Show that, for a gas with second virial coefficient B=a-b...

Show that, for a gas with second virial coefficient $B=a-b e^{c / T}$, where $a, b, c$ are constants, the maximum inversion temperature is a solution of the equation $a T=b(c+T) e^{c / T}$. Hence find the maximum inversion temperatures of helium and argon, for which $[a, b, c]=$ $[114.1,98.7,3.245 \mathrm{~K}]$ and $[154.2,119.3,105.1 \mathrm{~K}]$, respectively.

Key Concepts

Inversion Temperature

The inversion temperature is the specific temperature at which the Joule-Thomson coefficient changes sign, meaning that a gas will neither cool nor heat during an adiabatic expansion. The maximum inversion temperature represents the highest temperature at which a gas exhibits cooling upon expansion and is determined by setting the conditions that nullify the Joule-Thomson coefficient. This concept is crucial for understanding the limits of gas cooling in various thermodynamic processes.

Joule-Thomson Effect

The Joule-Thomson effect describes the change in temperature that occurs when a real gas expands adiabatically through a porous plug or throttling device without performing external work. This effect is central to gas liquefaction processes and refrigeration, as it depends on the interplay between the kinetic energy of the molecules and their intermolecular forces.

Virial Expansion

The virial expansion is a series representation that modifies the ideal gas law to account for intermolecular interactions by including additional terms dependent on the gas density or pressure. Each term in the series, such as the one involving the second virial coefficient, provides corrections that become increasingly important as the density increases or as the gas deviates from ideal behavior.

Second Virial Coefficient

The second virial coefficient quantifies the first correction to the pressure of a real gas due to pairwise interactions between molecules. Its value, and especially its temperature dependence, reflects the balance between attractive and repulsive molecular forces, offering important insight into the nature of these interactions and their impact on gas behavior.

Transcript

00:01 In the given question, we have to prove that for a monoatomic gas, which follows the equation, pv minus b is equals to rt, the ratio of temperature to the power 3 by 2 is equal to its molar volume minus b over second molar volume minus b.

00:25 Okay, here b is some constant and we have to prove this expression.

00:32 As we know that for adiabatic expansion, the heat is zero.

00:38 There is no transfer of heat.

00:40 That means change in internal energy is equal to work done.

00:45 Okay.

00:46 Now change in internal energy, we can write it to be equal to cvdt.

00:52 Okay.

00:52 While our work done is minus pdv.

00:56 Now from the given equation, we can write p equals to r over v minus v into temperature.

01:03 So we can replace this pressure with volume and temperature terms.

01:07 Okay.

01:08 So this will be r v minus b over temperature dv and this will be cv d t.

01:14 We can take this temperature on the left hand side.

01:17 Okay.

01:17 So this will be cv t d t to be equals to r 1 upon v minus b dv.

01:26 So to get the ratio of temperature and volume we need to integrate on both side.

01:30 Okay.

01:30 So we know this here the limit will be t1 to t2.

01:33 There is one minus sign also and here the limit is v1 to v2 and just one thing the temperature is in the numerator okay so this is p equals to r t v minus b and similarly this temperature will be in the numerator such that here the temperature will be in the denominator so cv can come out of the integration since it is a constant so we know that the integration of 1 upon x is l and x so here we will get l and t the limits will be t 1 to t 2.

02:04 While on the right end side, we get minus r ln, the limits are v1 to v2.

02:14 Okay, sorry, this will be v minus b...

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