How can you derive the ideal gas law?

Question

Peyton Adams · Accepted Answer

Ideal gas equation is arrived at from experimental evidence.

From Charles' law,

#V prop T# at #p# constant.

From Boyle's law,

#pV# is constant at #T# constant.

Also, from Avogadro's law that equal volumes of gases at the same temperature and pressure have equal number of molecules,

#V prop N# at constant #T# and #p#, where #N# is number of molecules.

The above results are combined immediately to obtain that,

#pV prop NT#
#implies (pV)/(NT) =#constant

Therefore,

#(pV)/(TN)=k_B# where #k_B# is a constant known as the Boltzmann constant.
Hence,

#pV = k_BNT#

But for #mu# moles of gas, the number of molecules is #N = muN_A# where #N_A# is the Avogadro's number.

Thus, #pV = muk_BN_AT#

Thus, we finally arrive at the ideal gas law,

#pV = muRT#

Where, #R = k_BN_A# is another constant known as the universal gas constant.

Noah Aldrich · Answer

Another way I recently came across is essentially a derivation. I'll illustrate below.

From statistical mechanics, the Helmholtz free energy is given as, #A = -k_BTln Z # where #Z# is the partition function. But, #zeta = sum g_ie^(-E_i/(k_BT))# If energy levels are closely spaced, #zeta = int g(E)e^(-E/(k_BT))dE# where #g(E)# is the density of states. However, #g(E)dE = (2piV)/h^3(2m)^(3/2)sqrtEdE# Therefore, the partition function of each molecule (sub-system), #zeta = int e^(-E/(k_BT))(2piV)/h^3(2m)^(3/2)sqrtEdE# This is evaluated by Gamma functions and is, #zeta = V/h^3(2pimk_BT)^(3/2)# Now, the partition function of the entire system consisting of #N# molecules that are indistinguishable, #Z = zeta^N/(N(N-1)(N-2)...1)# However, from Thermodynamics, #p = -((delA)/(delV))# at constant #T# #implies p = (Nk_BT)/V# Thus we get the ideal gas equation, #pV = Nk_BT# which can be rewritten as, #pV = muRT# where #N = muN_A# and #N_Ak_B = R#

Luna Chen · Answer

I want to give a derivation of statistical mechanics that goes all the way back to particle statistics, even though Aritra has a good one.

As a bonus, we get expressions for #G#, #mu#, #A#, and #S# in terms of the partition function and internal energy!

The statistical distribution of the majority of particles, or so-called corrected boltzons, is as follows (Norman Davidson, Statistical Mechanics, 1969):

#lnt = sum_i (N_i ln (g_i/N_i) + N_i)#

where #t# gives the number of microstates in the system with #N_i# particles in a state of energy #epsilon_i# of degeneracy #g_i#.

The well-known Boltzmann entropy formulation is as follows:

#S = k_Blnt#

We can first determine the absolute entropy from this known distribution as long as the energy levels in a system only depend on the volume and not the entropy itself:

#S = k_B sum_i (N_i ln (g_i/N_i) + N_i)#

A fixed number of particles and energy constrain a conservative system, meaning that

the total number of particles is constrained to #N = sum_i N_i#.

the total internal energy is constrained to #E = sum_i N_iepsilon_i#.

Moreover, the following represents the distribution of states:

#N_i/N = (g_ie^(-epsilon_i//k_BT))/(sum_i g_i e^(-epsilon_i//k_BT)#

where #q# is the microcanonical partition function, #q = sum_i g_ie^(-epsilon_i//k_BT)#, and #q# is a function of only temperature and volume.

Consequently, the absolute entropy can be expressed as follows:

#S = k_B [sum_i N_i ln (q/N e^(epsilon_i//k_BT)) + sum_i N_i]#

Since #q# and #N# are independent of which particle index is involved,

#S = k_B [ln (q/N) sum_i N_i + sum_i N_iln(e^(epsilon_i//k_BT)) + sum_i N_i]#

Based on the provided constraints, the absolute entropy is:

#barul(|stackrel(" ")(" "S = Nk_Bln(q/N) + E/(T) + Nk_B" ")|)#

(Fun fact: This equation can be used to derive the standard molar entropy that is typically found in the back of General Chemistry textbooks.)

From the idea that the Gibbs' free energy is given as a function of the enthalpy and entropy, #G = H - TS#.

Similarly, we also have that the Helmholtz energy is #A = E - TS#, where #E# is the internal energy. We then get:

#A = cancelE - (Nk_BTln(q/N) + cancelE + Nk_BT)#

#=> barul(|stackrel(" ")(" "A = -Nk_BTln(q/N) - Nk_BT" ")|)#

The chemical potential #mu# is found in the Maxwell Relation of many thermodynamic functions, but it is most convenient to recall it within the Helmholtz free energy:

#dA = -SdT - PdV + sum_i mu_idN_i#

#mu# is related to the molecular Gibbs' free energy and the Helmholtz free energy:

#mu = G/N = ((delA)/(delN))_(T,V)#

#= -k_BT(N cdot (delln(q//N))/(delN) + ln(q/N)) - k_BT#

Since #q# is a function of #T# and #V# but not of #N#,

#barul(|stackrel(" ")(" "mu = -k_BTln(q/N)" ")|)#

By comparison, since #Nmu = G = -Nk_BTln(q/N)#, this nicely compares with #A# to give:

#G = A + Nk_BT#

From thermodynamics, just like #H = E + PV#, since #G = H - TS# and #A = E - TS#, we also have that

#G = H - TS#

#= -TS + E + PV#

#= A + PV#.

The ideal gas law is thus:

#color(blue)(barul(|stackrel(" ")(" "PV = Nk_BT = nRT" ")|))#

where #N/nk_B = N_Ak_B = R#, with #n# being the mols of gas and #N# being the number of gas particles.

And according to Norman Davidson's 1969 book Statistical Mechanics, "the number of system states in the subsystems is equal to the number in the most probable distribution" indicates that this is valid.

Caleb Adams · Answer

undefined The ideal gas law can be derived from the combination of Boyle's law, Charles's law, and Avogadro's law. Boyle's law states that at constant temperature, the pressure of a gas is inversely proportional to its volume (P ∝ 1/V). Charles's law states that at constant pressure, the volume of a gas is directly proportional to its temperature (V ∝ T). Avogadro's law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas (V ∝ n).

Combining these three laws, we get the following equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature

This equation is known as the ideal gas law.