a 2.00L container of helium had a pressure of 0.75 atm what volume would be nessesary to increase the volume?
482843
Boyle's law\n\nP1xV1=P2xV2\n\n(.75 atm)x(2 L) = (2 atm)x(X)\n\n1.5=2x\n\ndivide both sides by 2\n\n.75 L would be the volume of helium if the container was at a pressure of 2 atm.
840037
PV = NRT\n\nN - Mols of argon\n\nR = .0821\n T = 26+273.15\n\nV  = 75/1000 to convert to L\n\nThen you will find your P, pressure, in atmospheres.\n\nconvert that to torr
798162
Timothy gives good advice.  If you are mixing helium with O2, then it must come through a SCUBA shop, else you risk disaster!\n\nFor other uses, none of those gases should be restricted for sale in any way.  A welding supply would hook you up.
670840
It depends on what the egg weighs.  The lift L (in dynes) of helium in air is equal to \n(rhoa - rhoh)*V*g  where\n\nrhoa = density of air (grams/liter)\nrhoh = density of helium (grams/liter)\nV = displaced volume at STP (Standard Temperature & Pressure)  (liters)\ng = gravitational constant = 980.665 cm/sec^2\n\nSolving this equation for V and putting in the density of air as 1.2928 g/l and helium as .1785 g/l gives\n\nV = L/1092.775\n\nSo, you gotta weigh the egg first and then find L by multiplying the weight in grams by 980.665 before using the formula above, which will then give the liters of He required to lift the egg.  A rough approximation is a bit under 1 liter per gram of egg.
116087
keeping with the same unit fractor harry is correct.\n \n .2 ml
173870
1.Since Volume is proportional to Temperature, it could also be written with a constant(k): V = k x T , V/T = k , k = 0.923L/ (25+273)K = 3.10x10^-3 L/K\n2.I'm not too sure about this one, but here is my best shot.\nUsing the ideal gas law PV=nRT, same Temp & Pressure, let Va be the volume of the first baloon and Vb be the volume of the second balloon. Va = 1.25Vb, Let na be the number of moles in baloon 'a' and nb be the number of moles in balloon 'b'.\n\nBalloon a: PVa = naRT\nBalloon b: PVb = nbRT\nsolve for P,T or R.\nThis case I solved for T:\nT = PVa/naR  & T = PVb/nbR, since same temp (or pressure or constant) which is PVa/naR = PVb/nbR. Divide by P & R, Va/na = Vb/nb\nVa x nb = Vb x na, Va = 1.25 Vb, @75%\n1.25(.75)Vb nb = 0.75Vb x na, divide by .75Vb\n1.25nb = na\nSo there is 25% more moles (or mass) in baloon A.
198092
Use:\n\npressure = density x height x g\nwhere g = 980 cm/s^2\n\nfor each liquid.  the answer (in dynes) will be the sum of the two pressures.
330764
Either drop the temperature (remove the enthalpic - heat - energy) or increase the pressure (the less space it has to move freely the closer to solid it becomes) or both
595563
Yes, it will weigh less. Fill ANY container with lighter-than-air gas, and it will weigh less.\n\nException: If the gas is under high pressure, such as in a pressurized tank, the container will actually weigh MORE.\n\nReason: Gas is compressible. If helium were half the weight of normal air, then helium under twice the pressure will weigh the SAME AS normal air under normal pressure. So if the pressure on the helium were increased, the container will weigh MORE.
741778
This is not tricky at all. The volume is 1-1/2 cylinders of "height" 50,000 meters and radius 4 meters.\n\nThe formula for the volume of a cylinder is\n\nV = pi * r^2 * h
196284
Take a quick sip of it and that air comes out when you speak.\n\nDon't try to do it a bunch of times in a row though. There's no oxygen in helium so i passed out once trying to do it for too long.
1074728
Helium is less dense than the air we normally breathe (Air is made up of approx 80% nitrogen and 20% oxygen)\n:\nWhen we talk air passes out from our lungs through our vocal chords and is caused to vibrate and you hear the words. When you mix helium with air the resulting mix is less dense so as you breathe out the timbre of your voice is altered. The spped of sound in a less dense helium/air mix is higher than in just air. You are increasing the speed of sound of your voice when breathing out this less dense mixture and this increase in speed will affect the frequencies that make up your 'voice'.\nIf the frequencies are affected the timbre is affected too. A more dense gas may have a very wierd affect.\nStrictly speaking although most people say it is the pitch that changes, it is the actually the 'timbre' which is changed, as your vocal chords produce the same shapes and therefore the same pitch.
200662
771+47=818 mmHg (It is a question of open manometer and the container side has higher pressure than atmosphere because the it is lower than the open end. So you just add the values.)\n\nTo convert it to atm and kPa use conversion factors below:\n\n760 mmHg = 1 atm = 101.325 kPa\n\n818/760=1.076 atm\n\n1.076*101.325=109.026 kPa or\n818*101.325/760=109.058 kPa
804421
The higher the balloon goes in the atmosphere the less pressure on the balloon. This causes the helium to expand its volume. If the balloon was filled to capacity it would *pop* at higher altitudes.
501025
you could use a blown up ballon for volume and a flat ballon for area.........cuz volume takes up space....and area is just how big something is
794147
You might work up some Q&A while showing this old demo of the effect of atm pressure...thereby adding some visual interest to the subject.\n\nAdd a few oz of water to a square-sided, metal, one-gallon can having a tightly fitted cap...bring to a vigorous boil (cap now off, please :-))..) and continue boiling until all air is expelled...screw cap in place and allow to cool...they will see the can collapse.\n\nWork up questions about the vacuum thus created (by the condensation of the water vapor) allowing atm pressure to exert its force.
937535
The helium is compressed when it is in the tank.  When it exits the tank, it expands in the balloons until it equalizes with atmospheric pressure.  Any gas can be compressed, as to occupy a smaller volume.  You can see this if you would take a plastic container fill it with air, and submerge it to the bottom of a swimming pool.  you would able to see that the container would deform because of the volume change of the container.