A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin.?
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
905957
20 celsius is 293.15 kelvin
358671
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.
602468
Absolute Zero or around 20 K.\n\nMelting point 14.01 K  (−259.14 °C, −434.45 °F) \nBoiling point 20.28 K  (−252.87 °C, −423.17 °F)
684353
P1V1T2=P2V2T1 \n\nRemember to express  temperature in Kelvin (with pressure in Pascals and volume in m^3)
980842
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
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.
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.
1076679
According to Boyle's Gas Law, the product of the pressure and volume of an ideal gas is a constant.\n\nTherefore, P1V1 = P2V2\nHere, P1 = 5 psi, V1 = 20 cubic inches, P2 = 10 psi.\n\nPutting into the equation,\n5*20 = 10*?\n\nAnswer is 10 cubic inches - B in the list of choices.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
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)
65568
18.8 degrees F., (-7.3 degrees C)
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.
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.
310335
You can never get to -273 C or 0 Kelvin.  But if you could, the pressure would be zero.  It would be difficult to measure since the pressure gauge itself is likely to emit heat (specifically thinking of an ion gauge).  If there was any pressure, then you would not be at 0 K since you would not be at the lowest energy state.  The only exception I can think of is if you only had one molecule in the container then the concept of temperature and pressure becomes ambigious.
849763
Water vapor.\n2 H2 + O2 = 2 H2O
552193
Can be solved by simply plugging values into the Ideal Gas equation without using two states.  Pressure must be converted to atmospheres, however, and temperature to Kelvin.  P = 1.243 atm, T = 306 K\n\nn = PV/RT = 1.243 * 2.74 / (0.08206 * 306) = 0.136 mol\n\nmass = moles * MM = 0.136 * 28.01 = 3.81 g