I'll give a slightly different answer: The escape velocity at the surface of our Moon is about 2.38 km/s. Derived from this paper, as a rule of thumb, an atmosphere can survive 4.5 billion years, if it's average molecule velocity is below 1/6 of the escape velocity of the planet/moon. Applied to our Moon, it's 2380 m/s / 6 = 396.67 m/s.
Carbon dioxide has a molecule mass of 44u.
The according temperature for an average molecule velocity for carbon dioxide is hence
$$mbox{temperature}=(v_{mbox{gas}}/157)^2 cdot mbox{molecule mass}mbox=(396.67/157)^2cdot 44mbox{ K}=280.87mbox{ K}.$$
Moon's mean surface temperature at the equator is about 220 K. That's well below the allowed 280.87 K. A carbon dioxide atmosphere could have survived 4.5 billion years according to these oversimplified assumptions.
When looking closer to the atmosphere, it turns out, that the higher layers of the thermosphere and the exosphere can reach more than 1000 K.
Even carbon dioxide can escape from Moon at this temperature over time:
The average molecule velocity for carbon dioxide at 1000 K is
$$157cdot sqrt{frac{1000}{44}}mbox{ m/s}= 748.5mbox{ m/s}.$$
This results in a Jeans escape parameter of
$$lambda_0=left(frac{v_{mbox{esc}}}{v_{mbox{gas}}}right)^2 =left(frac{2380}{748.5}right)^2= 10.1,$$
resulting in an escape rate of about $10^{-4}$ relative to a free molecular flow.
From this it isn't yet straightforward to calculate the period of time, Moon would loose a carbon dioxide atmosphere, since it's a function of the height-dependent temperature curve of the atmosphere. But it's at least clear, that Moon wouldn't loose a carbon dioxide atmosphere suddenly, since this is likely just for $lambda_0 < 3$.
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