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8.32. A gas sample has a temperature of 22°C with an unknown volume. The same gas volume of 456 mL when the temperature is 86 C, with no change in the pressure or amount of gas. What was the initial volume, in milliliters, of the gas? has

8.32. A gas sample has a temperature of 22°C with an unknown volume. The same gas volume of 456 mL when the temperature is 86 C, with no change in the pressure or amount of gas. What was the initial volume, in milliliters, of the gas? has

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Problem 8.32:** A gas sample has a temperature of 22°C with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86°C, with no change in the pressure or amount of gas. What was the initial volume, in milliliters, of the gas? --- This problem can be solved using Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure and the amount of gas remain constant. The formula is: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where: - \( V_1 \) is the initial volume - \( T_1 \) is the initial temperature in Kelvin - \( V_2 \) is the final volume (456 mL) - \( T_2 \) is the final temperature in Kelvin Convert the given temperatures from Celsius to Kelvin: - \( T_1 = 22°C + 273.15 = 295.15 \, K \) - \( T_2 = 86°C + 273.15 = 359.15 \, K \) Using Charles’s Law, solve for the initial volume (\( V_1 \)): \[ V_1 = \frac{V_2 \times T_1}{T_2} \]
Transcribed Image Text:**Problem 8.32:** A gas sample has a temperature of 22°C with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86°C, with no change in the pressure or amount of gas. What was the initial volume, in milliliters, of the gas? --- This problem can be solved using Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure and the amount of gas remain constant. The formula is: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] Where: - \( V_1 \) is the initial volume - \( T_1 \) is the initial temperature in Kelvin - \( V_2 \) is the final volume (456 mL) - \( T_2 \) is the final temperature in Kelvin Convert the given temperatures from Celsius to Kelvin: - \( T_1 = 22°C + 273.15 = 295.15 \, K \) - \( T_2 = 86°C + 273.15 = 359.15 \, K \) Using Charles’s Law, solve for the initial volume (\( V_1 \)): \[ V_1 = \frac{V_2 \times T_1}{T_2} \]
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