a balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa?
You need to know how much the temperature changes. For an altitude rise by that much, atmospheric temperature typically drops from 288 K to 217 K. (I got that info from a standard atmosphere table). You could also assume an adiabatic expansion, in which case
T2/T1 = (P2/P1)^[(g-1)/g)]
where g = 5/3 for helium
That would give you T2 = 163 K. This probably won't happen, because outside air heat transfer will warm the balloon. So will sunlight, if there is any.
Assume P1 V1/T1 = P2 V2/T2 and solve for V2.
V2 = V1*(P1/P2)*(T2/T1)
If they expect you to assume T is constant, you need a better teacher or text.
To find the volume of the helium when the pressure changes from 103 kPa to 25.0 kPa, we will use Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature. This can be written as:
P1 * V1 = P2 * V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this case, the initial pressure (P1) is 103 kPa, the initial volume (V1) is 30.0 L, and the final pressure (P2) is 25.0 kPa.
We can rearrange the equation to solve for V2:
V2 = (P1 * V1) / P2
Plugging in the given values:
V2 = (103 kPa * 30.0 L) / 25.0 kPa
Now we can calculate the final volume of the helium:
V2 = 123.6 L
Therefore, the volume of the helium when the balloon rises to an altitude where the pressure is 25.0 kPa is 123.6 L.