Chemistry · Ch 10 — States of Matter
Values of 'R' in different Units
Values of 'R' in different Units
The numerical value of depends entirely on which units are chosen for , and , and is worked out from STP conditions. In SI units -- pressure in Pa (N m), volume in m, temperature in K -- using the current IUPAC STP (1 bar Pa, molar volume m, K): Pa m K mol J K mol. In litre-atmosphere units -- pressure in atm, volume in L (or dm) -- using the old STP (1 atm, L, K): L atm K mol dm atm K mol. In calories, since cal J: $R = 8.314/4.184 = 1.98 …
Pressure (P) | Volume (V) | Moles (n) | Temperature (T) | Gas constant (R)
Pa (pascal) | m3 | mol | K | 8.314 J K-1 mol-1
atm | dm3 | mol | K | 0.0821 atm dm3 K-1 mol-1
atm | L | mol | K | 0.0821 L atm K-1 mol-1
The value used for R in any calculation with the ideal gas equation must always match the units chosen for pressure and volume in that same calculation; mixing units (e.g. pressure in Pa with R in L atm units) giv …
Worked out. 9.0 L of N2 gas at 300 K and 1.5 atm is compressed to 3.0 L while being heated to 600 K; find the new pressure. Using the combined gas law, P1V1/T1 = P2V2/T2, so P2 = (P1V1T2)/(T1V2) = (1.5 atm x 9.0 L x 600 K)/(300 K x 3.0 L) = 8100/900 = 9 atm. Both effects push pressure up together -- compressing the volume by a factor of 3 and doubling the absolute temperature both independently raise pressure, and combined they raise it by a factor of 6, from 1.5 atm to 9 atm. …
Worked out. Find the temperature (in Celsius) at which 1 mole of nitrogen gas has volume 10 dm3 and pressure 2.46 atm, using R = 0.0821 dm3 atm K-1 mol-1. From PV = nRT, T = PV/(nR) = (2.46 atm x 10 dm3)/(1 mol x 0.0821 dm3 atm K-1 mol-1) = 24.6/0.0821 = 299.63 K. Converting to Celsius: 299.63 K - 273.15 = 26.48 degrees C. …