**Problem Statement:**A balloon is filled to a volume of \( 6.65 \times 10^2 \) mL at a temperature of 15.0°C. The balloon is then cooled at constant pressure to a temperature of \( 1.07 \times 10^2 \) K. What is the final volume of the balloon?**Volume =** \_\_\_\_ mL---In this problem, the relationship between the initial and final volumes and temperatures of a gas is explored at constant pressure, which applies Charles's Law. This law can be expressed as:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Where:- \( V_1 \) is the initial volume.- \( T_1 \) is the initial temperature in Kelvin.- \( V_2 \) is the final volume.- \( T_2 \) is the final temperature in Kelvin. To solve the problem:1. Convert the initial temperature from Celsius to Kelvin: \[ T_1 = 15.0°C + 273.15 = 288.15 \, K \]2. Set up the equation using Charles's Law to find \( V_2 \): \[ \frac{6.65 \times 10^2 \, \text{mL}}{288.15 \, K} = \frac{V_2}{1.07 \times 10^2 \, K} \]3. Solve for \( V_2 \): \[ V_2 = \frac{(6.65 \times 10^2 \, \text{mL})(1.07 \times 10^2 \, K)}{288.15 \, K} \]Calculate to find the final volume of the balloon.

Answered: Image /qna-images/answer/f5db1ed7-4818-4b77-b0e9-c99270591d00.jpg

Á balloon is filled to a volume of 6.65 × 10´ mL at a temperature of 15.0°C. The balloon is then cooled at constant pressure to a temperature of 1.07 × 10“ K. What is the final volume of the balloon? Volume = mL

Á balloon is filled to a volume of 6.65 × 10´ mL at a temperature of 15.0°C. The balloon is then cooled at constant pressure to a temperature of 1.07 × 10“ K. What is the final volume of the balloon? Volume = mL

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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