= Solution
Write the gross <star formation rate> as $\psi=kM_g$, with $k>0$ constant. Use the <instantaneous recycling approximation>: a fraction $F$ returns promptly, while a fraction $1-F$ stays in long-lived <stars> and remnants. Assume $0\le F<1$, instantaneous mixing, and no <galactic gas inflow> or <galactic outflow>. If $M_s$ denotes the permanently locked stellar mass, <mass conservation> gives
$$
\dot M_s=(1-F)\psi,\qquad \dot M_g=-(1-F)\psi=-(1-F)kM_g.
$$
Set $\lambda=(1-F)k$. Integration, with $M_g(0)=M_0$, gives the <exponential closed-box gas consumption> law
$$
\boxed{M_g(t)=M_0e^{-\lambda t},\qquad \psi(t)=kM_0e^{-\lambda t}.}
$$
The locked mass is $M_s=M_0(1-e^{-\lambda t})$, so $M_g+M_s=M_0$ at every time. The decay time is $[(1-F)k]^{-1}$, longer than the gross gas-processing time $k^{-1}$ because returned gas can form subsequent generations. The formal $F=1$ limit retains all gas and locks no mass; a finite <stellar yield> defined per locked mass then needs separate treatment.
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