What's the Boyle's law,the equation&graphs?
362563
In the 1700's a number of people investigated gas behavior in the laboratory. Robert Boyle investigated the relationship between the volume of a dry ideal gas and its pressure. Since there are four variables that can be altered in a gas sample, in order to investigate how one variable will affect another, all other variables must be held constant or fixed. Boyle fixed the amount of gas and its temperature during his investigation. He found that when he manipulated the pressure that the volume responded in the opposite direction. For example, when Boyle increased the pressure on a gas sample the volume would decrease. Mathematically, PV = constant value if the gas is behaving as an Ideal Gas. A practical math expression of Boyle's findings is as follows: \n\nP1V1 = P2V2 \n\nwhere the variables with the 1 subscript mean initial values before the manipulation and the variables with the 2 subscript mean final values after the manipulation. \n\nA practical application illustrating Boyles Law would be the action of a syringe. When we draw fluids into a syringe, we increase the volume inside the syringe, this correspondingly decreases the pressure on the inside where the pressure on the outside of the syringe is greater and forces fluid into the syringe. If we reverse the actin and push the plunger in on the syringe we are decreasing the volume on the inside which will increase the pressure inside making the pressure greater than on the outside and fluids are forced out. \n\nA more life dependent example of Boyles Law is the ction of the diaphragm of our body. This is a muscle that is located just below the lungs. When we inhale the diaphragm moves downward allowing the lungs an increased volume. This decreases the pressure inside the lungs so that the pressure is less than the outer pressure. This results in forcing air into the lungs. When we exhale the diaphragm moves upward and decreases the volume of the lungs. This increases the pressure inside the lungs above the pressure on the outside of the lungs so that gases are forced out of the lungs. Of course, all of this is totally automatic and we take this important cycle which is performed hundreds of times a day for granted until we receive a sharp blow to that region that briefly paralyzes the diaphragm muscle. We say the wind was knocked out of us, but Boyles Law was not allowed to function.
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.
645900
Boyles Law PV/T I think. Anyway it has to do with the relationship of Pressure to Volume. If one increases the other decreases. Its useful in industry in making liquid Nitrogen and has endless applications in dealing with gasses in Chemisrty and Physics.
865453
Assuming temperature is static, I think Dalton's Law would apply:\nThe sum of all the partial pressures.\n\nP(gas 1)V(vol 1) = P(new gas 1)V(together)\n10 bars * 300L = ? bars * 600 L\nP(new gas 1) = 5 bars\n\nP(gas 2)V(vol 2) = P(new gas 2)V(together)\n3 bars * 400L = ? bars * 600 L\nP(new gas 2) = 2 bars\n\nP(together) = P(new gas 1) + P(new gas 2)\nP(together) = 7 bars
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.
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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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Gay-Lussac's Law states that for a fixed amount of gas (the number of moles stays constant throughout the problem) and a fixed (constant) volume, pressure and temperature are proportional, the formula is:\n\nP1/T1=P2/T2  \n\nThis means as temperature increases in a fixed volume, the gas exerts more pressure on the container, they expand and hit the container walls and push on them.  If the gas is cooled and kept in the same container, the particles exert less pressure on the container, they're more condensed.\nTemperature must be in Kelvin (K) not Celcius to use the formula.
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R is the gas constant (= 8.31 J/mol*K).  The value should be with the list of constants in your book.\n\nOk Nic, you seem to be a bit confused on this issue so lets go back and start from scratch.\n\nFirst you obtained you values for T and K (experimentally or otherwise).  MAKE SURE YOU CONVERT YOUR TEMP VALUES TO KELVIN.  It's important for the units of the constant (R) are in Kelvin.  So then you calculate 1/T and lnK and plot the values for each set to generate your graph.\n\nYou calculate slope by taking two of your points (x1,y1) and (x2,y2).  From the data you added, let's say we'll use your first (0.002976, -3.079) and last points (0.003165, -4.225).\n\nslope= [-4.225- (-3.079)]/[0.003165-0.002976]= -6063.49 kelvin\n\nSince slope = -delta H/R\nThen -delta H = slope * R\nSo -delta H = -6063.49kelvin * 8.31 J/molkelvin\nAnd -delta H= -50387.62 J/mol\nRemoving the negatives from both sides\nDelta H= 50387.62 J/mol\n\nFor the second part of the question, use one (x,y) experimental point to find the y-intercept.  They don't want you to determine it from your plot, they want you to calculate the exact value.\n\nThe equation of a line is y=mx+b where "m" is the slope and "b" is the intercept.  So take another experimental point from readings (ie 0.003012, -3.751) and calculate the intercept by using the slope value calculated above.\n\nIf y=mx+b, then b= y-mx\nPlugging in the values\nb= -3.751 - (-6063.49kelvin * 0.003012/kelvin)\nb= -3.751 - (-18.26)\nb= 14.51\n\nTo calculate delta S for that same point (for that's what is requested), first you have to calculate delta G for that point.  Use the  following equation;\n\ndelta G = -RTlnK\n\nYou know the value for R (8.31 J/molkelvin), and you have the temp value and lnK value for that point for it was expermientally determined\n\n1/T= 0.003012/kelvin so T=332.01 kelvin\nlnK= -3.751\n\nSo delta G= -8.31 J/molkelvin * 332.01 kelvin * -3.751\nand delta G = 10349.02 J/mol\n\nYou can then calculate delta S for\nintercept = delta S/ delta G (given in your original data)\n\nTherefore delta S = intercept * delta G\nSo delta S = 14.51 * 10349.02 J/mol = 150164.28 J/mol\n\nHope this clarifies things.
905377
According to Boyles Law, P1V1 = P2V2\nlet p = rou, the density of pure water (assuming tt the lake is pure)[1000 kg/m^3]\nTake g to be 10N/kg, and atmospheric pressure [P(atm)] to be 1.013x10^5 Pa \n\n[P(pgh) + P(atm)] x 1.55 = P(atm) x New Volume\n(1000x10x120 + 1.013x10^5) x 1.55 = 1.013x10^5 x new volume.\ntherefore, \nnew volume = 1210130 x 1.55 / 10130\n                     = 185.16cm^3\n\nHope this helps =)\n\n**The Boyles Law which i stated above states that the volume of gas in an enclosed area is inversely proportionate to its pressure. Initially under the water, the pressure on the bubble is taken to be atmospheric pressure + the weight of the column of water above it. As it rises, the pressure gradually decreases to only atmospheric pressure when it is at the surface. As such its volume increases proportionately, and tt's how i found out the ans.
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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
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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
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Boyle's law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.\n\nUsing P1 and V1 to stand for initial conditions and P2 and V2 to stand for new conditions results in the following equation.\n\nP1xV1=P2xV2\n\nSimple proportion...\n\nNow, the law relating to pressure and temperature would be Gay-Lussac's law which states the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.\n\nFor a given mass of gas at constant volume, the ratio P/T is the same for any set of pressure-temperature values.\n\nThe proportion for Guy-Lussac's law would be:\n\nP1/T1=P2/T2\n\nSo, to answer your question, for Guy-Lussac's law, not Boyle's law, a reduction of pressure WOULD cause a reduction in temperature.
978453
With Boyles law, doubling the pressure doubles the mass. No sure what the relative masses of air and helium are, but let`s say it`s 10 to 1, then 10x the pressure is what you need
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Use the ideal gas law: PV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nSolving for V: V=nRT/P\nYou have to convert pressure from atm to Pa. The conversion factor is 1 atm = 101325 Pa\n\nV=(2.44 mol)(8.314 J/mol K)(343 K)/243180 Pa\n\nV=0.0286 m^3
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Wow, this question is harder than I thought lol, I thought it would just be a plug and chug but its not u actually have to think!  I havent quite figured it out but I will give you a start:)  Maybe I'll come back to it.  \n\nSo the formula is: \nOsmotic Pressure in atm. = Molarity ( R ) (Kelvin Temp.)\n\nOsmotic Pressure = 3.05 atm\n\nR = 0.0821 liter-atm/mole kelvin \n\nConvert the temperature from Celsius to Kelvin. \nK = 10 + 273 = 283 K \n\nWe dont have the Molarity so first find that, then since we know Molarity = (g/ molecular weight)/liter and in the problem we have the relationship 4g/liter and 2g/liter.  Figure it out!:)\n\nGood luck!
818036
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
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'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3