why the application of ideal gas law is essential and what is a difference between ideal and real gas.?
112298
Ideal gas laws assume that all particles are spherically symmtric and engage only in elastic collisions.  They ignore all other types of molecular energy (rotation in other planes of symmetry or asymmetry, bond stretching/compressing/bending, rotation about a molecular bond--all these things can and do absord energy from a collision), and they negate the effects of all intermolecular forces, like dipole interations, hydrogen bonding, London dispersion forces, etc.\n\nIn gases like helium, the atoms are basically spherically symmetric, and most of these random forces are neglible.  Hence, ideal gas laws work really well for these--only at really ridiculously low temperatures and high pressures do these forces manage to liquify helium.  The larger you molecule is--and I'm not talking about mass so much as the number of chemical bonds--the less ideally it behaves.  The greater the heterogeneity of a gass, the less ideally it behaves.  The greater the polarity (eg seperation of charges) displayed by the molecule, the less ideally it behaves.
160259
if it is an ideal gas is definitely will\npV =nRT\np increase then T increase if V is stationary
382823
Boyle's law: P1V1 = P2V2. As gases get compressed their volume shrinks. As gases decompress their volume expands. Taking a 1 cubic foot balloon down in the ocean would cause it to shrink to 1/2 its size at 33'. Breathing at depth, ascending, and holding your breath may cause serious lung over-expansion resulting in arterial gas embolism, etc. \n\n\nHenry's law: A gas over a liquid will flow into the liquid until the pressure is equal. If you covered a lake with neon, the atoms of that gas would pass though the surface of the water until the neon in the water had the same pressure as what is left in the air. \n\n\nCharles's law - When you heat a gas in a container that is closed the pressure of the gas will rise. Likewise, if you cool the gas the pressure will drop. Likewise, if you decrease the pressure on a gas it will cool. If you take your scuba tank to have it filled and they fill it fast, the temperature will rise. As time goes by the air and tank will cool down. When that happens the pressure in the tank will drop. So, pressure, temperature, and  changing gas volume are all related. \n\n\nDalton's law: When you have a mixture of gases, each gas contributes to the total pressure. If you have a container of air that is at 2 atmospheres, 21% of the 2 atmospheres (0.42 atm.) is due to the oxygen, and 78% (1.56 atm) is due to the nitrogen. Both gases exert their own pressure and the two are combined to find the total pressure. \n\nThe gas laws are a set of laws that describe the relationship between thermodynamic temperature (T), pressure (P) and volume (V) of gases. Three of these laws, Boyle's law, Charles's law, and Gay-Lussac's law, may be combined to form the combined gas law\n\nP1V1/T1 = P2V2/2\n \nwhich with the addition of Avogadro's law later gave way to the ideal gas law. Other important gas laws include Dalton's law of partial pressures. The kinetic theory of gases, Graham's law of effusion and root mean square velocity all explain how individual molecules in a gas act and their relation to pressure, volume, and temperature.\n\nA gas that obeys these gas laws exactly is known as an ideal gas (or perfect gas). An ideal gas does not exist; however, some gases follow the laws more closely than others given standard conditions.\n\nThe most important gas law is the ideal gas law, which states that:\n\nPV = nRT \n \nwhere\n\nP is the pressure in Pascal \nV is the volume in Cubic meters \nn is the number of moles of gas \nR is the ideal gas constant (8.31 J/mol K) \nT is the temperature in kelvins. \nOther gas laws, such as van der Waals equation, seek to correct the ideal gas laws to reflect the behaviour of actual gases. The van der Waals equation alters the ideal gas law to reflect how actual gases function using a series of calculated values called van der Waals constants.\n\nIt is also possible to apply Boltzmann's analysis to determine further information about gases.
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.
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.
788371
Air has pressure, much like any gas.  By using this pressure it fills in cracks and crevices.
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
224663
According to Charle's Law, the volume and temperature of a gas are related as follows,\n\nV1 / T1 = V2 / T2\n\nSo if you half the initial temperature, the volume of the the gas would need to be halfed aswell in order for the above eqaution to stay true.\n\nWhen the temperature of gas is decreased (all other things remaining constant), its volume decreases aswell.  The same for when a gas's temperature increases, the gas expands.
1153522
The quantities you have listed are in fact related via the ideal gas law.  So, assuming you aren't in huge temperature or pressure extremes, you can use the ideal gas law with decent results.\n\nIdeal Gas Law\nPV = nRT\n\nwhere P = pressure (N/m^2)\nV = volume (m^3)\nn = number of moles of gas\nR = universal gas constant = 8.3145 J/(mol*K)\nT = temp (in Kelvins)\n\nSo, by looking at the equation you can not convert air pressure directly to volume without also knowing the temperature and number of moles of gas.
14337
i never really thought about it and so i was just going to say that temperature doesnt matter because its a vacuum and is dealing with the pressure... but when i thought about it... come to think of it, i do have an answer for you.  the temperature would be zero kelvin, absolute zero.\n\n in an absolute vacuum, the pressure would be 0 if there were no particles, because the more particles you cram into one place, the more pressure there is because p=F/A pressure equals the force of the particles divided by the area of the wall of the container, so with no force of particles(since there are no partiles), the pressure would be zero.  in an absolute vacuum, there are no particles.  Now, if the pressure is zero, then according to the formula PV=nRT (which is just the ideal gas law formula,) if P(pressure) equals 0, i think that means that T(temperature in Kelvin)would also be 0 which would mean that the temperature would be at absolute zero(which is defined as the temperature where all matter stops moving.. not that this would make a difference because there are no particles there because it's an absolute vacuum.  \n\nThis is what i suspect and to confirm my beliefs,  it is consistent with the fact that someone else said that outer space was 3K, and although it's very close to an absolute vacuum, there are a few particles and atoms of gas floating around there, and so it is not quite an absolute vacuum, and the temperature is 3K, meaning it's not quite at absolute zero.  so it stands to reason that since temperature decreases when pressure decreases, that in an absolute vacuum the temperature would be absolute zero, 0 kelvin.\n\ni hope this is a "good answer" and that i made sense while i took you through my reasoning :)
703096
Charles' law states that, under constant pressure, the ratio between a gas' volume and absolute temperature remains constant.  So if it heats up it expands and if it cools it contracts.  For example, you fill a balloon with helium at room temperature, and then put it in a refrigerator.  The helium will contract, and the balloon will shrink.
265643
the law state that P(pressure)=K*T(temperature), where K is a constant ,when other things are constant.Any closed vessel which is heated is a real life example of gays rule.But it is important to note that it is imposible to find a completely real example for this rule because it is valid only for ideal gases and for real gases there are some factors which leads to deviations from ideal behaviour
716268
PV = nRT\nn = PV/RT\n\nP = standard pressure(don't know which one you're picking)\nV = 2 L\nT = standard temp(again don't know which one you're picking)\nR = constant you can decide which one you want to use at\nhttp://en.wikipedia.org/wiki/Ideal_gas_constant
434788
well, the ideal gas law is PV = mRT.  You need to write this expression for the butane that escaped the lighter, because we don't have enough information about the butane still in the lighter to solve the problem (we don't know the pressure).\n\nFirst you need to convert all your units into the units found in your gas constant, R. looking at the information given, you can see that you'll need the gas constant with units of PV/(mT).\n\nR for an individual gas is the universal gas constant, R_u, divided by the molar mass, which is what we're trying to find. \n\nR = R_u/M\nR_u = 8.314 kPa*m^3/(kmol*K)   [I chose this form because it most closely matched the form I know I need]\n\nThis means that really, the ideal gas law is:\n\nPV = m(R_u)T/M\n\nConversions:\nvolume:  1m^3* = 1000 L\n           477ml = 0.477L = 0.000477 m^3\n-note: you need to know the volume of butane that escaped the lighter.  To do this, find the density of butane (mass/volume) of the origional amount of butatane, and then divide the mass that escaped by the density. You're assuming that the density is constant, because we don't know the volume of butatane that escaped the lighter, or the pressure inside the lighter.  If you have to list sources of error, this would be one.\n\ndensity  = (0.01311 kg/0.000477m^3) = 27.28 kg/m^3\n\nV = 0.00072 kg/27.28 kg/m^3 = 2.6 x 10^-5 m^3\n              (see how the kg units cancel, leaving m^3)\n             \nTemperature:\n23 degrees Celsius  = 23 kelvin\n\nPressure:\n100.33 kPa  - this would be about atmospheric pressure\n\nMass:\n1kg = 1000g\n\n13.11g = 0.01311 kg\n12.39g = 0.01239 kg \n0.72g = 0.00072 kg\n\n\nPutting the numbers in pv  = mrT, with units :\n\n(100.33 kPa)(2.6x10^-5 m^3)=0.00072 kg*(8.314 kPa*m^3/(kmol*K))*(23 K)/M \n\nI would write all that down on a piece of paper, so you can see it more clearly.  The key to getting this kind of problem right is making sure that all your units cancel, leaving you with only the units you want- in this case, kg/kmol.\n\nSolving for M, and canceling all the units, I got:\n\n0.0189 kmol/kg = 1/M\n\nM = 52.9 kg/kmol\n\n\nAccording to my tables, the actual value for Butane is 58.124 kg/kmol, so this is pretty close!\n\n\nHope this helped.  Feel free to email me if you need more help.
891922
Basically solids and liquids are not compressible because the molecules are so close together. A gas can be compressed because the molecules are far apart. Since there is still room betwen the molecules in a gas until it forms a liquid, there is space for absorbing impact. The molecules can get closer upon impact.
570434
when all parameters r given use ideal gas formula i.e;pv=nrt when temp is kept constant use boyles law ie pv=constant or while solving problems use the form p1.v1=p2.v2 and when pressure is kept constant use charles law and when volume is kept const use pressure temp law morover when u hav rms velocity(=c) given use kinetic gas equation ie pv=mn(c.c)/3 [m=mass of gas molecule,n=total no of gas molecule]
761748
I'll answer a few of these to get you started.\n\n1.) Heat is only one type of energy. If, for example, you just heat something up or cool it down, then there is just a change in heat. But if there is a reaction, then there is a change of energy without a change of heat.\n\n3.) Enthalpy is defined as the thermodynamic potential of a system, which means that it's a function of heat, pressure, and volume, not just heat.\n\n4.) Standard conditions: Usually Standard Temperature and Pressure, meaning 0C and 100 kPa.\n\n7.) Q = energy added by heat. W = work. Thus, the internal energy of a system is dependant on both the heat and the work done on a system. For example, you could heat a gas, or you could simply compress it (lowering volume, increasing pressure) - both of which would increase the internal energy of the system.\n\n8.) Related to 7: Changing the pressure and volume of a gas will do work upon it.\n\n9.) Work is a form of energy - it's the actual mechanical work that can be achieved when energy is expended. In an ideal system, energy is not lost, but in a real system, there are inefficiencies where energy is not completely converted to work, so while you may extract, say 100 J of energy, it might only do 90 J of work.