What do you do with the given R in the ideal gas law equation? Do you multiply it to your missing variable at the end once you find it? Like what purpose does it serve?

The units of constant is derived from equation
PV = nRT.
Where the pressure – P, is in atmospheres (atm) the volume – V, is in liters (L) the moles -n, are in moles (m) and Temperature -T is in Kelvin (K) as in all gas law calculations.
When we do the algebraic reconfiguration we end up with Pressure and Volume being decided by moles and Temperature, giving us a combined unit of atm x L / mol x K. the constant value then becomes 0.0821 atm(L)/mol(K).
If you choose not to have your students work in standard pressure unit factor, you may also use: 8.31 kPA(L)/mol(K) or 62.4 Torr(L)/mol(K).
Temperature must always be in Kelvin (K) to avoid using 0 C and getting no solution when students divide.
There is a variation of the ideal gas law that uses the of the gas with the equation PM = dRT.
Where M is the Molar Mass in g/mol and d is the Density of the gas in g/L.?
Pressure and Temperature must remain in the units atm and K and the Gas Law Constant remains R = 0.0821 (atm) L / (mol) K.
If a 100 g of nitrogen gas compressed in a rigid container at 1.5 atm and 25 C What volume should the container be?
P = 1.5 atm
V = ?
n = 100 g / 28g = 3.57 moles (28 g is the mass of Nitrogen gas ##N_2##
R = 0.0821 atm L / mol K
T = 25 C + 273 = 298K
PV =nRT becomes V = nRT/P
V = 3.57 mol (0.0821 atm L /mol K) (298 K / 1.5 atm
V = 58.2 L
Notice the gas law constant allows to cancel all the units except the liters (L).
I know this was long, but I hope it is helpful

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