Aballoon is filled with 8l/.28cf of air at 2.7bar/atm. pressure. if the balloon is taken to 1.3bar/atm, what will the new volume be?

That will depend on temperature change between the two places.
I can show you the answer if we assume that the temperature doesn't
change, but that's so unlikely between 2 places where the pressure
changes by 50% that it's going to be a very unhelpful answer.

We also have to assume that the air in the balloon acts like an "ideal gas".

For an ideal gas at constant temperature, the product of

(pressure) x (volume)

is a constant number.

The balloon starts with pressure = 2.7 bar and volume = 8 L.
The product is 21.6 bar-liter.
If the temperature doesn't change, then that number won't change either.

The balloon is moved to a place where pressure = 1.3 bar.
Since the temperature doesn't change, we know that (press) x (vol)
is still 21.6 bar-liter. 
So we can write . . .

(1.3 bar) x (volume) = 21.6 bar-liter

Divide each side by (1.3 bar) :

Volume = (21.6 bar-liter) / (1.3 bar)  =  16.62 liters

You get 6 for mechanical