Ok this is a chemistry question about the ideal gas law?
956827
First, you have to understand what an ideal gas is. Read your book.\n\nSecond, derive your ideal gas equation to get V for Q1, and P for Q2. This is easy.\n\nTo convert Celsius to Kelvin, add 273 to the temperature you have in Celsius. So, if you have 17C, its value in Kelvin is 290K.\n\nFor the gas constant, \nR = 0.08206 L*atm / mol*K\n\nPlug your values to the ideal gas equation. Good luck!
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.
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.
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
92170
The ideal gas law is another tool that we can use to determine the molar mass of a gaseous substance. To determine a molar mass, we need the mass of a given sample and also the number of moles in that sample. You recognize immediately that given the volume, temperature, and pressure of a gaseous sabstance, you can calculate the number of moles.\n\nfirst:\n \n V = Volume = 987.56g / 1g/cm^3 = 987.56cm^3 (1 liter/ 1000cm^3) = .987 liter\n\n T = temperature = 23 + 273 =  296K\n\n P = pressure = 97.7 kPa (  1 atm / 101.3 kPa ) = .964 atm\n \n r = .0821 L'atm/mol.K\n \n n = mole = ?\n     \nThe ideal gas law equation:\n \n   PV  =  nRT\n\nsolve for  n\n\n   n = PV /RT = .964*.98756 / .0821*296 = .039 mol \n\n The molar is simply the mass divided by the number of moles:\n\n  mass of gas in flask =  187.942g - 183.257g = 4.685g\n\nso, the answer is:\n\n            4.685g / .039 mol = 120.1 g/mol
712158
questioner didnt say the gas becomes ideal when volume is increased. he said it MORE CLOSELY approximates.\n\nthis is due to a pressure decrease at constant temperature. (boyle's law). less pressure, atoms further apart, less intermolecular attractions.\n\ncomplete absence of intermolecular forces is a required term in ideal gas principle. so reducing the force obviously creates a more ideal gas.
766934
Assume it is an ideal gas and use PV = nRT\n\nP is in atm 1 atm = 760 torr\n    225 torr / 760 torr = 0.296 atm\nV is in L so 5.7 L\nR is 0.0821 atm*L/mol*K\nT is in K so C + 273 = K\n     27 + 273 = 300 K\n\n(0.296)(5.7) = (0.0821)(300)n\nn = 0.0685 mols
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
895405
use dalton's law to get the pressure of CH4(total pressure - H2 pressure)\nthen use ideal gas law to get the moles of CH4(PV=nRT)\n*watch your units*\nchange moles of CH4 to grams of CH4(moles x molar mass)
865243
C3H6 +  9/2 O2 ---> 3 CO2 + 3 H2O\nassumed oxygen is ideal gas, so\nPV=nRT\nR = 82.057 cm3.atm/gmol.K\nV = 15 l = 15000 cm3 ; T = 37 C = 310 K\nn (mol) O2 = PV/RT = 2.4*15000/82.057*310 = 1.415 gmol\nfrom balanced reaction CO2 = 3/(9/2) * 1.415 gmol = 0.943 gmol\n                                         = 0.943 * 44 = 41.507 gram
542119
As described by the ideal gass laws, P1 V1 = P2 V2. This means that, as the pressure increases, the voulume of the contained vapor decreases and condences.
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....:)
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
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
609831
pv=nrt\nmeasure pressure, volume, and temperature to find the number of moles, n\n\nfind the mass of the gas\n\nmass of gas/ moles of gas= molecular mass (g/mol)
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.
1112212
Boyle's Law states that the product of the volume and pressure of a fixed quantity of an ideal gas is constant, given constant temperature.  Mathematically, PV = K, where K is a constant.  So, if the volume or the pressure is changed, the other changes accrodingly.  For example, let's assume the volume of gas is V1 and the pressure is P1.  If we change the pressure to P2, the volume will change to a new volume (V2) such that P1V1 = P2V2