Which of the following four statments are true for real gases and the corrections that must be made?
918606
Real gases usually have equations of states associated with them.  One of the earliest and simplest is the Van Der Waal real gas equation.\n[P + a(n^2/V^2)]*[V - nb] = nRT\n\nP is the total pressure\nV is the total volume \nn is the number of moles of gas\na is a measure of the attraction between the particles  \nb is the volume enclosed within a mole of the particles b = NAb', \nR is the gas constant \n\nc seems to be the only correct statement.\nas V-nb decreases, P + a*n^2/V^2 must increase to keep nRT a constant.
950681
Van derWaal's equation is this:\n(P+a(n/V)^2)(V-nb)=nRT\n\nThe ideal gas law is:\nPV=nRT\n\na) b will lower the total volume, accounting for the space taken up by the molecules.  True.\n\nb) a will increase the volume based on intermolecular repulsive forces.  True.\n\nc) The attraction will decrease the pressure.  False.\n\nd) The individual particles will stay a constant size; the volume and pressure don't change bond lengths or shell sizes, just the spacing between the particles.  True.
907905
At constant pressure it wont. Noble gases are unreactive, thus cannot chemically disturb the equilibrium. \n\nAt constant volume, however the noble gas will. Remember that you are introducing more particles in the system, thus at constant volume the pressure will increase. This will shift the equilibrium of the reacting substances to the side with less gas particles to reduce the pressure. This is different, as the gas is PHYSICALLY disturbing the equilibrium.\n\nNeed more explanation?
351293
it is based on arrhenius equation\nK=A e ^ (-Ea/RT)\nA- frequency factor\nEa- activation energy (proportional to Delta H)\nR -Gas constant\nT -Temperature\nSince you dont know A- fequncy fator, yu need to do a lengthy calculation\n\nThe modified arrhenium equation will be ( by considering two situations T1, T2 there will be two rates k1, k2 but A-frequency factor is smae for one reaction so it's a constant..)\n\nEa= (R X T1 X T2 X Ln (k1/k2) ) / (T1-T2)\n\nEa= delta H = 17,390 J\nR = 8.314 J/K\nK1 = 1.6 X 10E-4\nK2= ?\nT1 = 298\nT2= 273+97= 370K\n  After substituing all these values, you will get K2= 6.27 E-4\nwhich is in agreement with theory (At high Temperatures, rate should be high..!)\n\nIs this info will help you?\nGood luck!
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!
724209
Delta Hf = Delta Hf of the products - Delta Hf of the reactants\n\nThe Delta H's of pure species in their natural form are zero. That means that O2 and H2 will have delta Hf's of zero.\n\nThe other values you have to look up.\n\nOverall it will be...\n\nDelta Hf = 2*Delta H of CO - 2*Delta H of CH4\n\nThe twos are the stoichometric coefficients.
816266
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
168650
Ideal gas law: PV = nRT.\n\nno change in pressure (P), no change in ideal gas constant (R), no change in number of moles of gas (n). Temperature (T) was multiplied by 1.6, so volume (V) must also. The answer is d.
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
917314
A real gas is a form of matter that will conform to the sides of its container.  As it is compressed its temperature increases.  It is the only form of matter than can be compressed at all.  Gas is the least organized form of matter.
256088
Start with the ideal gas law:\nPV = nRT, where P is pressure, V is volume, n is the number of moles, R is a constant, and T is temperature.\n\nSay we have a fixed number of moles (n) and a fixed temperature (T). Then the entire right hand side of the ideal gas law is a constant (some number that we could compute if we wanted to). Under these circumstances, the pressure times the volume must therefore be constant, and therefore, all else being equal, pressure and volume are inversely proportional to each other (if you increase pressure, volume must decrease, and vice versa). This is Boyle's Law.\n\nIf you wanted to derive Charles' Law, you would do the same thing but this time hold pressure constant instead of temperature. So P, n, and R are now fixed. Doing some algebraic manipulation you get V/T = nR/P. Since all the terms on the right hand side are constant, the ratio of the volume of a gas to the temperature of a gas must be constant if we fix the pressure and thus volume is directly proportional to temperature (ie increasing temperature must increase the volume). This is Charles' Law.\n\nGay-Lussac's Law states that if we hold n and V constant, pressure will be directly proportional to temperature. You can prove this using similar logic. Dalton's Law holds T and V constant and reaches a conclusion about P and n.\n\nSo in short, you've got P, V, n, and T. Hold two of those values constant, solve for the other two, and you've got yourself a gas law.
895488
3. is correct.\n1. is not correct since formaldehyde has 3 sigma and one pi bond\n2. is not correct since H cannot form a pi bond\n4. is not correct since only one pi bond forms
948757
The concentrations of all the components: the reactants, the products, the catalyst.\nYou will also need the temperature.\nIf it's a gas-phase reaction, the pressure..\nIf it's a solid-liquid reaction, the surface area; or if the catalyst is a solid, the surface area of the catalyst.
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
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.
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.
260662
The ideal gas law shows the relationship between pressure and atoms / molecules.  PV=nRT.  P = pressure, V = volume, n = moles of the gas, R = gas constant, T = temperature.  Pressure goes up linearly as the number of atoms / molecules of a gas increase.  See the links for more info.