Reduce the atmospheric pressure
Actually, the atmosphere has a weight that is always and constantly pushing out to the ground (not confuse with gravity). Our atmosphere is about 101.325 kPa of pressure, that means 101,325 kN/m2 or 10.332 kgf/m2 of pressure in our bodies. Also, the atmosphere has a density (even low) which increase the friction of our body while we are moving.See at the bottom for water boiling point
Increase oxygen levels
If we increased our atmospheric oxygen level we would be able to perform more exhausting task breathing less and tiring us also less.
But, how we can reduce the atmospheric pressure and increase the oxygen level of it? It's very easy: just reduce the pressure while at the same time increase the oxygen percentage. Our atmosphere has 101.325 kPa of pressure (equal to 1 atm and almost 1 bar), reduce the overall atmosphere to only 21-35 kPa of 100% pure oxygen.
Remember that a normal oxygen partial pressure is around 21 kPa, and it must be lower than 50 kPa (oxygen toxicity) but higher than 16 kPa (hypoxia). At higher pressures, fewer you need to breathe.
Side note: a bit of CO2 is also necessary, just a bit to prevent hipocapnia. Also, please see at the bottom for water boiling point
Reduce the surface gravity
It's quite obvious that reducing our gravity (1 G = 9.8 m/s2) will make thing lighter for us. Remember that a lower gravitational force will also decrease the weight of the atmosphere reducing its pressure. However, lower gravitational force can have the risk of releasing the atmosphere.
Basically, your oxygen molecules must not move faster than your escape velocity. I have already done this kind of calculations in this answer but I'll do them again.
So if the RMS (Root-mean-square speed) velocity of the oxygen molecule in the atmosphere is equal or greater than the escape velocity of the planet then that gas will escape rapidly and will be absent.
$$\text{RMS} = \text{v}_{\text{rms}}=\sqrt{\frac{3\times\text{R}\times\text{T}}{\text{M}_{\text{m}}}}$$
- Where:
- $\text{Vrms}$ is the root mean square of the speed in meters per second.
- $\text{Mm}$ is the molar mass of the gas in kilograms per mole. $\text{O}_2 = 0.031998 \text{ kg/mol}$
- $\text{R}$ is the molar gas constant. $\text{R} = 8.3144598(48)\text{ J}\times\text{mol}^{-1}\times\text{K}^{-1}$
- $\text{T}$ is the temperature in degrees kelvin (K = °C + 273.15). I'll use 25°C (298.15 K), I think that is the "normal" temperature used in gas calculations where it's specified.
$$v_{rms}=\sqrt{\frac{3\times8.314459848\times273.15}{0.031998}} = 482.096 \text{ m/s}$$
Or simply use this online calculator.
So we already know that our escape velocity must be greater than 482.096 m/s.
To calculate the escape velocity we could use:
$$\text{v}_\text{e} = \sqrt{\frac{2\times\text{G}\times\text{M}}{\text{r}}} = \sqrt{2\times\text{g}\times\text{r}}$$
Where:
- $\text{G}$ is the gravitational constant. $\text{G} ≈ 6.67 \times 10^{11} \text{ m}^3 \times \text{kg}^{-1} \times \text{s}^{-2} ≈ 0.0000000000667$
- $\text{M}$ is the mass of the planet.
- $\text{r}$ is the radius of the planet in meters.
- $\text{g}$ is the surface gravity of your planets in meters per second squared
You can calculate the escape velocity either with your mass and radius or with your gravitational acceleration and radius because they can be exchangeable:
$$\text{g} = \frac{\text{G}\times\text{M}}{\text{r}^2}$$
I can't tell you your minimal surface gravity to hold oxygen because it's your decision decide how much will be the relation between mass and radius (density) in your planet (small but denser or big but lighter).
Water boiling point
As @Tyler S. Loeper suggest in comments, the reduction of the atmospherical pressure will reduce the boiling point of water. So I'll teach you how to calculate that:
$$\text{T}_\text{ebm} = \frac{\text{T}_\text{ebn} - \text{K}_\text{SY} \times 273.15 \times (\text{P}_\text{n} - \text{P}_\text{m})}{1 + \text{K}_\text{SY} \times (\text{P}_\text{n} - \text{P}_\text{m})}$$
Where:
- $\text{T}_\text{ebm}$ is the ebullition point of a given atmospherical pressure.
- $\text{T}_\text{ebn}$ is the ebullition point of normal atmospherical pressure.
- $\text{K}_\text{SY}$ is the Sidney-Young constant which is $\text{K}_\text{SY} = 0.0012$ for polar sustances (like water) and $\text{K}_\text{SY} = 0.00010$ for non-polar sustances
- $\text{P}_\text{n}$ and $\text{P}_\text{m}$ are the atmospherical pressures. $\text{P}_\text{n}$ is for the normal atmosphere and $\text{P}_\text{m}$ for the given atmosphere. Both must be in the same measure (mmHg, bar, pascal, atm, etc).
So, using a 100% oxygen atmosphere of 21 kPa and other of 35 kPa (as examples) they would be:
$$\text{T}_\text{ebm} = \frac{100 - 0.0012 \times 273.15 \times (101.325 - 21)}{1 + 0.0012 \times (101.325 - 21)} = 67.19 \text{°C at } 21 \text{kPa}$$
$$\text{T}_\text{ebm} = \frac{100 - 0.0012 \times 273.15 \times (101.325 - 35)}{1 + 0.0012 \times (101.325 - 35)} = 72.49 \text{°C at } 35 \text{kPa}$$
Any of both number would produce a danger in our body (that is because our body isn't at 67.19 °C!)