how to calculate the pressure of CO2 using van de waal's and ideal gas equation?
94455
Please take the time to\nvote for the best answer\n\nThank You!!\n\n\nWe derive the shape of the phase boundary for solid to gas transitions (sublimation), examples being dry ice (CO2) or ice at low pressure. We derive the van der Waals equation of state, which is an improvement on the ideal gas equation pV=nRT. The ideal gas equation is based on two assumptions: 1. Particles occupy zero volume, and 2. Particles do not interact. Allowing for particles to have a finite size, and also allowing for the fact that at close range, gas particles feel van der Waals attractions, we get the new improved van der Waals equation of state for a gas made of sticky but hard molecules. Van der Waals attractions work because at close range, atoms and molecules notice each other's dipole moments. The dipole moments are due to the fact that at any given instant, the electron cloud is not quite centered on the nucleus of the atom (although it will be centered on average). This instantaneous dipole moment causes atoms in the vicinity to arrange their instantaneous dipoles so as to lower their energy, which causes attraction. \nIt turns out that geckos can cling to walls and ceilings because of van der Waals attractions. Gecko feet have tiny hairs that split many times to make many very fine tips, giving the hairs a very large total surface area. The fine hairs are able to form many contacts with any surface, and the surface-to-hair contact is adhesive due to van der Waals forces. One gecko foot can support the weight of an entire human.\nVideo: Sticky gecko feet, and their van der Waals adhesive properties.
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water is incompresible, so ...\n\nVgas = Vtank - Vwater\n\nm(gas) = M(gas)*Niu(gas)\n\nNiu(gas) - no. of moles \nM(gas) - molecular mass of gas\n\nNiu(gas) = m(gas)/M(gas)\n\nwe have the following equation:\n\nPV = Niu*R*T, where:\n\nP - pressure (N/m^2)\nV - volume (m^3)\nNiu - no. of moles of gas\nR - universal gas constant (8.31 in these conditions)\nT - temperature (K)\n\nP = Niu(gas)*R*T/V(gas)\n\nP = [m(gas)/M(gas)]*R*T/[Vtank - [m(water)/(water density)]]\n\nin formula introduce Vtank in m^3 (not in liters 1 m^3 = 1000 dm^3 = 1000 liters), and the pressure will result in N/m^2\n\ntransform it in psia ... and the probleme is solved
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
795154
2CO(g) + O2(g) --> 2CO2(g)
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First you have to balence the equation.\n\n2LiOH + CO2 → Li2CO3 + H2O\n\n119.75 g LiOH x 1 mol LiOH     x    1 mol H2O  x    18.015 g H2O \n````````````````````` ----------------          --------------        ---------------\n``````````````````````23.948 g LiOH      2 mol LiOH        1 mol H2O\n= 45.04 g H2O (this is the theoretical yeild)\n\nto get % yeild you take auctual yeild (what you're looking for) and divide it up theoretical yeild (the above answer), then to turn it into a % you multiply it by 100. So % yeild= (auctual/theoretical)x100. \n\nsimple algebra gets auctual by itself. (%yeild x theoretical)/100\n\n91.72 x 45.04\n------------------ = 41.31 g H2O\n      100
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log10(P) = A − (B / (T + C))\n    P = vapor pressure (bar)\n    T = temperature (K)\n\nAntoine Equation Parameters\nTo get the References, go to the website below, search by compound name using Phase Change checkmarked, hyperlink to complete journal reference\n\nEthane: (two sets for two temperature ranges)\nTemperature (K) 91.33 - 144.13\nA  4.50706\nB  791.300\nC  -6.422\nReference: Carruth and Kobayashi, 1973\n\nTemperature (K) 135.74 - 199.91\nA  3.93835\nB  659.739\nC  -16.719\nReference: Loomis and Walters, 1926\n\n\nEthylene:\n\nTemperature (K) 149.37 - 188.57\nA  3.87261\nB  584.146\nC  -18.307\nReference: Michels and Wassenaar, 1950
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.
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See, first of all you need to convert your Celcius to Kelvin by the formula, 273+C=K. Therefore 20C becomes 293 K and -80C becomes 193. Now use this combined gas equation:\nP1V1/T1=P2V2/T2 where P1 is the initial pressure, V1 is the initial volume and T1 is the initial temperature. \nP1=335mm Hg\nV1=300ml\nT1=293K\n\nwhereas P2,V2 & T2 stand for final pressure, final volume (the one you have to find in this question), and final temperature.\nP2=760mm Hg, V2=x, T2=193K\n\nso substitute these values now,\n \n335*300/293 = 760*x/193\n\nx=335*300*193/760*293\nx=87.1ml
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first, u should find the no. of moles in 11 grams of CO2 gas\n\n1 mole = 44g\n\ntherefore, 11g = 11 divided by 44\nwhich is equal to 0.25 moles\n\n1 mole of co2 gas occupies 24000cm cubed.\n0.25 moles of co2 occupies x cm cubed\n\nx = o.25 times 24000\n\nwhich is equal to 6000 cm3\n\nthats the volume...\n\nby da way, r u kashyap from canada? my email id harshini317. contact me if u r kashyap sripada from canada
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giving some sample equations would help!\n\nyou have to make the left hand side balance wilh the right hand side...\n\nsuppose you have the unbalanced equation\n\nC4H10 + O2   =    CO2  +  H20\n\nin this equations the easiest thing to balance first would be the carbon so on the right hand side you put \n\nC4H10  + O2   =   4CO2  +  H2O\n\nThen the hydrogen,you have 10 Hydrogens on the left so you need 10 on the right\n\nC4H10  +  O2  =  4CO2  + 5H2O\n\nthus you will need 13 oxegen atoms on the left side.. since oxegen exists as O2 and 13 is an odd number you multiply everything by a factor of 2 to make it just right... so your final equation looks like....\n\n2C4H10  +  13O2  =  8CO2   +  10H2O
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)
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Use the gas law\nPV=nRT
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PV=nRT\nP=pressure\nV=volume\nn=moles\nR=constant(.08026atm*L/mol*K)\nT=temperature\n\nThis is the ideal gas law.
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There's a formula for volumes of gasses:\n\nPV = nRT.\n\nIf I remember right, P = pressure, V = volume, n = number of moles, T = temperature, and R = molar density, or something molar that is specific to the gas.
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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.
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1) look up standard temperature and pressure\n2) use PV = mRT with the given conditions to find the mass of helium\n3) use PV = mRT, with the mass you found above and the standard temperature and pressure to solve for V.\n\n*note:  Make sure you are using the correct gas constant, R, for helium.  Either look it up, or use R = R_univarsal/Molar weight.
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If you know the volume and density of the liquid you can calculate the number of moles n.\n\nOnce you have n and if you assume that the gas will behave like and ideal gas and you fix the temperature and pressure, the volume will be given by:\n\nV = nRT/P.