A weather balloon is filled with 0.295 m3 of helium on the ground at 18 °C and 756 torr.
What will the volume of the balloon be at an altitude of 10 km where the temperature is
48 °C and the pressure is 0.14 atm?
since PV=kT, PV/T is a constant.
So, using the given numbers,
(.14 atm)V/(271+48) = (756 torr)(.295)/(271+18)
You will need to reconcile atm and torr, and then solve for the new V
To find the volume of the weather balloon at the given altitude, we can use the combined gas law equation. The combined gas law equation is as follows:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ = pressure at the initial state (ground)
V₁ = volume at the initial state (ground)
T₁ = temperature at the initial state (ground)
P₂ = pressure at the final state (altitude)
V₂ = volume at the final state (altitude)
T₂ = temperature at the final state (altitude)
Let's plug in the given values into the equation and solve for V₂.
P₁ = 756 torr
V₁ = 0.295 m³
T₁ = 18 °C converted to Kelvin (T₁ = 18 + 273) = 291 K
P₂ = 0.14 atm
V₂ = volume at the final state (to be determined)
T₂ = 48 °C converted to Kelvin (T₂ = 48 + 273) = 321 K
Now we can substitute these values into the equation:
(756 torr * 0.295 m³) / 291 K = (0.14 atm * V₂) / 321 K
To simplify the equation, convert torr to atm:
1 atm = 760 torr
756 torr = 756 / 760 atm = 0.99474 atm
Plug in the values and solve for V₂:
(0.99474 atm * 0.295 m³) / 291 K = (0.14 atm * V₂) / 321 K
Simplify the equation:
0.003418 atm m³ / K = 0.000436 atm * V₂ / K
Rearrange the equation to solve for V₂:
0.003418 atm m³ / K * 321 K = 0.000436 atm * V₂
1.098258 atm m³ = 0.000436 atm * V₂
Divide both sides by 0.000436 atm:
V₂ = 1.098258 atm m³ / 0.000436 atm
V₂ ≈ 2520 m³
Therefore, the volume of the weather balloon at an altitude of 10 km will be approximately 2520 m³.