What is an example of an ideal gas in real life?

Hydrogen #("H"_2"(g))# is a pretty real-life, ideal gas at a temperature of #15^@ "C"# and a pressure of #"1.013 bar"#. Its compressibility factor #Z = (PV)/(nRT)# is #1.0006#, whereas that of an ideal gas is always #1.0000# in all conditions.

If a gas deviates far from ideality, you'd get significant inaccuracies in using the ideal gas law as opposed to, for example, the van der Waals, Redlich-Kwong, or Peng-Robinson Equations of State.

With hydrogen, it doesn't deviate from ideality that much at those conditions.

In fact, I get a #\mathbf(0.0693%)# difference in the volume per #"mol"# of #"H"_2(g)# when solving for it using the van der Waals equation as compared to the ideal gas law. If you want to know how to do that, I've written it out below!

Normally, the ideal gas law is:

#PV = nRT#

or

#\mathbf(P = (RT)/barV)#

with the molar volume #barV = V/n# is in #"L/mol"#.

VAN DER WAALS EQUATION OF STATE

Let's do an example calculation with the van der Waals equation of state, the simplest alternative:

#\mathbf(P = (RT)/(barV - b) - a/(barV^2))#

where #a# is a constant specific to the gas and is the average amount of attraction between gas particles, #b# is a constant specific to the gas and is the volume displaced by a #"mol"# of gas particles, and the rest of the variables mean the same as they do in the ideal gas law.

For hydrogen gas, we have the conditions and constants:

#P = "1.013 bar"#
#R = "0.083145 L"cdot"bar/mol"cdot"K"#
#T = "288.15 K"#
#a = "0.24646 L"^2cdot"bar/mol"^2# (McQuarrie, Table 16.3)
#b = "0.026665 L/mol"# (McQuarrie, Table 16.3)

CUBIC FORM OF THE EQUATION OF STATE

Let's solve for the molar volume by solving the equation for #barV#:

#P = (RTbarV^2)/((barV - b)barV^2) - (a(barV - b))/((barV - b)barV^2)#

#P(barV - b)barV^2 = RTbarV^2 - a(barV - b)#

#PbarV^3 - bPbarV^2 = RTbarV^2 - abarV + ab#

#\mathbf(Pcolor(green)(barV^3) - (bP + RT)color(green)(barV^2) + acolor(green)(barV) - ab = 0)#

NEWTON-RAPHSON METHOD FOR SOLVING CUBICS

Now, a straightforward way to solve this cubic equation of state is to use the Newton-Raphson method:

#\mathbf(barV_"new" = barV_"old" - (f(barV))/(f"'"(barV)))#

Now, if we let:

#barV = color(green)(x)#
#P = color(green)(A)#
#bP + RT = color(green)(B)#
#a = color(green)(C)#
#ab = color(green)(D)#

...then we may conveniently use the "store" function on, say, a TI-89 calculator to store each set of values in individual variables. If you don't have one, that's okay. It's not required, but it helps.

Anyways, for the purposes of typing this into a calculator potentially many times, we can now make this easier on the eyes:

#Ax^3 - Bx^2 + Cx - D = 0#

So the derivative is:

#3Ax^2 - 2Bx + C = 0#

For #barV_"old"#, you just type a guess for the molar volume. It requires a little bit of foreknowledge on what is realistic, but nevertheless, following the Newton-Raphson formula, you'd type out:

#[x_"old" - (Ax_"old"^3 - Bx_"old"^2 + Cx_"old" - D)/(3Ax_"old"^2 - 2Bx_"old" + C)] -> "store result into " x_"old"#

in one input line without yet pressing enter. Then press enter repeatedly until the answer stops changing. Each time you press enter, you update your #x_"old"#! :D

COMPARISON OF THE MOLAR VOLUMES

To save you the time, if you guess anything above #25# for #x_"old"#, when doing it correctly, you'd eventually get convergence, giving you:

#color(blue)(barV_"vdW" = "23.6672 L/mol")#

On the other hand, the ideal gas law is (evidently) much easier to solve, giving you:

#color(blue)(barV_"id") = (RT)/P#

#= (0.083145*288.15)/(1.013)#

#color(blue)("= 23.6508 L/mol")#

(either of which can be converted to be close to or equal to the more conventional mass density #color(blue)(rho = "0.00008517 g/mL")# at #15^@ "C"# and #"1.013 bar"# by noting that one #"mol"# of hydrogen gas is #"2.0158 g"#.)

These molar volumes are only different by about #0.0693%#; very negligible! Hence, hydrogen is quite ideal at #15^@ "C"# and #"1.013 bar"#.