= Solution
A <thermodynamic potential> packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=T\,dS-p\,dV+\mu\,dN$. <Legendre transforms> give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with <natural variables of a thermodynamic potential> $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed <particle number>, \b[<Helmholtz free energy> is minimized at fixed $T,V$; <Gibbs free energy> at fixed $T,p$].
A flow vessel is an open <control volume>, even if its walls are enclosed. For each inlet, <pressure> forces carry flow <work> $p\,dV$ in addition to <internal energy>. The steady first-law balance is
$$
\sum_{\rm in}\dot m(h+v^2/2+gz)+\dot Q-\dot W_{\rm shaft}
=\sum_{\rm out}\dot m(h+v^2/2+gz).
$$
With no <heat> exchange, shaft <work>, or significant kinetic and gravitational <energy> differences, the <enthalpy> fluxes are equal. <Mass> flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.
Use the <liquefaction yield with separate warm-gas and liquid outlets>. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing <liquid> to have that <temperature>. The two outlet streams need not be in mutual <thermal equilibrium>. Taking $H$ as the printed <specific enthalpy>, the <steady-flow energy balance> gives
$$
H(p)=\alpha H_\ell+(1-\alpha)H(p_0),\qquad
\boxed{\alpha=\frac{H(p_0)-H(p)}{H(p_0)-H_\ell}.}
$$
For fixed outlet enthalpies the greatest yield occurs at the smallest inlet <enthalpy>. \b[Among the tabulated <pressures> this is 30 atm.] Approximating the atmospheric-pressure gas <enthalpy> by the tabulated zero-pressure limiting value gives
$$
\boxed{\alpha\simeq\frac{8.74-7.18}{8.74-1.01}=0.202,\quad p\simeq30\,\mathrm{atm}.}
$$
This uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq8.651\times10^4\,\mathrm{J\,kg^{-1}}$ and $\alpha\simeq0.193$. The sparse table does not determine an exact continuous optimum between entries.
Atmospheric-pressure liquid <helium> must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.<Index of Semiconductor Process Gases - Helium by National Institute of Standards and Technology>[Helium-4 characteristic-property table: normal boiling point 4.224 K]{ref} A <counterflow heat exchanger> can return cold gas through the incoming stream before that gas leaves the overall <control volume>, while the collected <liquid> leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_\ell$ is an outlet-stream enthalpy difference, not the <latent heat> of vaporization at one common <temperature>.
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