its about ideal gas law...?
813197
0.0821 moleP1V1/T1=P2V2/T2\n375*4/293=760*V2/273.5\nV2=375*4*273.5/(293*760)\n22.4 litres contains 1mole\n375*4*273/293*760\n=375*4*273/293*760*22.4\n=0.0821 mole
757981
Ideal gas law\n\nPV=nRT\n\nP is pressure\nV is volume\nn is moles of gas\nR is the ideal gas constant\nT is temperature\n\nGood luck!
24946
For gases, we speak of "standard gas density." This is the density of the gas (expressed in grams per liter) at STP. If you discuss gas density at any other set of conditions, you drop the word standard and specify the pressure and temperature. Also, when you say "standard gas density," you do not need to add "at STP." STP is part of the definition of the term. It does no harm to say "standard gas density at STP," it just looks a bit silly. \n\nYou can calculate the ideal standard gas density fairly easily. Just take the mass of one mole of the gas and divide by the molar volume. \n\nMore explanation at the site linked below.
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.
782736
If ideal gas law is obeyed...\n\ndensity = mass/V == n/V\n\nPV = nRT\nP = (n/V)RT\n\nIf temp is held constant then increasing density increases pressure.
1080767
It depends on the pressure and temperature. 50 lbs. of O2 represent a fixed number of moles of CO2. From the ideal gas law PV=nRT. Cannot solve for volume V without knowing P and T.
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
956716
The first question is not a basic question. It is not even a good question. "Ideal" is what you decide to call your model. It is your definition and you may use it as you see fit. I doubt your teacher wants that particular answer to the first question, but it is a true one.\n\nYou have the right idea for the second one, but you could describe it better. First, you must heat alcohol up to turn it into a gas, so you know that the gas has more internal energy... because you added some to make it gaseous. The reason that gas molecules tend to be spaced farther apart from each other than liquids is that increased energy causes atomic motion to increase. The atoms jiggle more and more until they jiggle away into a gaseous state, so, really, the molecules are farther apart because of an increased energy, not the other way around.
736070
If you know the volume and density of the liquid you can calculate the number of moles n.\n\nOnce you have n and if you assume that the gas will behave like and ideal gas and you fix the temperature and pressure, the volume will be given by:\n\nV = nRT/P.
724196
Ideal gas is a gas that behaves according to the ideal gas equation,\nPV = nRT\nwhere P = pressure, V = volume, n = number of moles, R = ideal gas constant, T = temperature.\n\nGenerally, ideal gas is an imaginary gas that follows the assumption in which each gas molecule occupy no space and there is no attraction between any adjacent molecules. These assumption becomes valid to a real gas under the condition when the temperature is high or the pressure is very low, because in both cases, each molecule experience very little interaction with the adjacent molecules and thus behave very ideal.
865453
Assuming temperature is static, I think Dalton's Law would apply:\nThe sum of all the partial pressures.\n\nP(gas 1)V(vol 1) = P(new gas 1)V(together)\n10 bars * 300L = ? bars * 600 L\nP(new gas 1) = 5 bars\n\nP(gas 2)V(vol 2) = P(new gas 2)V(together)\n3 bars * 400L = ? bars * 600 L\nP(new gas 2) = 2 bars\n\nP(together) = P(new gas 1) + P(new gas 2)\nP(together) = 7 bars
193517
You don't provide enough data to determine this.  The gas law PV=nRT has variables you need to define.
979386
it is a ratio. \nBecause there are 2 moles of A for every 2 moles of D there is a 1:1 ratio. That means that for every 2 liters of gas A there are 2 liters of gas B.\n\nHope I helped
33029
Some of the basis laws that govern the behavior of gases follow:\n\nBoyle's Law= Volume is proportional to the inverse of pressure\n\nCharles' Law= Volume is proportional to temperature\n\nAvaogadro's Law= Volume is proportional to number of moles present\n\nThese together lead to the ideal gas equation:\n\nPV = nRT\n\nwhere P is pressure\n      V is volume\n      n is the number of moles present\n      R is the gas constant\n      T is the temperature\n\nThe value of R is 0.082057 L*atm/(mol*K).\nTemperature must be in kelvins for this relationship.  Temperature in kelvins = tempature in Celcius +273.15.
956387
Ideal gas law PV=nRT, which means Pressure x Volume = constant x Substance x Temperature.  The n remains constant and the Volume increase is so small that we'll say it is constant too.  That leaves us with P = RT, we know the R (substance) goes up since you are filling the tire with air, but we have two variables left, and depending on certain conditions the temperature or pressure can go up or down.  For that you would need to apply the second law of thermodynamics (entropy, simply stated if you put something hot next to something cold, the cold thing will get hotter and the hot thing will get colder, not vise versa) If you use hot air to fill the tire and the rubber from the tire is cold, the temp will go down (which will also decrease the pressure), if the air is colder than the tire the temp will increase (and so will the pressure).
414884
Are you sure the unit is micropascal? I think there is an error in you units, its too small.\n3141.59 micropascal = 0.000000031 atmosphere\n\nPressure in ideal monoatomic gases is directly proportional to temperature, so: \nP1/T1=P2/T2\nT(ºK)=42,000*(177.94+273)/.000000031\n\nCheck the units and resolve it the same way.