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Chapter 14: Problem 26

To suck lemonade of density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) up a straw to a maximum height of \(4.0 \mathrm{~cm}\), what minimum gauge pressure (in atmospheres) must you produce in your lungs?

Short Answer

Expert verified
0.00387 atm

Step by step solution

01

Understand the Problem

We need to find the minimum gauge pressure required to suck lemonade up a straw to a height of 4.0 cm. The lemonade has a density of \(1000 \, \text{kg/m}^3\). Gauge pressure refers to the pressure relative to atmospheric pressure.
02

Use the Hydrostatic Pressure Formula

The pressure needed to lift a column of liquid to a certain height is given by the hydrostatic pressure formula: \( P = \rho \cdot g \cdot h \).Explain the terms: \( \rho = 1000 \, \text{kg/m}^3 \) (density), \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity), \( h = 0.04 \, \text{m} \) (height in meters).
03

Calculate the Hydrostatic Pressure

Substitute the values into the formula to calculate the pressure:\[P = 1000 \, \cdot \, 9.81 \, \cdot \, 0.04 = 392.4 \, \text{Pa}.\]This is the gauge pressure required to lift the lemonade to 4 cm.
04

Convert Pressure from Pascals to Atmospheres

To convert the pressure from pascals to atmospheres, use the conversion factor: \(1 \, \text{atm} = 101325 \, \text{Pa}\). Calculate the pressure in atmospheres:\[P = \frac{392.4}{101325} \approx 0.00387 \, \text{atm}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauge Pressure
Gauge pressure is a concept used to measure the pressure in a system above atmospheric pressure. Unlike absolute pressure, which includes the pressure of the atmosphere, gauge pressure is solely the pressure difference between the system and the surrounding environment.

For example, if the atmospheric pressure is 101325 pascals (Pa) and a system has a total pressure (absolute pressure) of 102000 Pa, the gauge pressure is the difference between these two values: 675 Pa (102000-101325).

This understanding is important when discussing pressure in closed environments, like in the exercise where you must create a certain gauge pressure within your lungs to suck lemonade through a straw. This pressure must be sufficient to overcome atmospheric pressure and lift the liquid to the desired height.
  • Gauge pressure helps determine the actual pressure exerted by a fluid in a controlled system.
  • It is useful because it directly relates to the conditions we can manipulate, such as lung pressure in our example.
Density
Density is a measure of how much mass is contained within a specific volume of a substance. It is defined by the formula: \[\text{Density} \, (\rho) = \frac{\text{mass}}{\text{volume}}.\]
In the exercise, the density of the lemonade is given as 1000 kg/m鲁. This means that every cubic meter of lemonade has a mass of 1000 kilograms.

Density is crucial in calculating hydrostatic pressure because it determines how much pressure is exerted by a fluid at any given depth.
  • Higher density fluids will exert more pressure at a given depth.
  • Density is a key factor in the hydrostatic pressure formula, alongside the height of the liquid column and gravitational acceleration.


Understanding density helps you see why different fluids require different pressures to move or lift, such as sucking lemonade through a straw.
Pressure Conversion
Pressure conversion is necessary when we need to express pressure in different units. In physics, pressure can be represented in various units, with pascals (Pa) and atmospheres (atm) being common ones.

The exercise requires finding the pressure in atmospheres once it is calculated in pascals. Converting between these units is simple with the conversion factor: \[1 \, \text{atm} = 101325 \, \text{Pa}.\]

To convert the calculated hydrostatic pressure of 392.4 Pa into atmospheres, you use the conversion factor by dividing the pressure in pascals by the number of pascals in one atmosphere:
\[\frac{392.4}{101325} \approx 0.00387 \, \text{atm}.\]

Such conversions are essential for interpreting pressure in a way that aligns with the context or standard expectations.
  • Different units of pressure serve different practical applications.
  • Understanding conversion methods ensures accurate communication of measurements.

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Most popular questions from this chapter

(a) If \(m\) is a particle's mass, \(p\) is its momentum magnitude, and \(K\) is its kinctic energy, show that $$ m=\frac{(p c)^{2}-K^{2}}{2 K c^{2}} $$ (b) For low particle speeds, show that the right side of the equation reduces to \(m\). (c) If a particle has \(K=55.0 \mathrm{MeV}\) when \(p=\) \(121 \mathrm{MeV} i c,\) what is the ratio \(\mathrm{m} / \mathrm{m}_{e}\) of its mass to the clectron mass?

Ionization measurements show that a particular lightweight nuclear particle carries a double charge \((=2 e)\) and is moving with a speed of \(0.710 c\). Its measured radius of curvature in a magnetic field of \(1.00 \mathrm{~T}\) is \(6.28 \mathrm{~m}\). Find the mass of the particle and identify it. (Hints: Lightweight nuclear particles are made up of neutrons (which have no charge) and protons (charge \(=+e\) ), in roughly cqual numbers. Take the mass of cach such particle to be 1.00 u.) (See Problem 53.)

How much energy is released in the explosion of a fission bomb containing \(3.0 \mathrm{~kg}\) of fissionable material? Assume that \(0.10 \%\) of the mass is converted to relcased energy. (b) What mass of TNT would have to explode to provide the same energy release? Assume that each mole of TNT liberates \(3.4 \mathrm{MJ}\) of energy on exploding. The molecular mass of TNT is \(0.227 \mathrm{~kg} / \mathrm{mol}\). (c) For the same mass of explosive, what is the ratio of the energy released in a nuclear explosion to that released in a TNI explosion?

What is the momentum in MeV \(/ c\) of an electron with a kinetic energy of \(2.00 \mathrm{MeV} ?\)

In one observation, the column in a mercury barometer (as is shown in Fig. \(14-5 a\) ) has a mcasured height \(h\) of \(740.35 \mathrm{~mm}\). The temperature is \(-5.0^{\circ} \mathrm{C},\) at which temperature the density of mercury \(\rho\) is \(1.3608 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3} .\) The free-fall acceleration \(\mathrm{g}\) at the site of the baromcter is \(9.7835 \mathrm{~m} / \mathrm{s}^{2}\). What is the atmospheric pressure at that site in pascals and in torr (which is the common unit for barometer readings)?
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