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4. Exactly 4.00L of air at 223K is warmed and expands to 6.0OL of volume. What is the new temperature if the pressure remains constant?

4. Exactly 4.00L of air at 223K is warmed and expands to 6.0OL of volume. What is the new temperature if the pressure remains constant?

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 4:** *Exactly 4.00L of air at 223K is warmed and expands to 6.00L of volume. What is the new temperature if the pressure remains constant?* **Explanation:** This problem involves the application of Charles's Law in thermodynamics, which states that, for a fixed amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin. The relationship can be written as: V1/T1 = V2/T2 where: - V1 = initial volume of the gas - T1 = initial temperature of the gas - V2 = final volume of the gas - T2 = final temperature of the gas Given: - Initial volume (V1) = 4.00 L - Initial temperature (T1) = 223 K - Final volume (V2) = 6.00 L To find the final temperature (T2), we rearrange the formula to solve for T2: T2 = (V2 * T1) / V1 Plugging in the given values: T2 = (6.00 L * 223 K) / 4.00 L = 1338 K / 4 = 334.5 K Therefore, the new temperature, T2, is 334.5 K.
Transcribed Image Text:**Problem 4:** *Exactly 4.00L of air at 223K is warmed and expands to 6.00L of volume. What is the new temperature if the pressure remains constant?* **Explanation:** This problem involves the application of Charles's Law in thermodynamics, which states that, for a fixed amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin. The relationship can be written as: V1/T1 = V2/T2 where: - V1 = initial volume of the gas - T1 = initial temperature of the gas - V2 = final volume of the gas - T2 = final temperature of the gas Given: - Initial volume (V1) = 4.00 L - Initial temperature (T1) = 223 K - Final volume (V2) = 6.00 L To find the final temperature (T2), we rearrange the formula to solve for T2: T2 = (V2 * T1) / V1 Plugging in the given values: T2 = (6.00 L * 223 K) / 4.00 L = 1338 K / 4 = 334.5 K Therefore, the new temperature, T2, is 334.5 K.
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