If the absolute pressure of a gas is 55.280 kPa, its gage pressure is what?
543227
55.280 kPa - 101.325 kPa = -45.045 kPa\n\nAbsolute pressure is gauge pressure plus atmospheric pressure which is 101.325kPa.
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
412802
p=F/A \n\nSo, you have the pressure of the atmosphere pluss the weight of the water above you.  Which is usually figured as a cylinder.  Hope that helps.
687925
P = Pressure in atmospheres\nV = Volume in Liters\nn = number of moles\nR = ideal gas constant, which in this equation is 0.08206 L * atm * mol^-1 * K^-1\nT = temperature in Kelvin (degrees Centigrade + 273.15)
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.
1105154
pressure is: \n p(density)gh / gc =y(specific weight)h\nflow of liquid flowing inside the pipeline is:\n use the bernoulli equation:\n [p1/p(density)g  +  v1^2/2g  +  z1]  =  [p2/p(density)g  +  v2^2/2g  +  z2  +  hf]\n\nnote: p(density) is rho and y(specific weight) ir gamma.
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.
1002151
14.7 PSI is the atmospheric pressure at sea level. It has nothing to do with gravity.
786443
mass * gravity / area\n\n79.5 * 9.81 / (2 * 0.018) = average pressure\n\nunits:\n\nkg * (m/s^2) / (m^2) = N/m^2 = Pa (Pascals)
926935
PV = nRT\n\n685 mmHg / 760 = .901 atmospheres\n\n18.6 mmHg / 760 = .0245 atm\n\n.901-.0245 = .8765 atm\n\n21 degrees c = 294 K\n\n.8765 atm x 7.80 L = n x .08206 x 294 K\n\nsolve for n then use the molar mass of H2 (2.016) to convert to grams.\n\nmy calculator's out in the car or else i'd do it for you.
1149287
pV = n R T => if pressure is constant => p = n R T / V = const. => p / n R = T/ V = const.\n=> T / V = const.\nT1 = 333 K\nT2 = 533 K\nV1 = 0.2 m3\n\nT1 / V1 = T2 / V2 =>\nV2 = V1 x T2 / V2 = 0.2 x 533 / 333 = 0.32 m^3\nanswer D is correct
856297
V=55.0L\nP[H2 = 2.00 atm\nP[O2 = 3.00 atm\nT[1 = 25°C = 298 K\nT[2 = 125°C = 398 K\nR = 0.082057 L · atm / (K · mol)\n\nIdeal gas law: PV = nRT\n\nAt 25°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 298) mol = 4.498 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 298) mol = 6.748 mol\n\nAt 125°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 398) mol = 3.368 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 398) mol = 5.052 mol\n\nThe Reaction that occurs is:\nH2 + 1/2 O2 = H2O\n\nSo, hydrogen is the limiting reactant, leaving water and oxygen as the resulting composition.\n\nAt 25°C, 4.498 mol of H2 and 2.249 mol of O2 form 4.498 mol of H2O, leaving 4.499 mol of O2.  The total number of moles (n) is 8.997 mol\n\nAt 125°C, 3.368 mol of H2 and 1.644 mol of O2 form 3.368 mol of H2O, leaving 3.408 mol of O2.  The total number of moles (n) is 6.776 mol\n\nUsing the ideal gas law again:\nPV=nRT\n\nP = nRT/V\n\nAt 25°C:\nP = 8.997 mol * 0.082057 L · atm / (K · mol) * 298 K / 55 L = 4.00 atm\n\nAt 125°C:\nP = 6.776 mol * 0.082057 L · atm / (K · mol) * 398 K / 55L = 3.00 atm
134712
This question is based on the gas law PV=nRT where \n\nP= pressure in atm (1atm=101.3kPa)\nV= volume in litres\nn= # of moles\nR= .08206\nT= temp. in Kelvin (273.15K=1*C)\n\nand the equation\n\nP1V1=P2V2
381756
You need to use Boyle's Law to calculate this.  Pressure is directly related to volume.  PV(inside) = PV(outside)  If the pressure of your tank is 2500PSI and holds 50L and atmospheric pressure is 14.7PSI we can calculate the volume once all the gas is released.  Since we can not get the last 14.7PSI out of the tank due to atmospheric pressure we have to subtract that from our tank pressure.  So we have...\n(2485.3PSI)(50L) = (14.7PSI)(X)  solving for X we would have a total volume of 169.07L.  Divide this by the flow rate and you can determine how long it should last.
169390
kilo pascals a unit of pressure which is 1000 N/m^2
202815
100g O2 x (1 mole O2)/(32g O2) x (22.4 L O2 @ STP)/(1 mole O2 @ STP) x (1/1.5) x (298/273)\n\n= 50.9 L\n\nThe 22.4 L = 1 mole is only true @ STP.  The higher pressure reduces that volume and the higher temperature increases it.
800141
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.