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(II) When using a mercury barometer, the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 $\mathrm{mm}$ -Hg. At sea level, the height $h$ of mercury in a barometer is about 760 $\mathrm{mm} .$ (a) If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? (b) What is the percent error? (c) What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?

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(a) $P_{\text {atm }}-P_{\text {vapor }}<P_{\text {atm }}$(b) $\left(-2.0 \times 10^{-4}\right) \%$(C) $0.603 \%$

Physics 101 Mechanics

Chapter 18

Kinetic Theory of Gases

Temperature and Heat

Cornell University

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster University

Lectures

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okay. And this problem, we are asked. You compare values from a barometer. So we know that at room temperature the pressure of mercury is 0.15 millimeters murky and at sea level. The hide in a barometer is about 760 millimeters, and Cordelia's prom Asif neglected if we just neglect, ah, neglected the mercury with the atmospheric pressure be greater than lesson or equal to this value of mercury were then exactly computer percent Harris's quantitative and then finally were asked for a present air. If instead, we use thes standard temperature pressure, uh, pressure of water, a lot of paper. So to start, For Bernet, true atmospheric pressure will be greater than the reading from the barometer. So we look at figure 13-11 if there's a per pressure at the top of the tube and the vapor pressure is going to be equal to atmospheric pressure minus row G H, which is the pressure of air pushing down. So the reading for the barometer rejection to see is ro D. H, which is just gonna be the difference between atmospheric pressure and vapor pressure, which is necessarily going to be less than atmosphere because pressure from the vapor is not zero in this case. So just a little bit of conceptual difficulty there. But the actual math is this around to be straightforward. Now you get a cent error. We just take row G H minus that atmospheric pressure over atmospheric pressure. But of course, the animal to apply that by 100. No problem. And this is the same. And that number. This difference is the same thing as vapor pressure by definition in party. So what we're dealing with is 0.15 millimeters mercury, and we have to get atmospheric pressure millimeters mercury, which it turns out to be 760 which you just look that up. So our value is going to be about two times 10 the negative 4% so pretty, really. It's a very small percentage, with meaning that we it's justified to neglect to this pressure of the mercury vapor pressure. Okay, now let's look at it for saturated water vapor pressure. So at STP, we can consult Table 18 dash, too. It's been for room temperature. We would see the water vapor pressure PV we'll call it is 6 11 Pascal. So once again we can get a percent error percent difference, which would once again be PV over p a t m amateur pressure. Except we just change the value of the vapor pressure state of 6 11 Pascal. And then we get atmosphere in Pascal's, which is roughly one times 10 to the five. You can really see that this is gonna be a much small, larger than the previous answer, but still really small because we're dividing by 10 to the five and we end up with, Oh, of course, multiplied by 100. We end up with 1000.6 of 3% so less than a percent. So that's more significant than mercury, but still not particularly significant. So that's it. Consulting some tables, Uh, and that's it.

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