Marcus95
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Homework Statement
a)Helium enters a closed system as a flow of compressed gas at a temperature
of 14 K and at pressure p (enthalpy conserved). A fraction α emerges as liquid and the rest as gas at 14 K, both at atmospheric pressure p0. Find an expression for α in terms of the enthalpy H(p) of He gas and the enthalpy Hℓ of the liquid. [6]
b)From the data below, find the maximum value of α and the initial pressure p for
which this occurs. [4]
p/[atm] | H/[##10^4##J/kg] at 14 K
0 | 8.74
10| 7.85
20 | 7.31
30 | 7.18
40 | 7.26
[The enthalpy Hℓ of liquid He at p0 is ##1.01 × 10^4## J kg−1.]
Homework Equations
Enthalpy: ##H = U + pV ##
The Attempt at a Solution
a) In a closed flow process, enthalpy is conserved. Hence we must have:
$$ H(p) = \alpha H_l + (a-\alpha) H(p_0) $$
which simplifies to:
$$ \alpha = \frac{H(p) - H(p_0)}{H_l - H(p_0)} (\star) $$
to write this as a function of ##H_l## and ##H(p)## only, we somehow need to express ##H(p_0)## in the other functions.
The definition of enthalpy: I thought that if we take the helium to be an ideal monoatomic gas we simply have: $$H = \frac{3}{2} nRT + nRT = \frac{5}{2} nRT $$ or $$H = \frac{5}{2} RT/M $$ per kg (as that seems to be the prefferred unit), M being the molar mass.
However, the problem is that ##T = T_0 = 14 K##, hence the expression above would suggest that ##H(p_0) = H(p)## and ##\alpha = 0##. This must clearly be wrong and the helium cannot be modeled as an ideal gas.
Does anybody know how ##(\star)## can be simplified? Considering that 60% of the marks (old exam question) are awarded for this part there must be something more to it than this.
b) Nevertheless, I have tried to progress with the question here by simply interpolating H(p_0) linearly from the data given. It leaves me with:¨
$$\alpha = \frac{8.651 - H(p)}{7.641} $$
and this is a linear function without a maximum...