= Solution
{c}
Use the <instantaneous recycling approximation> with a well-mixed gas reservoir, constant <stellar yield> $p$, no gas inflow, and an irreversible <galactic outflow>. Let $M_s$ be the mass locked into long-lived stars and remnants after prompt recycling, and put $\psi(t)=\dot M_s(t)\geq0$. The yield $p$ is the newly synthesized mass of heavy elements returned to the gas per unit increase of $M_s$. Thus $M_h$ in $Z=M_h/M_g$ denotes the heavy-element mass in the gas, excluding metals already locked into stars.
Assume $M_s(0)=0$, $M_g(0)=M_0$, and $Z(0)=Z_0$; initially pristine gas means $Z_0=0$. Take the outflow rate to be $\alpha\psi$ with constant $\alpha\geq0$, and assume its metallicity equals the current <gas-phase metallicity> $Z$. This is the well-mixed <leaky-box model of galactic chemical evolution>. The parameter $\alpha$ is a <mass-loading factor> relative to the net rate of locking mass into stars. If a gross <star formation rate> $\Psi$ and a prompt returned fraction $\mathscr R$ are used instead, $\psi=(1-\mathscr R)\Psi$ and a wind rate $\eta\Psi$ corresponds to $\alpha=\eta/(1-\mathscr R)$.
<Conservation of mass> gives
$$
\dot M_g=-(1+\alpha)\psi,
\qquad
M_g(t)=M_0-(1+\alpha)M_s(t),
$$
and the baryonic mass remaining in the box is $M_g+M_s=M_0-\alpha M_s$. Newly made metals enter the gas at rate $p\psi$, while pre-existing metals are locked into stars at rate $Z\psi$ and leave in the wind at rate $\alpha Z\psi$. Hence
$$
\frac{dM_h}{dt}=[p-(1+\alpha)Z]\psi.
$$
Using $M_h=ZM_g$ and the product rule,
$$
M_g\dot Z+Z\dot M_g=[p-(1+\alpha)Z]\psi
\quad\Longrightarrow\quad
\boxed{\dot Z=\frac{p\psi}{M_g}.}
$$
The cancellation expresses the fact that a well-mixed wind removes gas and metals in the same proportion, while fresh stellar production raises the abundance in the remaining gas.
Divide this equation by $\dot M_g=-(1+\alpha)\psi$ along the evolving reservoir, or integrate in time through intervals with no <star formation>. The resulting <gas-phase metallicity> is
$$
\boxed{Z(t)=Z_0+\frac{p}{1+\alpha}\ln\frac{M_0}{M_g(t)}
=Z_0+\frac{p}{1+\alpha}\ln\left[\frac1{1-(1+\alpha)M_s(t)/M_0}\right].}
$$
Equivalently, its fully explicit dependence on a prescribed net <star formation rate> is
$$
Z(t)=Z_0+\frac{p}{1+\alpha}
\ln\left[\frac{M_0}{M_0-(1+\alpha)\int_0^t\psi(t')\,dt'}\right],
$$
valid while $M_g(t)>0$. With pre-existing stars, replace $M_s(t)$ in the gas-mass relation by $M_s(t)-M_s(0)$. Without specifying $\psi(t)$ or a gas-consumption law, the model fixes $Z$ as a function of gas mass but does not determine a unique function of time.
For the <closed-box model of galactic chemical evolution>, set $\alpha=0$. Then $M_g=M_0-M_s$ and
$$
\boxed{Z_{\rm closed}=Z_0+p\ln\frac{M_0}{M_g}.}
$$
At the same remaining fraction $g=M_g/M_0$ of the initial gas reservoir, the leaky box has the logarithmic coefficient $p/(1+\alpha)$, lower than the true <stellar yield> by the wind factor. This coefficient is often called an <effective yield>, but the definition of the gas fraction must be specified.
In particular, the observable <gas fraction of a galaxy> is often $\mu=M_g/(M_g+M_s)$, using the mass still present, rather than $g$. In our initially star-free model,
$$
\mu=\frac{(1+\alpha)g}{1+\alpha g},\qquad
g=\frac{\mu}{1+\alpha(1-\mu)}.
$$
Therefore
$$
\boxed{Z_{\rm leaky}(\mu)=Z_0+\frac{p}{1+\alpha}
\ln\left[\frac{1+\alpha(1-\mu)}\mu\right],\qquad
Z_{\rm closed}(\mu)=Z_0+p\ln\frac1\mu.}
$$
For $0<\mu<1$ and $\alpha>0$, the leaky-box enrichment is smaller at the same $\mu$: $\ln[1+\alpha(1-\mu)]\leq\alpha(1-\mu)<\alpha\ln(1/\mu)$. If <effective yield> is defined observationally as $(Z-Z_0)/\ln(1/\mu)$, it is consequently not simply the constant $p/(1+\alpha)$ at all $\mu$.
As an explicit time example, choose a <linear gas-consumption law> $\psi=\epsilon M_g$, with constant $\epsilon>0$. The mass and metallicity equations give
$$
M_g=M_0e^{-(1+\alpha)\epsilon t},\qquad
M_s=\frac{M_0}{1+\alpha}\left(1-e^{-(1+\alpha)\epsilon t}\right),\qquad
\boxed{Z(t)=Z_0+p\epsilon t.}
$$
The closed box with the same $\epsilon$ has the same metallicity at a given time, while the leaky box exhausts its gas faster. Thus a comparison at fixed time depends on the chosen <star formation rate>, even though the comparison at fixed gas fraction above is unambiguous.
Finally, the assumption about wind metallicity is essential. If the expelled gas has abundance $Z_w$ instead, its metal-loss rate is $\alpha Z_w\psi$, and the same bookkeeping yields
$$
M_g\dot Z=[p+\alpha(Z-Z_w)]\psi.
$$
A wind enriched relative to the ambient gas lowers the enrichment rate; the logarithmic solution derived above applies to $Z_w=Z$.
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