ideal gas law and hot air ballons....?
535624
PV=nRT\n\nas the air is heated the average kinetic energy of the molecules increases. ie P increases.But V is constant. so some gas molecules escape through the bottom. Hence the density decreases and becomes lesss than that of surrounding air.Thus the balloon rises.
167746
At STP, 1 mole of any ideal gas will have a volume of 22.4 Liters.\nUsing this logic, we can deduce that at STP, and equal volumes of (what we will assume are ideal gasses) gasses will each contain the same number of moles.\nIf each flask contains the same number of moles, then they much each contain the same number of molecules.\n\nBy saying that the flasks are at STP, we are given all the information we need to solve this.  STP (Standard temperature and pressure) implies a temperature of 273 K and a pressure of 1 atm.
165328
According to the Kinetic Theory of Gases, the velocity of a gas molecule will be directly proportional to the square root of the temperature. So, taking the ratios at the temperatures you mentioned, v(@373K)/v(@273K) = root(373/273) = 1.169
800141
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
1167777
D.  Based on the ideal gas law, PV=nRT\n\nSolve for R to get: R = (P*V)/(n*T)\n\nfor two different conditions, then, you would have:\n\n(P1*V1)/(n1*T1) = R = (P2*V2)/(n2*T2)  or,\n\n(P1*V1)/(n1*T1) = (P2*V2)/(n2*T2)\n\nor P1*V1*n2*T2 = P2*V2*n1*T1\n\nNote how V1 and T1 can never be in the numerator together?  \n\nIn order for the ratio to work, one must be the numerator, the other the denominator, because volume is directly proportional to temperature -- the higher the temperature (keeping number of moles and pressure equal), the larger the volume.
170709
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
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.
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
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)
129942
the ideal gas law is:\n \n PV = nRT \n\nSO,\n\n P= .95 atm\n n=  .87\n T=  280K\n R=  .0821\n V=  ?\n\n solve for V\n\n     V = nRT/P =(.87* .0821* 280) /.95 = 21 liters
635000
The easiest way to work this out is from the Ideal Gas Law, PV=nRT. Assuming that they start at the same temperature, this is even easier because both R and T will be constants.\n\nLets make the first gas 1, the second gas 2 and the final mix f so that :\n\nPressure Gas 1 =P1, Volume Gas 1 = V1, Temperature Gas 1 = T1\nPressure Gas 2 =P2, Volume Gas 2 = V2, Temperature Gas 2 = T2\nPressure final mix =Pf, Volume final mix = Vf, Temperature final mix = Tf\n\nthen :  P1V1/RT1+P2V2/RT2=PfVf/RTf\n\ntake out the constant R\n\nP1V1/T1+P2V2/T2=PfVf/Tf\n\nIf T is different for each mix\n\nPf=(Tf*(P1V1/T1+P2V2/T2))/Vf\n\nif T is constant (which I assume from your question)\n\nP1V1+P2V2=PfVf\n\nPf=(P1V1+P2V2)/Vf\n\nOkay?
606777
the actual equation is:  (p + b·n^2/V^2)(V - a·n) = nRT\n\na:- every mol of particles of the gas has a tiny finite volume of a\n\nb:-modification is due to the particles sticking to each other, a phenomenon called cohesion\n\nR- rydberg's constant
355529
V(gas)=0,2 m3,T1=333 K,T2=533 K,V2=?\nidel gas law: pV=nRT,p=const. so nRT1/V1=nRT2/V2\nT1/V1=T2/V2, V2=T2V1/T1\nV2=533 K x 0,2 m3 /333 K =0.32 m3
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.
827958
The first part is correct.  If the numbers in the second part are correct, it is true only for an ideal gas.  Other materials have various densities in their liquid and solid states.
940323
Reynold's number if the ratio of intertial force to viscous forces:\n\nRe = rho * v * L / mu\n\nrho: fluid density\nv: fluid velocity\nL: characteristic length\nmu: fluid viscosity\n\nNow you are given a scenario where the model is 1/30 the size of its prototye.  That means L2=L1 / 30.  Velocity is the same and the fluid air is the same, which means mu is the same.  We know that density will be different because we have to change pressure (rho2 not equal to rho1).  Here's what the problem looks like:\n\nRe = rho1 * v * L1  / mu = rho2 *v*L2 / mu\n\nWe know that L2 = L1/30 so let's plug that in:\n\nRe = rho1 *v *L1/ mu = rho2 * v * (L1/30)/mu\n\nRearranging:\n\nrho2/rho1 = 30\n\nNow use the perfect gas law.  Let's manipulate it to get a form that is right for this problem.\n\nPV = NRT = m (R/MW) T\nP = rho * (R/MW)*T\nrho=(P/T)*(MW/R)\n\nWrite the last eqn above as ratio rho2/rho1, with constant T and MW (molec. weight)\n\nrho2/rho1 = P2/P1\n\nTherefore, P2/P1 =30.  P2 is 30 times P1.
415248
The ideal gas law (not sure if it has another name) establishes the relationship between volume and temperature and other things.  Its form is PV = nRT.  There is also Charle's Law and a few others.  Charle's law shows the direct relationship between temp and volume in V/T = k where k is a constant.  It shows that when T increases V must also increase in order for k to remain constant.