I'll start from the premise that a helium-filled party balloon has risen through the atmosphere to the point of buoyant equilibrium without otherwise changing significantly, and we then let the gas escape in order to propel the empty balloon toward the moon.
To impact the moon from a stationary (w.r.t. the ground) starting point in the upper atmosphere, we need virtually all of the 11 km/s of Earth's escape velocity. That is, we need 11 km/s of $\Delta v$. We'll approach this by trying to calculate the $\Delta v$ of the balloon when the throat (assumed previously held closed) is opened.
To calculate $\Delta v$ we will use the Tsiolkovsky rocket equation:
$$\Delta v = v_e \ \mathrm{ln} \frac{m_0}{m_f}$$
where $v_e$ is the effective exhaust velocity, $m_0$ is the initial mass of the balloon plus gas, and $m_f$ is the final mass of the empty balloon.
We'll start by calculating $v_e$. If we assume the throat of the balloon forms a nozzle with perfect efficiency, then $v_e$ when the gas first starts escaping will be similar to the speed of a molecule inside the balloon. The root mean square speed of a molecule of an ideal gas is (see Physics of Music: Speed of Sound in Air and Wikipedia: Speed of Sound):
$$v_\mathrm{rms} = \sqrt{\frac{3 k_B T}{m}}$$
where $k_B$ is the Boltzmann constant, which is about $1.4 \times 10^{-23} \frac{\mathrm{J}}{\mathrm{K}}$, $T$ is the temperature in Kelvin, which we'll just assume is $300 \ \mathrm{K}$ (room temperature), and $m$ is the mass of a single molecule of gas, which for helium is 4 atomic mass units or about $6.8 \times 10^{-27} \ \mathrm{kg}$. Plugging in those numbers yields:
$$v_e \approx v_\mathrm{rms} \approx 1400 \frac{\mathrm{m}}{\mathrm{s}}$$
(To two significant figures, that's a specific impulse of 140 s. Peter Cordes observes that Wikipedia, citing Nguyen et al., gives a theoretical 179 s for helium at 298 K, but I don't know how that was derived. Perhaps my $v_e \approx v_\mathrm{rms}$ assumption needs refinement, but I'll proceed with it anyway.)
Next, according to Aerodynamics of a Party Balloon, the mass of the empty balloon alone is about:
$$m_f \approx 1.3 \ \mathrm{g}$$
Finally, $m_0$ is $m_f$ plus the mass of the helium. If the balloon was filled to a volume of 5 liters at one atmosphere of pressure and $300 \ \mathrm{K}$, it will contain about $0.8 \ \mathrm{g}$ (based on $4 \ \frac{\mathrm{g}}{\mathrm{mol}}$ and the ideal gas law). Adding $m_f$, we get:
$$m_0 \approx 2.1 \ \mathrm{g}$$
Plugging in all the numbers, we estimate (with very generous assumptions):
$$\Delta v \approx 670 \ \frac{\mathrm{m}}{\mathrm{s}}$$
This is well short of the needed $11 \ \frac{\mathrm{km}}{\mathrm{s}}$, so we can safely conclude that a party balloon cannot reach the moon by releasing its trapped helium.
Of course the practical $\Delta v$ of a helium balloon would be much less than this, for a variety of reasons, including:
- The opening is not a good nozzle.
- The gas cools as it escapes.
- There is no attitude stabilization so it will spin more than accelerate linearly.
- The balloon will have undergone changes in temperature and pressure during the rise through the atmosphere to a point of buoyant equilibrium (where we assume the throat is opened). It might burst due to pressure, or freeze and shatter, etc.
- The point of buoyant equilibrium is still well inside appreciable atmosphere, so some $\Delta v$ will be lost due to drag, assuming it accelerates in the right direction.
