how do you do this problem:"400 ml of a gas at 1atm will have a final pressure at 2000ml of ___atm?
595268
Use the formula: \n\nPV(1)=PV(2) where P is pressure and v is volume.\n\n400(1)=(2000)x(P)\n\nSolving, P =.2atm\n\nPS: If you ever forget the formula, use common sense. When you place any gas under pressure, what happens to its volume? It goes down, because the pressure is forcing the gas into a smaller volume. So when you increase your gas volume by 5 (400x5=2000) you must lower that pressure by the same amount. So the pressure goes down by 5 (1/5=.2).
190939
ehm...aren't you suppose to substact 200psi of residual pressure from that 1300psi 1st?  giving you 265.7 minutes left.
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
397497
I would use the ideal gas law...more specifically...the combined ideal gas law...\n\nP1*V1/T1 = P2*V2/T2
847416
With only 2 gases, the partial pressure of the O2 is 0.440 atm - 0.160 atm = 0.280 atm O2\n\nThe temperature is 40°C, or (40°C + 273°K) = 313°K\n\nThe volume is 478 mL or 0.478 L\n\nYou will use the ideal gas law to calculate the number of moles (n) of O2:\n\nPV = nRT\n\nUse the partial pressure for P.\n\n0.280 atm * 0.478 L = n * R * 313°K\n\nR = 8.2057 E-2 L * atm / (mol * °K)\n(this will probably be in your textbook's appendix, but you can also search the internet for "gas constant")\n\n0.280 atm * 0.478 L = n * 8.2057 * 10^-2 L * atm / (mol * °K) * 313°K\n\nn = 0.280 * 0.478 / (.082057 (1/mol) * 313)\n\nn = 0.005211 mol\n\nThe molecular weight of O2 is (16*2) = 32 g/mol\n\nSo, the mass is 32 g/mol * 0.005211 mol = 0.1667 g or 166.7 mg\n\nYou might want to work it out yourself to be sure you can do it.
825631
Change grams to moles: 0.254 g/122.5g/mole = 0.00207 moles\n\nChange to moles O2 (ratio from equation is 3/2)\n0.00207 x 3/2 = 0.00311 moles O2\nCalculate vol. from relationship 22.4L/mole for any gas at STP\n22.4L/mole x 00311mole = 0.0697 L at STP\n\nUse Combined Gas Law to change to vol. at 24.0 degrees and 729.1mm. Reduce pressure by the water vapor pressure.\n\nP1 = 760.mm V1 = 0.0697 T1 = 0 + 273\nP2 = 729.1-22.4 V2 = x T2 = 24 + 273\n\nx = (0.0697)(760)(297)/(273)(706.7)... = 0.0815 L or 81.5 mL
709574
I get 1.69522 g/L, using constants with the most significant figures I can find:\n\nN/V = P/(R*T)\nP = 101.325e3 Pa\nV = 1 m^3\nR = 8.314472\nT = 273.15 K\n\nN/m^3 = 44.6149 mol/m^3 = 0.0446149 mol/L\n1 mol of F2 = 37.9968 g\n0.0446149*37.9968 = 1.69522 g/L\n\nIf I use 100 KPa (1 bar) as standard pressure, I get 1.67305 g/L\n\nThe difference in the values given in the answer and the "Handbook of Chem. & Physics" is apparently due to the shift from 1 atm to 1 bar as the definition of "standard pressure."\nI think the slight difference in my calculated value and the published one may be due to the fact that when you start getting out to that many decimal places even "ideal gases" aren't totally ideal....maybe I should try the Redlich-Kwong EOS...\n\nOooooops!  I didn't want to do your homework for you...maybe I should erase all this....:)
792353
P1V1/T1=P2V2/T2\n\nV2=P1V1T2/(T1P2)\n\nV2=25 Liters* 1 atm* 298 Kelvin/( (19.71/760)atm*373 kelvin)\n\nV2= 770 Liters\n\nuse temperature in kelvin and convert mm Hg to atm  760 mm Hg in 1 atm
940427
Convert Liters of CH4 into moles of CH4 with PV=nRT. n= (2 L x 1 atm)/(.082058 x 273.15 K) = .08923 moles CH4. Then, using the mole ratio of 2 moles of O2 per 1 mole of CH4, you multiply .08923 by two to get .17846 moles O2. Plug that into the PV=nRT. V= (.17846 x .082058 x 273.15 K)/1 atm, which equals 4.
688699
No it's not right...\n\n1 atm = 101.325 kPa = 760 mm Hg = 760 Torr = 14.7 psi\n\nall those are equal to eachother
580275
This is using VP=VP, Boyle's Law of Gases.  By the way, mm Hg is the same thing as torr.  \n400 mL * 500 mm Hg = V * 760 torr\nV=263.2 mL, so your answer checks.  You may want to check your significant figures, though.
1097483
P/T=P/T\n1/(20+273)=P/(600+273)\nP=30 atm\n\nThe new pressure would be 30 atms, but this is considering that the volume remains constant.
602334
PV = nRT\n\nV = 3 L\nn = g CO2/MW CO2 + g O2/MW O2\nT = 273 K\nR = 0.082057 L-atm / mol-K\n\nsolve for P
837288
Use the formula M1V1=M2V2\n\nThat is the first Molarity times the first volume are equal to the second Molarity times the second Voume.\n\n.2098M x70mL = .662M x V2\n\nSolve for V2.\n\n(.2098 X 70)/.622 = V2\n\nV2 = 23.61mL NaOH to titrate the solution to neutrality.
558434
dont use PV=nRT... you dont need it for these problems... just requires more work because you have to figure out more than one variable\n\nfor everything here it seems you need different laws\n\nBoyles Law is P1V1=P2V2\nCharles Law is V1/T1=V2/T2\nGuy Lussac's Law is P1/T1=P2/T2\n\nall of those laws are just proportions.. just substitute in the numbers you need.  for temperature problems you need to convert them to kelvin... just add 273 to the celcius temperature and you have the temperature in kelvin.\n\nfor example, look at number one...just plug in all the numbers\n\nsince you're dealing with volume and pressure, you'd use Boyle's Law.\n\nP1V1=P2V2\n\n(3atm) * (15mL) = (2.5atm) * (x)\n\nsolve for x\n\n3 times 15 is 45...  2.5 times x is 2.5x\n\n45=2.5x\n\ndivide both sides by 2.5\n\n45/2.5 = 18\n\nthe volume will increase because there is less pressure.  the volume would increase by 3 mL.\n\nlook for what you are given, find out which law you need, and just plug in the numbers\n\nfor number 4.... since you're dealing with pressure and volume again you'd use Boyle's Law.  Given pressure is STP\n\n(15L) * (1atm) = (x) * (3atm)\n\nsolve for x...\n\n15 times 1 is 15... x times 3 is 3x\n\n15=3x\n\ndivide by 3 on both sides...\n\nx=5L of methane gas.\n\nfor number 5 you're dealing with volume and temperature.  with that we would have to use Charle's law.\n\ncharle's law = V1/T1=V2/T2\n\nnow plug in the numbers...\n\n(450L)/(30+273K)= (x)/(90+273K)\n\n(450L)/(303K) = (x)/(363K)\n\n1.485 = x/363\n\nmultiply 363 on both sides\n\n539.01 = x\n\nthe volume of ethane would be 539.01 L\n\ngood luck
94789
Use Boyle's Law for that. Assuming the temperature is always constant, the product of pressure and volume is always constant.\n\nP * V = k\nor\nP1 * V1 = P2 * V2\n\nIf temperature is included in the equation, you could either use the combined gas law or the ideal gas law.\n\nNote: Elevation plays a significant role in Boyle's Law so it will have a difference in the gauge readings if you relocate it from sea level to the mountaintop.
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