partial pressure

Examples of partial pressure in the following topics:

• Dalton's Law of Partial Pressure

  • Dalton's Law of Partial Pressure states the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of each individual gas.
  • Dalton's Law (also called Dalton's Law of Partial Pressures) states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
  • where P1, P2 and Pn represent the partial pressures of each compound.
  • What is the partial pressure of He?
  • Our measurements demonstrate that the partial pressure of N2 as part of the gas PN2 is 0.763 atm, and the partial pressure of O2 as part of the gas PO2, is 0.215 atm.

• Basic Principles of Gas Exchange

  • Partial pressure (Px) is the pressure of a single type of gas in a mixture of gases.
  • For example, in the atmosphere, oxygen exerts a partial pressure, and nitrogen exerts another partial pressure, independent of the partial pressure of oxygen (Figure 1).
  • Total pressure is the sum of all the partial pressures of a gaseous mixture.
  • A gas will move from an area where its partial pressure is higher to an area where its partial pressure is lower.
  • Partial pressure is the force exerted by a gas.

• Dalton's Law of Partial Pressure

  • Dalton's law of partial pressures states that the pressure of a mixture of gases is the sum of the pressures of the individual components.
  • Dalton's law states that the total pressure exerted by the mixture of inert (non-reactive) gases is equal to the sum of the partial pressures of individual gases in a volume of air.
  • Mathematically, the pressure of a mixture of gases can be defined as the sum of the partial pressures of each of the gasses in air.
  • Because gasses flow from areas of high pressure to areas of low pressure, atmospheric air has higher partial pressure of oxygen than alveolar air (PO2=159mmHg compared to PAO2=100 mmHg).
  • Infer from Dalton's law of partial pressure the sum of partial pressures in alveoli

• Henry's Law

  • Henry's law states that the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure of that gas.
  • In addition, the partial pressure is able to predict the tendency to dissolve simply because the gasses with higher partial pressures have more molecules and will bounce into the solution they can dissolve into more often than gasses with lower partial pressures.
  • The amount of oxygen that dissolves into the bloodstream is directly proportional to the partial pressure of oxygen in alveolar air. 
  • Recall that the difference in partial pressures between the bloodstream and alveoli (the partial pressure gradient) are much smaller for carbon dioxide compared to oxygen.
  • Oxygen has a larger partial pressure gradient to diffuse into the bloodstream, so it's lower solubility in blood doesn't hinder it during gas exchange. 

• Gas Pressure and Respiration

  • The pressure for an individual gas in the mixture is the partial pressure of that gas.
  • The partial pressure of any gas can be calculated by: P = (Patm) (percent content in mixture).
  • Patm, the atmospheric pressure, is the sum of all of the partial pressures of the atmospheric gases added together: Patm = PN2 + PO2 + PH2O + PCO2= 760 mm Hg.
  • The pressure of the water vapor in the lung does not change the pressure of the air, but it must be included in the partial pressure equation.
  • At high altitudes, there is a decrease in Patm, causing the partial pressures to decrease as well.

• Expressing the Equilibrium Constant of a Gas in Terms of Pressure

  • For gas-phase reactions, the equilibrium constant can be expressed in terms of partial pressures, and is given the designation KP.
  • For gas-specific reactions, however, we can also express the equilibrium constant in terms of the partial pressures of the gases involved.
  • Our equilibrium constant in terms of partial pressures, designated KP, is given as:
  • The reason we are allowed to write a K expression in terms of partial pressures for gases can be found by looking at the ideal gas law.
  • Write the equilibrium expression, KP, in terms of the partial pressures of a gas-phase reaction

• Really Little Sound Waves

  • $\displaystyle \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho {\bf V}) = \frac{\partial \rho'}{\partial t} + \rho_0 \nabla \cdot {\bf V}' = 0 $
  • $\displaystyle \frac{\partial P'}{\partial t} + \rho_0 \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla \cdot {\bf V'} = 0$
  • $\displaystyle \frac{\partial ^2 P'}{\partial t^2} + \rho_0 \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla \cdot \frac{\partial \bf V'}{\partial t} = 0.$
  • $\displaystyle \frac{\partial ^2 P'}{\partial t^2} - \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla^2 P' = 0.$
  • Let's take a solution to this equation for the pressure,

• Specific Heat for an Ideal Gas at Constant Pressure and Volume

  • An ideal gas has different specific heat capacities under constant volume or constant pressure conditions.
  • Specific Heat for an Ideal Gas at Constant Pressure and Volume
  • The heat capacity at constant pressure of 1 J·K−1 ideal gas is:
  • where the partial derivatives are taken at: constant volume and constant number of particles, and at constant pressure and constant number of particles, respectively.
  • For an ideal gas, evaluating the partial derivatives above according to the equation of state, where R is the gas constant for an ideal gas yields:

• Real Sound Waves

  • In this case, we can express the pressure in terms of the density alone.
  • $\displaystyle \frac{\partial \rho}{\partial t} + \frac{(\partial \rho v)}{\partial x} = 0, ~~~ \frac{\partial v}{\partial t} + v \frac{\partial v}{\partial x} + \frac{1}{\rho} \frac{\partial P}{\partial x} = 0$
  • Using the relationships between the pressure, velocity and density we can obtain,
  • $\displaystyle \frac{\partial \rho}{\partial t} + \frac{(d\rho v)}{d\rho} \frac{\partial \rho}{\partial x} = 0, ~~~ \frac{\partial v}{\partial t} + \left ( v + \frac{1}{\rho} \frac{dP}{dv} \right ) \frac{\partial v}{\partial x} = 0.$
  • We can use this result to express the density and pressure in terms of the fluid velocity

• Steady Supersonic Flow

  • $\displaystyle v \frac{\partial v}{\partial t} = - \frac{1}{\rho} \frac{\partial p}{\partial t} = -\frac{c_s^2(\rho)}{\rho} \frac{\partial \rho}{\partial t}$
  • If the area of the pipe decreases (nozzle) in the direction of the flow, the velocity increases and the pressure and density decrease.
  • If the area of the pipe increases (diffuser) in the direction of the flow, the velocity decreases and the pressure and density increase.
  • If the area of the pipe decreases (nozzle) in the direction of the flow, the velocity decreases and the pressure and density increase.
  • If the area of the pipe increases (diffuser) in the direction of the flow, the velocity increases and the pressure and density decrease.