Chem HW 60 Help

Question

A gas‑filled weather balloon has a volume of 59.0L at ground level, where the pressure is 769mmHg and the temperature is 21.7.After being released, the balloon rises to an altitude where the temperature is -4.44 and the pressure is 0.0635 atm. What is the weather balloon's volume at the higher altitude?

Answer 1

Use the combined gas law P₁V₁/T₁ = P₂V₂/T₂

You did not provide units so I'm going to assume that the temperature is in degrees C.

1.) First convert degrees C to K

K = degrees C + 273.15

T₁ = 21.7C + 273.15 = 294.85 K

T₂ = -4.44C + 273.15 = 268.71K

2.) Set up your known equation by laying out your known variables to see if you need to convert any other units

P₁ = 769 mmHg, V₁ = 59.0 L, T₁ = 294.85 K, P₂ = 0.0635 atm, V₂ = ?, T₂ = 268.71 K

3.) We see that the units of pressure are not the same. Convert both to either mmHg or atm. In this case, I will put both units in atm. Use the conversion factor 1 atm = 760 mmHg

P₁ = 769 mmHg x (1 atm / 760 mmHg) = 1.01 atm

4.) Set up your equation by plugging in variables and solve!

P₁V₁/T₁ = P₂V₂/T₂

(1.01 atm)(59.0 L) / 294.85 K = (0.0635 atm)(V₂) / 268.71 K

(1.01 atm)(59.0 L)(268.71 K) / 294.85 K = (0.0635 atm) (V₂)

V₂ = 855.23 L

Hope this helps! Double check my calculations to make sure you are getting the same math but the logic here should be correct.