What is 1 Bar of Atmosphere in terms of water pressure?
105685
1 bar =  1x10^5 Pa and 1 atm = 101325 Pa\ntherefore 1 bar = 0.987 atm\n\n_______\nOK, got it.  Just for future reference, you can clarify a question using the "add details" button in the stick note in the top right rather than posting it as the answer to a different question :)\n\nWith a signficant increase in height, water pressure will change signficantly.  Basically, that extra 33-foot column of water weighs quite a bit and is pushing down on things under it.\n\nThe density of water is 1 g/cm^3 = 0.001 kg/cm^3.  33 feet = approx. 1,000 cm.  A 1,000 cm high column of water with one cm^2 cross-secional area would weigh about 1kg, adding 1kg/cm^2.  1kg/cm^2 = 1.097 bar.  So being three stories up would add just about 1 bar of water pressure.\n\nI'm not sure how these water tank assemblies work, but it seems to me that if you put the tank three stories up and didn't pressurize it you would have about 1 bar of water pressure at the bottom.  If you pressurized the tank so you had 1 bar of water pressure at the top, you'd have 2 bars down where your dishwasher is.
989194
If you are talking about the limit on the suction side of a pump then the answer is about 34 feet. This is because due to atmospheric pressure that is all the water can be raised without the portion at the bottom of the column starting to vaporize. \n\n14.7 psia (atmospheric pressure) equals 33.9 feet of water.
1034583
For example, the density of water is 1000 kilograms per cubic meter (in SI units), so the weight of a cubic meter of water is 1000 kg times g, the acceleration of gravity (9.8 m/sec2), or 9800 newtons. This force is exerted over 1 m2, so the pressure produced by a 1-meter depth of water is 9800 pascals (the Pa is the SI unit of pressure, equal to 1 newton per square meter). \n\nThe unit of pressure used in atmospheric work on Earth is the hectopascal; 1 hPa = 100 Pa. So the pressure 1 m below the surface of water (ignoring the pressure exerted by the atmosphere on top of it) is 98 hPa. Standard atmospheric pressure is 1013.25 hPa, so it takes 1013.25/98 = 10.33 meters of water to produce a pressure of 1 atmosphere. (That's about 34 feet, for those who like obsolete units.) \n\nThe pressure in the ocean increases by about 1 atmosphere for every 10 meters of depth. The average depth of the ocean is about 4 km, so the pressure on the sea floor is about 400 atmospheres.
963360
A possible reason this is incorrect is when you plugged your numbers into PV=nRT, you only had 1 mmHg while the problem reads 17.5 mmHg (2333.14 Pa). Also, you are to subtract this from the barometric pressure to get your actual pressure, not multiply (according to the link mei gave). I don't remember doing problems exactly like this, but it appears otherwise you have set it up correctly. The only other thing I can suggest is to double check your units on R. You didn't list them.
856212
There is a standard model of the atmosphere which is used for calibrating instruments such as aircraft altimeters.  The standard sea level pressure is 1.01325 bars, or about 760 mm Hg.  In a vacuum, the pressure is reduced sufficiently to facilitate whatever process the vacuum was intended to enable; in laboratories, vacuums of 1E-8 bar or lower are attainable.
1145151
A liquid will boil when its vapor pressure at the given temperature equals the [atmospheric] pressure above the liquid.\n\nI could not find a vapor pressure value of water at 90 degrees, but,\nAt 86 degrees F, the vapor pressure of water is about 31.8 mmHg.\nIf you lower the pressure above water to 31.8 mmHg at 90 degrees F, the water will boil.\n\n31.8 mmHg is about .042 atmospheres.
978221
p = 136(1 - e^(-0.51t))\n\n(i) Find the initial value of the pressure\n\nInitial always means at t = 0\nWhen t = 0,\np = 136(1 - e^(-0.51*0))\n= 136(1 - e^(0))\n= 136(1 - 1)\n= 0 bars\n\n(ii) Find the value of the pressure after 2.39 minutes\n\nWhen t = 2.39\np = 136(1 - e^(-0.51*2.39))\n= 136(1 - e^(-0.51*2.39))\n= 136(1 - e^(-1.2189))\n≈ 136(1 - 0.295555)\n≈ 95.80 bars\n\n(iii) Find the value of the pressure once it has settled after a long period of time\n\nAs t gets large  e^(-0.51t) = 1/e^(0.51t) = 1/(large nimber) and so gets smaller and smaller.\nTherefore (1- e^(-0.51t) gets closer and closer to 1\nSo p=136(1- e^(-0.51t))\ngets loser and closer to 136 * 1\nie it will settle on p = 136 bar\n\neg when t = 30 minutes (½ hour)\n\np = 136(1 - e^(-0.51*30))\n= 136(1 - e^(-15.3))\n≈ 136(1 - 2.266 x 10^(-7))\n≈ 136
1002151
14.7 PSI is the atmospheric pressure at sea level. It has nothing to do with gravity.
146352
A bar is a linear shoaling landform feature within a body of water. Bars tend to be long and narrow (linear) and develop where a current (or waves) promote deposition of particles, resulting in localized shallowing (shoaling) of the water. Bars can appear in the sea, in a lake, or in a river. They are typically composed of sand, although could be of any particulate matter that the moving water has access to and is capable of shifting around (for example, soil, silt, gravel, cobble, shingle, or even boulders). The size of the particles comprising a bar is related to the size of the waves or the strength of the currents moving the material, but the availability of material to be worked by waves and currents is also important. A vessel run aground on a bar is arguably worse off than one run aground on stationary rocks, because of the destructive capacity of the shifting action.\n\nThe term bar can apply to landform features over a considerable range in size, from just a few meters in a small stream to marine depositions stretching for hundreds of kilometres along a coastline, often called barrier islands.
875826
From Wikipedia:\n\nThe bar and millibar are defined as:\n\n1 bar = 100 000 pascals (Pa) = 1 000 000 dynes per square centimetre \n1 mbar = 0.001 bar = 100 Pa = 1 000 dyn/cm² \n(A pascal is one newton per square meter.)\n\n\n[edit] Origin\nThe word bar has its origin in the Greek word βάρος (baros), meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.\n\nThe bar and millibar were introduced by Sir Napier Shaw in 1909 and internationally adopted in 1929.\n\nAtmospheric air pressure is often given in millibars where "standard" sea level pressure is defined as 1013.25 mbar (hPa), equal to (1.01325 bar). Despite millibars not being an SI unit, they are still used locally in meteorology to describe atmospheric pressure.\n\nThe article goes further to explain that 1 mbar = 1 hPa, or 1 millibar = 1 hectoPascal.\n\nIn terms of inches Hg, 30.00 has long been accepted as the standard, or long-term average pressure at sea level.  30.04 in. is slightly above standard pressure, as 1017hPa is slightly above 1013.25.  In weather terms this would indicate clear or clearing skies, with winds shifting from counter-clockwise to clockwise.\n\nThe popularity of the millibar (now hPa) unit seems to stem from the coincidence that 1 bar is almost equal to one atmosphere.
998158
An elongated area of high atmospheric pressure that is associated with an area of maximum anticyclonic circulation. The opposite of a trough. An elongated area of low atmospheric pressure that is associated with an area of minimum cyclonic circulation. ISOBARS are lines drawn on a synoptic chart joining together areas of equal pressure. The further apart they are, the lighter the wind. When they are close together the pressure is usually low and the wind is strong
339441
1 atm of pressure is about 14.7 psi,\nso 2175 psi would be about 148 atm.
635522
It depends on the liquid AND the pressure. 100C is for water at atmospheric pressure.
41200
If it is a high-volume, low pressure pump, one of the easiest ways is to get a long piece of clear tubing and make a U-tube manometer.  the pressure will cause the water to climb in the tube vented to the atmosphere, while the other end is connected to the pump outlet.  You measure the height of the water column, and can get the head directly in feet of water, a popular unit of measurement in the pump industry.  If it is a high pressure pump, a simple pressure gage from the local plumbing supply or hardware store will read in PSI.  In rough terms, 14.7 psi is approximately equal to 32.2 Feet of water column pressure.
186344
At 20 degrees C, the vapor pressure is 17.54 mmHg.  At 25 degrees C, it's 23.76 mmHg.  You'd get close if you interpolate between these two points, even though the curve, strictly speaking, is not linear.  \n\nOr, if you're ambitious, use the Antoine equation:  \n\nAntoine Equation Parameters\n\nlog10(P) = A − (B / (T + C))\n\n    P = vapor pressure (bar)\n    T = temperature (K)\n\nFor temps in the range 273. - 303. Kelvin, A = 5.40221, B = 1838.675 and C = -31.737.
380084
The answer is Pressure\n\n10 pts plz! im only level 1!
132266
Ignore harsha.  A 1 atm decrease in pressure would mean zero pressure at 100 meters above sea level, which is absurd.\n\nPressure decreases as you rise because the total mass of the atmosphere weighing down on you decreases.  High pressure means you need more energy to make water boil, so at sea level the boiling point will be higher than it will be on a mountain top.