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Chapter 5: Problem 63

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 is approximately 1.97 L.

Step by step solution

01

Convert temperatures to Kelvin

First, convert the Celsius temperatures into Kelvin: Initial Temperature (T1) = \(23^{\circ} \mathrm{C} = 23 + 273.15 = 296.15\) K Final Temperature (T2) = \(-31^{\circ} \mathrm{C} = -31 + 273.15 = 242.15\) K
02

Convert pressure units to atmospheres

Given the pressure in torr, we need to convert it into atmospheres: Final Pressure (P2) = \(220\) torr \(= \frac{220}{760} = 0.289\) atm
03

Apply the Combined Gas Law

The combined gas law is given as: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We're given P1, V1, T1, P2, and T2 and need to find V2. Rearrange the formula for V2: \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\)
04

Calculate the final volume (V2)

Substitute the given values into the equation: \(V_2 = \frac{(1.00 \mathrm{~atm})(1.00 \mathrm{~L})(242.15 \mathrm{~K})}{(0.289 \mathrm{~atm})(296.15 \mathrm{~K})}\) \(V_2 \approx 2.97\) L
05

Calculate the change in volume

Now that we have the final volume, we can calculate the change in volume: Change in volume = V2 - V1 Change in volume = \(2.97\) L - \(1.00\) L Change in volume \(\approx 1.97\) L The change in volume of the balloon as it ascends is approximately 1.97 L.

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