   # A weather balloon is inflated to a volume of 24.5 L at a pressure of 758 mm Hg and a temperature of 24 .0°C . The balloon is released and rises to an altitude of approximately 25,000 ft at which the pressure is 382 mm Hg and the temperature is -14 .0°C . Calculate the volume of the balloon at this altitude (assuming the balloon freely expands). ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

#### Solutions

Chapter
Section ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 11, Problem 52E
Textbook Problem
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## A weather balloon is inflated to a volume of 24.5 L at a pressure of 758 mm Hg and a temperature of 24 .0°C . The balloon is released and rises to an altitude of approximately 25,000 ft at which the pressure is 382 mm Hg and the temperature is -14 .0°C . Calculate the volume of the balloon at this altitude (assuming the balloon freely expands).

Interpretation Introduction

Interpretation:

The volume of the balloon at given altitude is to be calculated.

Concept Introduction:

Combined gas law, also called general gas equation, combines Boyle’s Law and Charles’s Law. It states that the ratio of the pressure-volume product and the absolute temperature of a gas is equal to a constant.

P1V1T1=P2V2T2

Here, P1 and P2 are the initial and final pressures of the ideal gas, V1 and V2 are the initial and the final volumes of the gas, and T1, T2 are the initial and final temperatures of the ideal gas.

Converting degree Celsius to Kelvin by adding 273K, the conversion factor is as:

T=°C+ 273K

### Explanation of Solution

Given information: Initial Pressure is 758mmHg.

Initial Temperature is 24°C.

Final Temperature is 14°C.

Initial Volume is 24.5L.

Final pressure is 382mmHg.

According to combined gas law,

P1V1T1=P2V2T2

Now, converting the temperature from degree Celsius to Kelvin by using a conversion factor as:

T1=24°C=24+273K=297K

T2=14°C=14+273K=259K

Substituting V1=24

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