Question

Calculate the standard entropy of vaporization of argon at its normal boiling point ?

Answer 1

I get approximately `"73.50 J/mol"cdot"K"`.

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The only way I can think of doing this with known data is to first calculate `DeltaH_(vap)` from its phase diagram.

On this diagram, the normal boiling phase transition occurs at `"87.35 K"` and `10^(-4) "GPa"` (or `"1 bar"`); this is right on the liquid-vapor coexistence curve.

We'll need another temperature on the liquid-vapor curve, and we can consider the triple point for that (`T_(tp) = "83.80 K", P_(tp) = "0.6875 bar"`, or `6.875 xx 10^(-5) "GPa"`).

Therefore, we have enough info to find the slope:

`(dP)/(dT) ~~ (DeltaP)/(DeltaT) = ("1 bar" - "0.6875 bar")/("87.35 K" - "83.80 K") = "0.08803 bar/K"`

However, in the following diagram,

we see that we don't quite get a straight line here. So we re-estimate the pressure to be `"0.64 bar"`, to give a proper tangent line.

`(dP)/(dT) ~~ (DeltaP)/(DeltaT) = ("1 bar" - "0.64 bar")/("87.35 K" - "83.80 K") = "0.1014 bar/K"`

This slope is the left side of the Clapeyron Equation:

`(dP)/(dT) = (DeltabarH_(vap))/(T_bDeltabarV_((l)->(g)))`

where `DeltabarH_(vap)` is the molar enthalpy of vaporization, `T_b` is the boiling point, and `DeltabarV_((l)->(g))` is the change in molar volume due to going from liquid to gas.

Assuming argon remains an ideal gas, `DeltabarV ~~ barV_((g))`, so that

`T_bDeltabarV_((l)->(g)) ~~ T_b barV_((g)) ~~ (RT_b^2)/P`

Therefore,

`1/P(DeltaP)/(DeltaT) ~~ (DeltabarH_(vap))/(RT_b^2)`

`=> DeltabarH_(vap) ~~ (RT_b^2)/P((DeltaP)/(DeltaT))`

`= (("8.314472 J")/("mol"cdotcancel"K") cdot (87.35 cancel"K")^cancel(2))/(cancel"1 bar") cdot (0.1014 cancel"bar")/cancel"K"`

`=` `"6433.31 J/mol"`

`=` `"6.43 kJ/mol"`

(NOTE: the difference in the slope would give a `~~15%` error in the `DeltabarH_(vap)`.)

At a phase transition, there is a phase equilibrium such that `DeltaG_("transition") = 0`. Therefore,

`cancel(DeltabarG_(vap))^(0) = DeltabarH_(vap) - T_bDeltabarS_(vap)`

`=> color(blue)(DeltabarS_(vap)) = (DeltabarH_(vap))/T_b`

`= "6433.31 J/mol"/"87.35 K"`

`= color(blue)("73.50 J/mol"cdot"K")`

The actual value is `"74.53 J/mol"cdot"K"`, and the error is in the estimation of the slope.