Stephen,

You are asking what this statement means:

>  At 10,000 meters the air is roughly four times less dense, so the balloon
>  will be 4 times larger than at sea level.

By "four times larger", I mean larger in volume.
The ideal gas law says that if the amount of gas and the temperature
are both held constant, then the volume will be inversely proportional
to any change in pressure.  So if the pressure increases by a factor
of four, the volume decreases by a factor of four.

Likewise, if the temperature changes but the pressure is held constant, 
then the volume changes proportional to the change in temperature 
expressed in degrees Kelvin.  Zero degrees Kelvin is -273 degrees Centigrade,
so an increase in temperature from -50 C (-223 K) to 20 C (293 K) with pressure
held constant will cause the volume of our gas to increase by a factor of 293/223 = 1.31.
If we had 1 liter of gas at -50 C, it expands to 1.31 liters at 20 C.

Generally, all three can change simultaneously, for example compressing a gas causes it to heat up. 
Both relationships can be expressed simultaneously by the formula    PV = nRT,
where P is pressure, V is volume, T is temperature, and you can think of
the nR factor as a constant for a given amount of gas.
Digging a bit deeper, the n is a count of the number of molecules of gas,
and the R is a physical constant called "the ideal gas constant".
    https://en.wikipedia.org/wiki/Ideal_gas_law


The first line of this passage might be a bit confusing:
<  If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
<  then the pressure of the helium at altitude will be greater than the "just enough" case
<  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
<  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

best to edit it as follows:
>  If at 10,000 meters the balloon has more helium than what fits without stretching the mylar 
>  then the pressure of the helium at altitude will be greater than the "just enough" case
>  and thus it is more dense.  The lift will be less at altitude than it is at sea level.  
>  The balloon will rise to that altitude at which the lift is equal to the weight of the payload.

As Joe, WB9SBD, has also stated here, the lifting power of the balloon remains the same as
the balloon rises so long as the balloon can expand without exerting extra pressure on our lifting gas.
Think of a big floppy weather balloon about to be set free, the hydrogen has plenty of room to expand
before the balloon envelope reaches its maximum size and starts pressing back.
Once it gets high enough that the balloon envelope is tight and causes the hydrogen to be at a
higher pressure than the ambient air, the lifting power starts to drop. 

Jerry


On Sun, May 26, 2019 at 10:11 AM, Stephen Farthing G0XAR JO92ON97 wrote:

I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area? 

It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty.