100 ml of a gas is measured at 150 degrees K. if the pressure is kept constant, what volume will the gas?
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
700450
Letters A, B, D.\nIn not sure about C, its wording is tricky.\nIdeal gas law : PV = nRT\nReal gas law : (P + (a(n / V)^2)) * (v - nb) = nRT\nThe corrected pressure times the corrected volume.
1063841
W = PV = RTn, where P = pressure, V = volume, R = gas constant = 8.31 J/mol.K, T = temperature, n = moles of gas involved.\n\n1) Change in P = 4 x 10^5 Pa, then:\n\nChange in W = 4 x 10^5 Pa x 0.250 m^3 = 10^5 Pa.m^3 = 1 x 10^5 Joules.\n\n2) Change in V = 17.00 x 10^(-2), then:\n\nChange in W = 6 x 10^5 Pa x 17.00 x 10^(-2) m^3 = 102.0 x 10^3 Pa.m^3 = 1 x 10^5 Joules (note: there is only one significant number in result)
963774
Use the fact that PV/T = a constant, we find the constant by pluging in the values we have, T = 20C, but we must convert that to K first so 20C = (273 + 20) = 293K.  And 1 atm should be converted to physicall units of Nt/m^2 for an easier solution.  1 Atm = 101325 Nt/m^2.    So our constant is (101325)(4.5)/293 = 1.336 x 10^7 so now you have one equation as\n\nPV/T = 1.336 x 10^7 setting T = 260C or 533K gives\n1)  PV = 7.121 x 10^10\n\nThe next equation that we need is the force on the cylinder end given the pressure, which would be\n\n2)  .01P = F\n\nThe spring will stretch a distance 3)  d = F / (2.00 x 10^3)  \n\nThe volume changes by  4)  V = 4.5 + d * .010   \n\nThis gives you 4 equations which you can now solve to find the solution for V, d, P and F
224663
According to Charle's Law, the volume and temperature of a gas are related as follows,\n\nV1 / T1 = V2 / T2\n\nSo if you half the initial temperature, the volume of the the gas would need to be halfed aswell in order for the above eqaution to stay true.\n\nWhen the temperature of gas is decreased (all other things remaining constant), its volume decreases aswell.  The same for when a gas's temperature increases, the gas expands.
882010
The volume decrease due to temperature is:\n\n(22+273) / (37+273) = 295 / 310\n\nand the volume increase due to the change in pressure is:\n\n760 mmHg / 675 mmHg\n\n780 mL * (295/310) * (760/675) = 835.7 mL
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.
476691
As long as you are not changing the temperature or the number of moles of gas in the system, the following equation applies:\n\nP1V1 = P2V2\n\nthus (400 mL)*(500 torr) = (X mL)*(760 torr)\n\nSolve for X.
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.
1153522
The quantities you have listed are in fact related via the ideal gas law.  So, assuming you aren't in huge temperature or pressure extremes, you can use the ideal gas law with decent results.\n\nIdeal Gas Law\nPV = nRT\n\nwhere P = pressure (N/m^2)\nV = volume (m^3)\nn = number of moles of gas\nR = universal gas constant = 8.3145 J/(mol*K)\nT = temp (in Kelvins)\n\nSo, by looking at the equation you can not convert air pressure directly to volume without also knowing the temperature and number of moles of gas.
317634
Pressure will increase if temperature increases, assuming that the container remains at a constant volume. Because the molecules are moving faster, they collide more often with the walls of the container, so the pressure goes up.\n\nThis is demonstrated by Gay-Lussac's Law, which states that that for a fixed amount of gas (fixed number of moles) at a fixed volume, the pressure is proportional to the temperature, so:  P/T = constant.\n\nIn case you need to know this for a test, and easy mnemonic  is "Pete and Tom are Gay" (Pressure and Temperature are proportional according to Gay-Lussac).
821966
Not sure where the 600 came from, but it sounds like the question wants the total pressure of all the gases combined.  So you would just add together all the partial pressures, and the total pressure is 330 mm Hg.\n\nDividing the total pressure by the number of gases would give you the average partial pressure, but I don't think that's very useful.
167746
At STP, 1 mole of any ideal gas will have a volume of 22.4 Liters.\nUsing this logic, we can deduce that at STP, and equal volumes of (what we will assume are ideal gasses) gasses will each contain the same number of moles.\nIf each flask contains the same number of moles, then they much each contain the same number of molecules.\n\nBy saying that the flasks are at STP, we are given all the information we need to solve this.  STP (Standard temperature and pressure) implies a temperature of 273 K and a pressure of 1 atm.
690917
1) By definition, one mole of any gas at STP occupies 22.4 liters.  So, if you have 100 liters of CO2 at STP, you have 100/22.4 = 4.464 moles of CO2\n\n2) 28 grams of N2 = one mole of N2\nuse the ideal gas law: PV = nRT, so V = nRT/P\nn = 1 mole, \nT = 180C + 273.16 (to convert to Kelvins) = 453.16\nP = 620 kPa = 620 kPa * (1 atm/101.325 kPa) = 6.11 atm\n\n3) P1*V1/T1 = P2*V2/T2\nor V2 = P1*V1*T2/(P2*T1)\nP1 = 1 atm, P2 = 3.8 atm; T1 = 293K, T2 = 346K, V1 = 80L\nsubstitute those values into the equation to get V2\n\n4) P1*V1 = P2*V2, or (P1*V1)/P2 = V2; T1 = T2\nP1 = 1 atm, V1 = 66L, P2 = 6atm, and substitute into the equation\n\n5) 19 moles of any gas at STP occupies 19moles*22.4L/mole = 425.6 liters
1151439
Ideal Gas Law: PV = nRT\n\nYou are holding n and T constant, and R is a constant, so:\n\nP1V1 = P2V2\n\n(750 torr)(500ml) = (x torr)(1500 ml)\n\nx torr = (750 torr)*(500/1500) = 750 torr *1/3 = 250 torr
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
716138
mole of Argon = 100 / 40 = 2.5 mole\n\nBy using ideal gas theorem:\nP * V = n * R * T\n80 / 76 * 40 = 2.5 * 0.082 * T\nT = 205.39 K\nT = -67.75 C