a quantity of a gas has an absolute pressure of 400 kPa & an abso lute temp.of 110' Kelvin.?
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
414884
Are you sure the unit is micropascal? I think there is an error in you units, its too small.\n3141.59 micropascal = 0.000000031 atmosphere\n\nPressure in ideal monoatomic gases is directly proportional to temperature, so: \nP1/T1=P2/T2\nT(ºK)=42,000*(177.94+273)/.000000031\n\nCheck the units and resolve it the same way.
759551
V1/T1 = V2/T2\n\nFirst convert temperature to Kelvin. 198+ 273 = 471\n\nNext plug into equation\n\n4.81 L / 471 K = 2.49 L / T2\nT2 = 243.8\nthen put it back into C 243.8 - 273 = -29.2 C
216745
One unit kelvin, or 5/9 degree farenheit
797227
You have Pressure (being asked) You have the volume (5.43L). You have the number of moles (1.82)You have temperature (69.5C or 342.5K). I assume you know R.\nSo yes it is PV=nRT
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
412124
Do you know the Ideal Gas Law:\n\nPV = nRT?\n\nP = pressure in kPa\nV = volume the gas occupies\nn = the number of MOLES of gass\nR = 8.381\nT = Temperature in Kelvins\n\nYou might need to check for the units for volume and pressure, but that's pretty much it.  Solve for n, and use the density given to find the mass (assuming you know the MW for methane [16.04 g/mol])\n\nI'm not gonna do it for ya :-p, but it should be self explanatory
653606
44000*0.00012*56 = 26.4\n\n=44026.5Gal\n\nBe aware that actually the thermal coefficient by Volume (B) stricly speaking varies with temperature
840037
PV = NRT\n\nN - Mols of argon\n\nR = .0821\n T = 26+273.15\n\nV  = 75/1000 to convert to L\n\nThen you will find your P, pressure, in atmospheres.\n\nconvert that to torr
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
105211
Pressure * volume = number of mol * temperature (in ^oK) * 8.314\n\nYou have all the information you need right there. \nFirst solve for n for the initial state. \nSince you haven't added any more material you can then use the same value of n for the second state and solve for P.\n\nThis is sixth grade algebra. Anyone being asked to solve gas equations had better be able to do it standing on their head or they are in real trouble.
1149287
pV = n R T => if pressure is constant => p = n R T / V = const. => p / n R = T/ V = const.\n=> T / V = const.\nT1 = 333 K\nT2 = 533 K\nV1 = 0.2 m3\n\nT1 / V1 = T2 / V2 =>\nV2 = V1 x T2 / V2 = 0.2 x 533 / 333 = 0.32 m^3\nanswer D is correct
765956
A) 1\nB)2 moles of gas. ,1 mole of gas takes up 22.4 liters, so 44.8 L
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
978594
Acetone       2.13kJ/kg K or  0.51  btu/lb F\nWater, fresh 4.19kJ/kg K or  1.0   btu/lb F\n\nsee also http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html
598817
Absolute zero is the zero point of the ideal gas temperature scale, denoted by 0 degrees on the Kelvin and Rankine temperature scales, which is equivalent to −273.15°C and −459.67°F.\n\nFor most gases there is a linear relationship between temperature and pressure i.e., gases contract indefinitely as the temperature is decreased. Theoretically, at absolute zero the volume of an ideal gas would be zero and all molecular motion would cease. In actuality, all gases condense to solids or liquids well above this point. Although absolute zero cannot be reached, temperatures within a few billionths of a degree above absolute zero have been achieved in the laboratory. At such low temperatures, gases assume nontraditional states, the Bose-Einstein and fermionic condensates.