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Chapter 5: Problem 151

A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is $25^{\circ} \mathrm{C}$ and the barometric pressure is 730 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to 6000 ft?

Short Answer

Expert verified
The change in volume of the weather balloon as it ascends from sea level to 6000 ft is 167.32 L.

Step by step solution

01

Convert temperatures into Kelvin

To work with the gas laws, we should convert the given temperatures in Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperatures: T1 = \(25^{\circ} \mathrm{C} + 273.15 = 298.15\ \mathrm{K}\) T2 = \(15^{\circ} \mathrm{C} + 273.15 = 288.15\ \mathrm{K}\)
02

Convert pressures into atmospheres

To simplify our calculations, we should convert the given pressures from torr to atmospheres (atm). To do this, we divide the torr values by 760: P1 = 730 torr / 760 = 0.9605 atm P2 = 605 torr / 760 = 0.7961 atm
03

Setup the combined gas law equation

Now, we have T1, T2, P1, and P2 in the appropriate units. We can now use the combined gas law equation, remembering that V1 = 855 L: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)
04

Solve the equation for the final volume (V2)

Now we need to solve for V2. We can do this by rearranging the combined gas law equation and plugging in the known values: \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) Plug in all values: \(V_2 = \frac{(0.9605\ \mathrm{atm})(855\ \mathrm{L})(288.15\ \mathrm{K})}{(0.7961\ \mathrm{atm})(298.15\ \mathrm{K})}\)
05

Calculate the final volume (V2)

Perform the above calculation to find V2: \(V_2 = 1022.32\ \mathrm{L}\)
06

Calculate the change in volume

The change in volume is the final volume minus the initial volume: ΔV = V2 - V1 = 1022.32 L - 855 L = 167.32 L As the balloon ascends from sea level to 6000 ft, its volume increases by 167.32 L.

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