Question #eded3

1 Answer
May 30, 2017

Answer:

See Explanations Section

Explanation:

The Empirical Gas Law relationships are:

Boyles' Law: #"Pressure" prop=(1/"Volume")#; mass & Temperture remain constant.
=> #"P"prop"(1/V)"# => #P=k(1/V)# => #k = (PV)#
=> #k_1=k_2# => #P_1V1 = P_2V_2#

Charles' Law: #"Volume "prop" f(Temperature)"#; Pressure & mass remain constant.
=> #VpropT# => #V=kT# => #k = (V/T)#
=> #k_1=k_2# => #(V_1/T_1)=(V_2/T_2)#

Gay-Lussac Law: #"Pressure "prop" f(Temperature)"#; mass & Volume remain constant.
=> #PpropT# => #P=kT# => #k = (P/T)#
=> #k_1=k_2# => #(P_1/T_1)=(P_2/T_2)#

Volume-Mass Law : #"Volume "prop" f(mass)"#; Pressure & Temperature remain constant.
=> #Vpropn# => #V=kn# => #k = (V/n)#; n = moles
=> #k_1=k_2# => #(V_1/n_1)=(V_2/n_2)#

Pressure-Mass Law: #"Pressure "prop" f(mass)"#; Pressure & Temperature remain constant.
=> #Ppropn# => #P=kn# => #k = (P/n)#; n = moles
=> #k_1=k_2# => #(P_1/n_1)=(P_2/n_2)#

Combined Gas Law
=>#PVpropnT# => #PV=knT# => #k = ((PV)/(nT))#
=> #k_1=k_2# => #((P_1V_1)/(n_1T_1)) =((P_2V_2)/(n_2T_2))#

Ideal Gas Law => Assumes one of the P,V,n,T condition sets of the Combined Gas Law is at Standard Temperature - Pressure conditions (STP). The other set of P,V,n,T conditions are Non-Standard Conditions.

#STP# => #(P,V, n, T)# #(1.0Atm, 22.4L, 1"mole", 273K)#

=> #((P_1V_1)/(n_1T_1)) = ((1Atm)(22.4L))/((1mol)(273K))# = #0.08206((L)(Atm))/((mol)(K))#

= #"Universal" "Gas" "Constant" (R)#

Therefore, substituting 'R' into Combined Gas Law
=> #R = ((PV)/(nT))# => #PV = nRT#

Note:
One can also use the same logic for deriving relationships for ...

Henry's Law of Gas Solubility => #" Solubility of a gas" prop "Applied Pressure#

Graham's Law of Gas Effusion Rates => #"Effusion Rate" prop (1/sqrt(mol wt))#