A helium balloon has a volume of 2.50L at STP. If it's released and rised to an elevation at which pressure is 0.285 atm, and the temperature is -35 degrees. What's the new volume?

(P1V1/T1) = (P2V2/T2)

T must be in kelvin.

To find the new volume of the helium balloon at the given conditions, you need to use the combined gas law.

The combined gas law equation is:

(P₁V₁)/(T₁) = (P₂V₂)/(T₂)

where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume (which we need to find)
T₂ = final temperature

Now let's plug in the given values into the equation:

P₁ = 1 atm (at STP)
V₁ = 2.50 L
T₁ = 273 K (at STP, which is 0°C or 273 K)
P₂ = 0.285 atm
T₂ = -35°C (or 273 - 35 = 238 K)

(P₁V₁)/(T₁) = (P₂V₂)/(T₂)

(1 atm * 2.50 L)/(273 K) = (0.285 atm * V₂)/(238 K)

Now, we can solve for V₂:

(2.50)/(273) = (0.285 * V₂)/(238)

Cross multiplying, we get:

(0.285 * V₂) = (2.50)/(273) * 238

Now, divide both sides by 0.285 to isolate V₂:

V₂ = (2.50)/(273) * 238 / 0.285

Calculating this expression, the new volume, V₂, is approximately 5.519 L.

Therefore, the new volume of the helium balloon at an elevation where the pressure is 0.285 atm and the temperature is -35 degrees is approximately 5.519 L.