**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/T2where:- V1 = initial volume of the gas- T1 = initial temperature of the gas- V2 = final volume of the gas- T2 = final temperature of the gasGiven:- Initial volume (V1) = 4.00 L- Initial temperature (T1) = 223 K- Final volume (V2) = 6.00 LTo find the final temperature (T2), we rearrange the formula to solve for T2:T2 = (V2 * T1) / V1Plugging in the given values:T2 = (6.00 L * 223 K) / 4.00 L = 1338 K / 4 = 334.5 KTherefore, the new temperature, T2, is 334.5 K.

Ideal gas follows the kinetic gas theory assumptions and laws that explain the nature of an ideal…

Answered: 4. Exactly 4.00L of air at 223K is…

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

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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Question

**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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4. Exactly 4.0OL of air at 223K is warmed and expands to 6.00L of volume. What is the new temperature if the pressure remains constant?