a weather balloon is filled with helium to a volume of 250 l at sea level, where pressure is 760 mm hg. assuming constant temperature, what will be the volume when the balloon raises to a height of 25 km, where the pressure is only 10 mm hg
Why did the weather balloon decide to take a trip to the sky? Because it heard that life up there is just "uplifting"!
Alright, now let's calculate the volume of the weather balloon as it rises to a height of 25 km. We can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
So, we have:
Initial volume (V1) = 250 L
Initial pressure (P1) = 760 mmHg
Final pressure (P2) = 10 mmHg
According to Boyle's Law: P1 * V1 = P2 * V2
Substituting the values:
760 mmHg * 250 L = 10 mmHg * V2
Now, let's solve for V2, the final volume:
V2 = (760 mmHg * 250 L) / 10 mmHg
V2 = 19000 L
So, when the weather balloon reaches a height of 25 km, the volume of the balloon will be approximately 19,000 liters. That's one lofty balloon!
To determine the final volume of the weather balloon at a height of 25 km, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, assuming constant temperature.
Boyle's Law equation: P1 * V1 = P2 * V2
Where:
P1 = Initial pressure (sea level pressure)
V1 = Initial volume (250 L)
P2 = Final pressure (pressure at 25 km height)
V2 = Final volume (the required volume we want to find)
Let's now substitute the given values into the equation:
P1 = 760 mmHg
V1 = 250 L
P2 = 10 mmHg
760 mmHg * 250 L = 10 mmHg * V2
Now, let's solve for V2:
(760 mmHg * 250 L) / 10 mmHg = V2
190,000 / 10 = V2
19,000 L = V2
Therefore, when the balloon raises to a height of 25 km where the pressure is 10 mmHg, the volume of the balloon will be 19,000 L.
To calculate the volume of the weather balloon at a different pressure, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the temperature is assumed to be constant, we can rewrite the equation as:
P₁V₁ = P₂V₂
Where:
P₁ = initial pressure (sea level)
V₁ = initial volume (sea level)
P₂ = final pressure (25 km altitude)
V₂ = final volume (unknown)
Now, let's plug in the given values:
P₁ = 760 mmHg
V₁ = 250 L
P₂ = 10 mmHg
V₂ = ?
We need to convert the pressures to the same unit. In this case, let's convert mmHg to atm:
1 atm = 760 mmHg
Now we can rewrite P₁ and P₂:
P₁ = 760 mmHg / 760 mmHg = 1 atm
P₂ = 10 mmHg / 760 mmHg ≈ 0.0132 atm
Now let's solve for V₂:
P₁V₁ = P₂V₂
1 atm × 250 L = 0.0132 atm × V₂
250 L = 0.0132 atm × V₂
To isolate V₂, divide both sides of the equation by 0.0132 atm:
V₂ = (250 L) / (0.0132 atm)
Using a calculator, we can find:
V₂ ≈ 18939 L
Therefore, the volume of the weather balloon at a height of 25 km will be approximately 18939 liters.
You really need to find the caps key on your keyboard and use it. L is liters, not l, Hg is mercury, not hg, and sentences start with a cap letter.
Use P1V1 = P2V2
Post your work if you get stuck.