2.19 More on Pressure
Pressurepis defined as the force (F) exerted by a fluid per unit area (A), p=F/A, in the limit of infinitesimal area. Pressure is isotropic, that is the force on a surface is independent of the orientation of that surface.
mg p0
p Fig. 2.9 A piston resting on a liquid
A piston of mass m and cross section A rests on a liquid in a cylinder, as depicted in Fig. 2.9; the atmospheric pressure is p0. We determine the pressurepof the liquid at the piston.
The piston is at rest, in mechanical equilibrium, which implies that all forcesFi on the piston add up to zero,
Fi = 0. The acting forces are the weightmg of the piston, whereg= 9.81ms2 is the gravitational acceleration, and the pressure forces due to atmospheric and liquid pressure,p0AandpA, respectively. With the proper signs for the forces we have
mg+p0A−pA= 0 =⇒ p=p0+mg
A . (2.19)
The system pressure,p, balances the external pressure,p0, and the weight of the piston.
Gravitation leads to variation of water pressure with depth, and of air pressure with height. We compute both following Fig. 2.10. The water-air interface is at the locationz= 0 where the pressure is p0. The insert shows a small layer of substance, air or water, of thicknessdzand cross sectionA.
The mass of the layer, dm, follows from the mass density ρ and the layer volumedV =Adz asdm=ρAdz.
The forces acting on the layer are the contributions of the pressures below, p(z)A, and above,p(z+dz)A, and the weightg dm=ρgA dz. We assume the fluid (air or water) is at rest, so that the forces balance,
p(z+dz)A+ρgAdz−p(z)A= 0. (2.20) For infinitesimaldzwe can use Taylor’s formulap(z+dz) =p(z) +dp(z)dz dz and find a differential equation for pressure,
dp(z)
dz =−ρg . (2.21)
p(h) z= 0
z
dz p(z)
p(z+dz)
dm¢g
h
water air
p0
g
Fig. 2.10 On the computation of pressure variation in the gravitational field
To proceed, we have to differentiate between water and air. First we con- sider water: Water can be assumed in good approximation to be an incom- pressible substance, that is the water density is constant, with the well-known value ofρH2O ≃1000mkg3. Integration of (2.21) is straightforward, and gives, together with the conditionp(z= 0) =p0,
p(z) =p0−ρgz . (2.22)
This is the hydrostatic pressure law, which is often written in terms of depth h=−z as
p(h) =p0+ρgh . (2.23)
This relation is valid for all incompressible liquids, where the appropriate mass density ρ must be used. For water, depth increase by ∆h = 10.33 m increases the pressure by about 1 atm. Hydrostatic pressure depends only on depth, not on the actual weight of liquid above. This implies that hydrostatic pressure is independent of the geometry of the container, see Fig. 2.11 for an illustration.
Air, on the other hand, is compressible, it obeys the ideal gas lawp=ρRT. Using this to eliminate density from the differential equation for pressure (2.21), we find
dp(z)
p =− g
RT(z)dz . (2.24)
Integration is only possible when we have additional information on the tem- peratureT(z) as a function of heightz.
2.19 More on Pressure 29
p0
p(h) h
Fig. 2.11 Hydrostatic pressure depends only on depth
When the atmospheric temperature is constant, T(z) = T0, integration of (2.24) gives, together with the boundary condition p(z= 0) = p0, the barometric formula
p=p0e−RTgz0 . (2.25)
This formula describes the exponential decrease of pressure with height in an isothermal atmosphere.
In the actual atmosphere, however, the temperature is not constant, but decreases with height approximately as T(z) = T0−αz with α = 10kmK. Then, integration of (2.24) gives
p=p0 1−αz T0
αRg
. (2.26)
Figure 2.12 compares both pressure functions for air andT0= 293 K. For the non-isothermal atmosphere the pressure decreases slightly faster, but at moderate heights the difference is almost not noticeable. The Canadian town of Banff is located at an altitude of 1463 m above sea level. When the sea level pressure isp0= 1 atm we compute from (2.25) and (2.26) local pressures of 0.843 atm and 0.839 atm, respectively. Note that most weather forecasts do not present the actual local pressurep, but the normalized pressure, that is the corresponding sea level pressurep0. For instance, when the forecast gives the pressure for Banff as 990 kPa, based on (2.26) the actual pressure in town will be 831 kPa.
The example shows a marked influence of gravitation on pressure for heights on the kilometer scale. Most engineering devices are relatively small, at most on the scale of several meters, and the variation of gas pressure can be safely ignored. Therefore it is sufficient to assign just one value of pressure to a gaseous system in equilibrium.
Gas pressure results from the momentum change of those gas particles that hit the wall and bounce back, and thus exert a force. When a gas filled container is put on a scale, the scale will show the total weight of containerand gas, although most of the gas particles are not in contact with the container
0 2 4 6 8 10 0.3
0.4 0.5 0.6 0.7 0.8 0.9 1.0
z km@ D
pp0÷
variable temperature isothermal
Fig. 2.12 Atmospheric pressure over height for constant and variable temperature
walls. Indeed, it is the small variation with pressure between top and bottom of a container which puts the weight of the gas on the scale.
Problems
2.1. Ideal Gas I
A 5 litre container holds helium at a temperature of 25◦C and a pressure of 2 atm. Determine the mass of gas in the container, the number of moles, and the number of particles.
2.2. Ideal Gas II
A cylinder with radius 5 cm contains 5 g of pure oxygen at a temperature of 200◦C. The cylinder is closed with a freely moving piston, which in equilib- rium rests at a height of 50 cm. Determine the mass of the piston.
2.3. Ideal Gas Thermometer
An ideal gas thermometer holds a fixed gas volume of 1000 cm3.For calibra- tion, the thermometer is brought into contact with melting ice and boiling water, both at 1 atm, where the pressures measured arep1 = 51.6 kPa and p2= 70.5 kPa.
1. For Celsius temperature, assume a linear relation of the form T(◦C) = a+bpand determine the constantsaandb.
2. Determine the mole number of particles enclosed.
3. Careful measurement shows that the mass of gas enclosed is 1 g. Find the molar mass—what gas is it most likely?
2.4. Ideal Gas and Spring
The following process is done in a room at a temperature of 20◦C and a pressure of 100 kPa: A container with quadratic base of 10 cm side length is closed by a piston of massmp= 100 g. Initially, the piston is fixed at a height ofH0= 10 cm, and the cylinder is filled with 2.5 g of carbondioxide. A spring
2.19 More on Pressure 31
is attached to the piston from above, so that at the initial state the spring is at its rest length. When the fixing of the piston is removed, the piston moves up, and the spring is compressed. The system comes to an equilibrium state such that the piston has moved up by∆H= 3 cm.
1. Determine the spring constantk.
2. The gas in the container is now heated to 120◦C. Determine the final displacement of the piston.
2.5. Climbing a Mountain
In an atmosphere where the temperature depends on heightzasT =T0−αz, with T0 = T(z= 0) = 15◦C and α = 8.5kmK , a climber located at height z1= 500 m fills a piston-cylinder device with air, so that the device contains 2000 cm3of air. At this height, the climber measures an atmospheric pressure of p1 = 0.95 bar. The device is closed by a freely moving piston with mass mp= 300 g and surface area ofAp= 40 cm2. The climber carries the system to the top of the mountain, atz2= 4810 m.
1. Determine temperature and pressure of the atmosphere atz1,z2.
2. Determine the pressure inside the system atz1, and the mass of air in the system.
3. The climber reaches the top of the mountain atz2. After the system has established equilibrium with the surrounding air, determine the pressure in the system, and the system volume.
2.6. Ascent of a Balloon
The volume of a closed balloon shell is Vf = 800 m3, if completely filled.
The mass of the balloon, including basket, but without the gas filling, is mB = 500 kg. The temperature of the environment and of the gas filling is 5◦C, and remains constant during the ascent. Initially, the balloon is filled withV0= 500 m3of helium at the ground level pressure ofp0= 0.98 bar. As the balloon rises, its volume increases until it reachesVf.
For the solution of the following questions, assume that the air pres- sure depends on height z according to the barometric formula p(z) = p0exp
−RairgzT0
where Rair is the specific gas constant for air, and z is the height above ground.
Helium can be considered as an ideal gas withMHe= 4kmolkg . 1. Compute the mass of the helium filling.
2. As the balloon ascents, the volume of the filling increases. Compute the vol- ume of the balloon as function of height. Above which height is the balloon completely filled? (Hint: the pressures of helium filling and the surrounding air are equal as long the balloon is not filled completely).
3. Compute the buoyancy for V < Vf, and show that it is independent of height. The buoyancy force is given as FB = ρair(z)gV where V is the actual balloon volume, and ρair(z) the density of the surrounding air at heightz.
4. Set up an equation for the buoyancy for the case that the balloon is com- pletely filled. How high will the balloon rise?
2.7. An Experiment
As you accelerate in a car, you are pressed in the seat—what happens to a helium filled balloon? Think about it, or try it, then explain!