Question

1. Hydrogen gas is used in weather balloons because it is less expensive than helium. Assume that 5.57 g of H2 is used to fill a weather balloon to an initial volume of 67 L at 1.04 atm. If the balloon rises to an altitude where the pressure is 0.047 atm, what is the new volume in Liters? Assume that the temperature remains constant. 2. A weather balloon having an initial volume of 67 L and 1.04 atm is weighted down and dropped into the ocean. Every 10.1 m, water pressure increases by 1 atm. What is the volume of the balloon in Liters after it sinks 64.0 m below the water's surface? Assume that the temperature remains constant. 3. A 2 L soda bottle is pressurized with 4.5 atm of CO2 at 298 K. If the temperature is increased to 317 K, what is the pressure in atm of the CO2.

          1. Hydrogen gas is used in weather balloons because it is less
expensive than helium. Assume that 5.57 g of H2 is used to fill a
weather balloon to an initial volume of 67 L at 1.04 atm. If the
balloon rises to an altitude where the pressure is 0.047 atm, what
is the new volume in Liters? Assume that the temperature remains
constant.
2. A weather balloon having an initial volume of 67 L and
1.04 atm is weighted down and dropped into the ocean. Every 10.1 m,
water pressure increases by 1 atm. What is the volume of the
balloon in Liters after it sinks 64.0 m below the water's surface?
Assume that the temperature remains constant.
3. A 2 L soda bottle is pressurized with 4.5 atm of
CO2 at 298 K. If the temperature is increased to
317 K, what is the pressure in atm of the CO2.

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Added by Dawn S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ashley Brady

05/15/2022

Step 1

04 atm to 0.047 atm. Given: \(P_1 = 1.04 \, \text{atm}\) \(V_1 = 67 \, \text{L}\) \(P_2 = 0.047 \, \text{atm}\) Using the formula: \(1.04 \, \text{atm} \times 67 \, \text{L} = 0.047 \, \text{atm} \times V_2\) Solving for \(V_2\): \(V_2 = \frac{1.04 \times Show more…

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1. Hydrogen gas is used in weather balloons because it is less expensive than helium. Assume that 5.57 g of H2 is used to fill a weather balloon to an initial volume of 67 L at 1.04 atm. If the balloon rises to an altitude where the pressure is 0.047 atm, what is the new volume in Liters? Assume that the temperature remains constant. 2. A weather balloon having an initial volume of 67 L and 1.04 atm is weighted down and dropped into the ocean. Every 10.1 m, water pressure increases by 1 atm. What is the volume of the balloon in Liters after it sinks 64.0 m below the water's surface? Assume that the temperature remains constant. 3. A 2 L soda bottle is pressurized with 4.5 atm of CO2 at 298 K. If the temperature is increased to 317 K, what is the pressure in atm of the CO2.

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Transcript

00:01 Okay, so for this problem, there are actually three different calculations that need to be done.

00:04 So the first and second, we're dealing with the same balloon.

00:08 The third, we're dealing with a different bottle.

00:10 So the first one, we have, well, the first and second one, we are going to use the law that relates pressure and volume.

00:18 So that is p1b1 equals p2 v2.

00:23 We're given the initial pressure and volume conditions.

00:27 We're given the final pressure condition.

00:28 So we're just going to plug in and solve.

00:30 So it's going to be 67 liters times 1 .04 atmosphere equals 0 .047 atmosphere times volume.

00:45 So we're going to divide both sides by 0 .047.

00:56 So when we do that, 67 times 1 .04 divided by 0 .047.

01:03 We end up with a new volume that is much larger.

01:07 So the final volume equals 1 ,483 liters.

01:17 So that is for number one.

01:19 Number two, then, we're going to put the balloon down in the ocean.

01:22 So we're going to put it 64 meters under the ocean.

01:26 And we're giving this conversion that every 10 .1 meters, we increase the pressure by an atmosphere.

01:32 So if we go 64 times 1 divided by 10 .1, we end up with a new pressure of 6 .34 atmospheres.

01:50 So then we're going to do the same thing as we did up above using the p1b1 equals p2b2...

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