1) Yes, they mean pressure at the interface. They normally assume pressure is the same throughout the gas. (As you learned from studies of Archimedes' principle there are pressure gradients, especially in the liquid, and the pressure changes with depth). ## \\ ## In the case of a 3rd phase=solid, they generally would assume the liquid phase, (and the solid phase), isn't thick enough to create very much additional pressure. (In the phase diagrams, sometimes the pressure, (gas pressure=which is assumed to be the pressure everywhere in the system, even in the liquid and solid), can include pressures of 10 or 20 or even 100 atmospheres and more).## \\ ## 2) Molar volume is the inverse of the density.Jahnavi said:
Charles Link said:
Charles Link said:
1) The increase in the pressure from the depth in the liquid in a sample in a bottle or other container may be .05 atmospheres at most, and the change in gas pressure may be .01 atmospheres or less. On a scale where they are measuring things that may include pressures from 0 to 10 atmospheres, these changes are very small, e.g. the point on the graph at P=1.0 atmospheres is not affected significantly, and any differences caused by the effects of gravity to change the pressure can normally be ignored. ## \\ ## 2) Molar volume is the volume of 1 mole of material. If density is measured as ## \delta ## (grams/cm^3), then ## \frac{\delta}{M.W.}=\frac{1}{molar \, volume} ##.Jahnavi said:
Charles Link said:
?In very precise terms, the pressure is the same at the interface, and will vary with height, due to gravity, throughout both the gas and liquid. Sometimes, it can be important to take these pressure differences into account. In calculations where the barometric pressure is computed as a function of altitude, it is of prime importance. ## \\ ## In most ordinary laboratory calculations, gravitational effects on the system are usually ignored, but a good researcher will point them out, and include them, if they have a significant effect on the system being considered.Jahnavi said:OK .
Don't you think this is what the book meant when it says that the pressure is same throughout the system , and NOT that the pressure is same at the interface (even though it is ) to which you have agreed
It would appear they are referring to a result that follows from Statistical and Thermal Physics, that the density of a gas as a function of energy will be of the form ## \rho(E)=\rho_o \, e^{-E/(k_B T)} ## where ## E=+mgh +(\frac{1}{2})mv^2 ##. This calculation is somewhat advanced. In addition, the average of ## \frac{1}{2} mv^2=\frac{3}{2}k_B T ##, so that you can compute the effect that the ## mgh ## term will have on the density. ## \\ ## If ## mgh<<k_B T ## , the height ## h ## will have very little effect on the density. ## \\ ## Edit: The result of the above is basically that the density as a function of height ## h ## will be ## \rho(h)=\rho(0) \, e^{-mgh/(k_B T)} ##. If ## mgh <<k_BT ## for all ## h ##, then the density is nearly uniform. And if ## mgh <<k_B T ##, you can also say that ## mgh << \frac{1}{2}mv^2 ##.## \\ ## The book's author apparently is familiar with this type of calculation, but it is really a somewhat advanced upper level undergraduate topic. ## \\ ## ................................## \\ ## And it may be worthwhile to relate this to Archimedes' result: force per unit volume from the pressure gradient must offset the gravitational force per unit volume, so that ## -\nabla P=\rho g \, \hat{z} ##. ## \\ ## Now ## PV=nRT ## and ## \rho=\frac{n \, M.W. }{V} ##, so that ## P=\frac{\rho RT }{ M.W.} ##. ## \\ ## The result is ## (\frac{RT}{M.W.}) \frac{d \rho}{dz}=-\rho g ##. ## \\ ## Solving this differential equation, we get ## \rho=\rho_o \, e^{-(M.W.) (g z)/(RT)} ## . Here the gas constant ## R=N_o \, k_B ##, where ## N_o ## is Avagadro's number (particles/mole). Thereby, ##\frac{ M.W. (in \, grams/mole)}{N_o}=m ##, where ## m ## is the mass of the gas molecule. ## \\ ## This gives ## \rho=\rho_o \, e^{-mg z/(k_B T)} ##. ## \\ ## These results from Archimedes and the ideal gas law are consistent with our formula above ## \rho(h)=\rho(0) \, e^{-mgh/k_B T} ##, which can also be written in the form ## P(h)=P_o \, e^{-mgh/(k_BT)} ##. Note that although the pressure has a somewhat complicated form as a function of height ## h ##, (more complicated than for a liquid), Archimedes principle still works with a gas as well: The buoyant force is equal to the weight of the gas that is displaced. ## \\ ## It may also be of interest, that for an uncompressible liquid, the pressure varies with depth ##d ## as ## P(d)=P(0) +\rho_o \, g \, d ##. Gases are compressible, so that the equation describing pressure as a function of height is much different. ## \\ ## .............................. ## \\ ## And let's try some gas molecules at room temperature: ## \\ ## ## k_B T=1.381 \cdot 10^{-23} (300) \approx 4.1 \cdot 10^{-21} \, joules ##, ## \\ ## and ## mgh ## for a height of ##h= 1.0 \, meters## , (## M.W.=30 ##), is ## \\ ## ## mgh \approx 30 (1.67 \cdot 10^{-27}) (9.8)(1.0)=4.9 \cdot 10^{-25} \, joules ##. ## \\ ## It looks like ## k_BT>> mgh ##. ## \\ ## The equation also gives the result that the atmospheric pressure should be ## P(h) \approx 0.5 \, atm ## at ## h=5800 \, meters ## altitude. ## \\ ## Edit: Note that the atmosphere is 78% ## N_2 ## with M.W.=28, and 21% ## O_2 ## with M.W.=32, so that using a number of M.W. ## \approx ## 30 for the atmosphere should give us reasonably accurate results for the pressure as a function of altitude. (Temperature is also known to vary somewhat with altitude=we're just looking for approximate results for the pressure vs. altitude calculation).Jahnavi said:
. The purpose of verifying the answer to PV/T at A is to ensure that the calculated values for pressure, volume, and temperature are accurate and reflect the real-world conditions at point A. This verification process is important in order to validate the results and make sure they can be relied upon for further analysis or experimentation.
PV/T at A is calculated by using the ideal gas law, which states that the product of pressure and volume is directly proportional to the absolute temperature. This equation is written as PV = nRT, where n is the number of moles of gas and R is the gas constant. By rearranging this equation, we can solve for T and determine the temperature at point A.
Comparing T2 and T1 in PV/T at A allows us to see if there is a change in temperature between the initial and final states of the gas. This can provide insight into the behavior of the gas and help us understand how it responds to changes in pressure and volume.
There are several factors that can affect the accuracy of the answer to PV/T at A, including experimental errors, variations in gas behavior, and limitations of the ideal gas law. It is important to carefully control and measure these variables in order to obtain the most accurate results.
No, PV/T at A, or any other calculation based on the ideal gas law, is only applicable to ideal gases. Real gases may deviate from ideal behavior under certain conditions, such as high pressures or low temperatures, and therefore, the results from PV/T at A may not accurately predict the behavior of all gases. It is important to consider the properties of the specific gas being studied when using this calculation.
