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1. The pressure on 2.50L of an enclosed gas changes from 105kPa to 40.5kPa. If the temperature remains unchanged what is the new volume?

1. The pressure on 2.50L of an enclosed gas changes from 105kPa to 40.5kPa. If the temperature remains unchanged what is the new volume?

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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**Boyle's Law Application** **Problem Statement:** The pressure on 2.50L of an enclosed gas changes from 105 kPa to 40.5 kPa. If the temperature remains unchanged, what is the new volume? **Solution:** To solve this problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature (isothermal conditions), the volume of the gas is inversely proportional to its pressure. Mathematically, this is represented as: \[ P_1 \times V_1 = P_2 \times V_2 \] where: - \( P_1 \) = initial pressure = 105 kPa - \( V_1 \) = initial volume = 2.50 L - \( P_2 \) = final pressure = 40.5 kPa - \( V_2 \) = final volume (which we need to find) Rearranging the formula to solve for \( V_2 \): \[ V_2 = \frac{P_1 \times V_1}{P_2} \] **Calculation:** \[ V_2 = \frac{105\ \text{kPa} \times 2.50\ \text{L}}{40.5\ \text{kPa}} \] \[ V_2 = \frac{262.5}{40.5} \] \[ V_2 \approx 6.48\ \text{L} \] **Conclusion:** The new volume of the gas when the pressure changes to 40.5 kPa, while keeping the temperature constant, is approximately 6.48 L.
Transcribed Image Text:**Boyle's Law Application** **Problem Statement:** The pressure on 2.50L of an enclosed gas changes from 105 kPa to 40.5 kPa. If the temperature remains unchanged, what is the new volume? **Solution:** To solve this problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature (isothermal conditions), the volume of the gas is inversely proportional to its pressure. Mathematically, this is represented as: \[ P_1 \times V_1 = P_2 \times V_2 \] where: - \( P_1 \) = initial pressure = 105 kPa - \( V_1 \) = initial volume = 2.50 L - \( P_2 \) = final pressure = 40.5 kPa - \( V_2 \) = final volume (which we need to find) Rearranging the formula to solve for \( V_2 \): \[ V_2 = \frac{P_1 \times V_1}{P_2} \] **Calculation:** \[ V_2 = \frac{105\ \text{kPa} \times 2.50\ \text{L}}{40.5\ \text{kPa}} \] \[ V_2 = \frac{262.5}{40.5} \] \[ V_2 \approx 6.48\ \text{L} \] **Conclusion:** The new volume of the gas when the pressure changes to 40.5 kPa, while keeping the temperature constant, is approximately 6.48 L.
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