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Chapter 5: Problem 61

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

Short Answer

Expert verified
The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert both the given temperatures from Celsius to Kelvin by adding 273.15 K to each temperature: Temperature T1: \(23^\circ C + 273.15 K = 296.15 K\) Temperature T2: \(-31^\circ C + 273.15 K = 242.15 K\)
02

Convert pressure to atm

Next, we need to convert the given pressure (P2) from torr to atm. We know that 1 atm = 760 torr, so we can use the conversion factor: Pressure P2: \(\frac{220 \, torr}{760 \, torr/atm} = 0.2895 \, atm\)
03

Apply the combined gas law

Now, we can apply the combined gas law equation to find the final volume (V2) of the balloon: \(\frac{1.00 \, L}{296.15 \, K} \times \frac{242.15 \, K}{0.2895 \, atm} = V2\) Solve for V2: \(V2 = 3.29 \, L\)
04

Find the change in volume

Now that we have the initial volume of the balloon (V1) and its final volume (V2), we can find the change in volume: Change in volume = \(V2 - V1\) Change in volume = \(3.29 \, L - 1.00 \, L\) Change in volume = \(2.29\, L\) The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

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