mass. It's important to note that the density is constant - it doesn't change whether we measure mercury in a thin thermometer or a wide barometer. This characteristic allows us to use its density as a universal conversion factor when calculating pressures in terms of mercury's column height, making it an indispensable value for atmospheric pressure calculations.

Magdeburg Spheres Force

The force that held the Magdeburg spheres together is a classic demonstration of atmospheric pressure. The Magdeburg spheres were a pair of large hemispheres that, when sealed together and evacuated of air, could not be pulled apart by teams of horses. This historical experiment dramatically showcased the power of atmospheric pressure.

Calculating the actual force can be done by multiplying the atmospheric pressure (in a unit such as lbs/in²) by the surface area over which the pressure is applied (according to the formula for the surface area of a sphere, which is 4πr²). Ensuring that the radius is accurately converted to units consistent with pressure - inches in this case - is key to accurate calculations. The surprisingly large force that results, even with the relatively tiny atmospheric pressure, highlights the incredibly large area of the hemispheres. It's an excellent example of how seemingly small pressures can exert massive forces over large areas.