The event described is historically true. The mountain of this experiment exists though the scientists have died.

Pascal's Experiment: In 1648, a year after Torricelli died, Blaise Pascal convinced his brother-in-law, Perier, to conduct a test of a mercury barometer by carrying one from the base to the top of the French volcanic formation, Puy de Dome. "Tell me," Blaise asked, "What difference in mercury height you observe."

The density of atmospheric air is about 0.075 pounds per cubic feet. The peak of Puy de Dome is 4803 feet above its base. Approximately what difference of mercury height in the barometer would Perier have observed?

♦  The sketch shows the barometer readings as Perier observed them. The pressure in the top, seemingly empty space, of a mercury barometer is not zero; at 25°C it is about 20 Pascals. That is so small a pressure that it is usually set equal to zero.

The hydrostatic equation relates pressure changes with elevation change and fluid density along a selected path through the fluids. The path we will follow begins at point A in the mercury vapor at the top of the barometer at ground level. The pressure there is about 20 Pa.

We begin at "A" in the barometer at the base of the volcano. From the void of this barometer, pressure increases in passing to the mercury/air interface, "B."

Our next task is to estimate the pressure change associated with change of position (in air) from the base to the top of the volcano. Pressure changes along that path and so does density. The relation we have is:

The pressure difference of our interest is p_C - p_B. To obtain this we integrate the differential of pressure from p_B to p_C.

We see the integral requires that we know how the density of air and the local acceleration vary over the elevation change. We know gravity changes very little. For the air, we assume density is constant at an approximayted surface value of 0.075 lbm/ft³. Thus, in effect, we use the Mean Value of Calculus to perform the integration.

The logic that applied to the barometer at ground level is applied to the barometer on top.

We can combine i), ii), and iii) to determine: H₁ - H₂

There is an easier way to solve hydrostatics problems. We recognize the two known pressures (at the top of the barometers). So we write one equation combining the pressure differences.

Observe: Pressure at A increased by column of mercury (going down, therefore "+") decreased by column air (going up "-") decreased by column of mercury (going up "-") equals pressure at D.