a balloon with a volume of 1.375 L is released from earths surface at sea level. what will the balloon occuy at an altitude of 20.0km, where the pressure is 10 kpa?

pleasee help!!
show work please

P1V1 = P2V2

P1 = 101.325 kPa (sea level pressure = 1 atm = 101.325 kPa).
V1 = 1.375 L
P2 = 10 kPa
Solve for V2.
Check my work.

would the answer be in kpa??

Yes. If P1 is in kPa, then P2 will be in kPa. You could have used 1 atmosphere for P1 and converted 10 kPa to atmospheres.

10 kPa x (1 atm/101.325 kPa) = ?? atm.
OR you can calculate from the original to find P2 in kPa, then divide by 101.325 to convert that way. You should get the same answer either way.

What is the answer

The answer would be 13.93 Liters you would not put it into kPa.

Well, that balloon must be quite the adventurer, floating all the way up to 20.0km! Let's see what it gets up to.

To solve this, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. The formula is:

P1 x V1 = P2 x V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Now, we know the initial volume is 1.375 L, and the initial pressure is the sea level pressure on Earth. Typically, that's around 101.3 kPa, but the question doesn't mention anything specific, so we'll stick with that.

So, P1 = 101.3 kPa and V1 = 1.375 L.

We're also given that the final pressure is 10 kPa and the question asks for the final volume, which we'll call V2.

Plugging these values into the formula, we get:

101.3 kPa x 1.375 L = 10 kPa x V2

Now, let's solve for V2:

V2 = (101.3 kPa x 1.375 L) / 10 kPa

V2 = 139.4375 L

So, at an altitude of 20.0km where the pressure is 10 kPa, our adventurous balloon will occupy approximately 139.4375 liters of volume.

Hope that helps, and may your balloon have a safe and hilarious journey up there!

To solve this problem, we can use the ideal gas law equation:

PV = nRT,

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's find the number of moles, n, of the gas in the balloon at sea level. We'll assume that the gas in the balloon behaves ideally and that the conditions at sea level are at room temperature (around 298 K):

n = (PV) / (RT).

At sea level, the pressure, P, is 101.3 kPa (standard atmospheric pressure) and the volume, V, is given as 1.375 L. The ideal gas constant, R, is 0.0821 L·atm/(mol·K).

Calculating n:
n = (101.3 kPa * 1.375 L) / (0.0821 L·atm/(mol·K) * 298 K)
= 4.858 moles.

Now, let's consider the new conditions at an altitude of 20.0 km, where the pressure is 10 kPa. We need to find the new volume, V', that the balloon occupies at this altitude.

Using the ideal gas law equation again, but with the new pressure, P', and the same number of moles, n:

PV = nRT

V' * P' = nRT.

Rearranging the equation to solve for V':

V' = (P' * V) / (n * R).

Substituting the values:

V' = (10 kPa * 1.375 L) / (4.858 moles * 0.0821 L·atm/(mol·K))
≈ 3.790 L.

Therefore, the balloon will occupy approximately 3.790 L at an altitude of 20.0 km, where the pressure is 10 kPa.