R is called the gas constant. It was first discovered, as part of the discovery in the mid-1830's by Emil Clapeyron of what is now called the Ideal Gas Law. Sometimes it is called the universal constant because it shows up in many non-gas-related situations. However, it is mostly called the gas constant or, sometimes, the universal gas constant. Depending on the units selected, the "value" for R can take on many different forms. Here is a list. Keep in mind these different "values" represent the same thing.
V = nRT / P
V = [(2.34 g / 44.0 g mol¯ 1 ) (0.08206 L atm mol¯ 1 K¯ 1 ) (273.0 K)] / 1.00 atm V = 1.19 L (to three significant figures)
n = PV / RT
n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯ 1 K¯ 1 ) (273.0 K)] n = 2.50866 mol (I'll keep a few guard digits)
2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)
T = PV / nR
T = [(1.95 atm) (12.30 L)] / [(0.654 mol) (0.08206 L atm mol¯ 1 K¯ 1 )] T = 447 K
Since one mole of gas occupies 22.414 L at STP, the molecular weight of the gas is 30.6 g mol¯ 1
11.2 L at STP is one-half molar volume, so there is 0.500 mol of gas present. Therefore, the molecular weight is 80.0 g mol¯ 1
(1.00 atm) (19.2 L) = (n) (0.08206) (273 K) n = 0.8570518 mol (I'll keep a few guard digits)
12.0 g / 0.8570518 mol = 14.0 g/mol
(19.2 L / 12.0 g) = (22.414 L / x ) 19.2x = 268.968 x = 14.0 g/mol
n = PV / RT n = [(700.0 mmHg / 760.0 mmHg atm¯ 1 ) (48.0 L)] / [(0.08206 L atm mol¯ 1 K¯ 1 ) (293.0 K)] n = 1.8388 mol
96.0 g / 1.8388 mol = 52.2 g/mol
n = PV / RT n = [(79.97 kPa / 101.325 kPa atm¯ 1 ) (4.167 L)] / [(0.08206 L atm mol¯ 1 K¯ 1 ) (303.0 K)] n = 0.13227 mol
20.83 g / 0.13227 mol = 157.5 g/mol
8.31451 J mol¯ 1 K¯ 1 / 0.0820574 L atm mol¯ 1 K¯ 1 = 101.3255 J L¯ 1 atm¯ 1 This means that 1 L atm = 101.3255 J
0.0820574/8.31451 = 0.00986918 (try putting the units in as was done just above) This means that 1 J = 0.00986918 L atm You could have also done this: 1 / 101.3255 = 0.00986918
(0.0820574 atm L/mol K) (101.3255 J/L atm) = 8.31451 J/mol K and (8.31451 J/mol K) (0.00986918 L atm / J) = 0.0820574 L atm / mol K
5.600 g / 44.009 g/mol = 0.1272467 mol
(P) (4.00 L) = (0.1272467 mol) (0.08206 L atm mol¯ 1 K¯ 1 ) (300 K) P = 0.7831 atm (to four sig figs)
Notice that we have pressure, volume and temperature explicitly mentioned. In addition, mass and molecular weight will give us moles. It appears that the ideal gas law is called for. However, there is a problem. We are being asked to change the conditions to a new amount of moles and pressure. So, it seems like the ideal gas law needs to be used twice.
P 1 V 1 = n 1 RT 1 This equation will use the 2.035 g amount of H 2 as well as the 1.015 atm, 5.00 L, and the −211.76 °C (converted to Kelvin, which I will do in a moment). P 2 V 2 = n 2 RT 2 This second equation will use the data in the second sentence and T 2 will be the unknown. What I need to do is set the two equations equal to each other. First, I rearrange a bit.
P 1 V 1 = n 1 RT 1 leads to: P 1 V 1 R = ––––– n 1 T 1 and P 2 V 2 = n 2 RT 2 leads to: P 2 V 2 R = ––––– n 2 T 2
R = R, therefore: P 1 V 1 P 2 V 2 ––––– = ––––– n 1 T 1 n 2 T 2
Since the volume never changes, we can eliminate it from the equation: P 1 P 2 ––––– = ––––– n 1 T 1 n 2 T 2
P 1 n 2 T 2 = P 2 n 1 T 1
P 2 n 1 T 1 T 2 = ––––– P 1 n 2
T 2 = P 2 n 1 T 1 / P 1 n 2
Each of the mole amounts would be arrived at by dividing the grams by the molar mass (in this case, H 2 ). However, notice the molar masses will cancel, being the same numerical value and one in the nominator and one in the denominator. After cancelling, this is what we wind up with: P 2 mass 1 T 1 T 2 = ––––––––– P 1 mass 2
(3.015 atm) (2.035 g) (61.24 K) T 2 = ––––––––––––––––––––––––– (1.015 atm) (4.134 g) T 2 = 89.546867 K
°C = 89.546867 K − 273.15 (I decided to use 273.15 rather than 273.) Using four sig figs gives −183.6 °C for the final answer
(P) (22.414 L) = (2.00 mol) (0.08206 L atm / mol K) (T)
T/P = 22.414 L / [(2.00 mol) (0.08206 L atm / mol K)] T/P = 136.57 K/atm Any T/P combination that gives 136.57 will be an answer.
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