question archive 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## is in moles (mol) and temperature ##T## is in Kelvin (K) as in most gas law calculations
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## is in moles (mol) and temperature ##T## is in Kelvin (K) as in most gas law calculations
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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## is in moles (mol) and temperature ##T## is in Kelvin (K) as in most gas law calculations.
##R## is the Ideal Gas Law constant and is derived through algebraic reconfiguration we end up with Pressure and Volume being decided by moles and Temperature, giving us a combined unit of ##(atm xx L) / (mol xx 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)##.
