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A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a problem a. missing variable b. change in conditions C. 22.414 L/1 mol d. density = molar mass/22.414 L e. partial P second: we need to solve for using the equation f. P g. P1 h. P2 i. V j. V1 (not V) k. V2 (not V) I. n m. n1 (not n) n. n2 (not n) o. T p. T1 (not T) q. T2 (not T) r. R= 8.205 x 10-2 L atm/K mol s. 1 atm = 760 torr t. Ptotal = P1 + P2 + .... u. PV = nRT v. P¡V1 P2V2 ------ ------ finally: The answer is v. 4.000 L w. 2.250 L x. 0.7500 L у. 1.333 L

A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a problem a. missing variable b. change in conditions C. 22.414 L/1 mol d. density = molar mass/22.414 L e. partial P second: we need to solve for using the equation f. P g. P1 h. P2 i. V j. V1 (not V) k. V2 (not V) I. n m. n1 (not n) n. n2 (not n) o. T p. T1 (not T) q. T2 (not T) r. R= 8.205 x 10-2 L atm/K mol s. 1 atm = 760 torr t. Ptotal = P1 + P2 + .... u. PV = nRT v. P¡V1 P2V2 ------ ------ finally: The answer is v. 4.000 L w. 2.250 L x. 0.7500 L у. 1.333 L

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: A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon? ### Step-by-Step Solution: #### First: Identify the type of problem: **This is a:** - [ ] a. missing variable - [x] b. change in conditions - [ ] c. 22.414 L/1 mol - [ ] d. density = molar mass/22.414 L - [ ] e. partial P #### Second: Determine what needs solving: **We need to solve for:** - [x] k. V₂ (not V) **Using the equation:** - v. \[ \frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2} \] ### Information and Formulas: - **Symbols and values:** - f. P - g. P₁ - h. P₂ - i. V - j. V₁ (not V) - l. n - m. n₁ (not n) - n. n₂ (not n) - o. T - p. T₁ (not T) - q. T₂ (not T) - r. R = 8.205 x 10⁻² L atm/K mol - s. 1 atm = 760 torr - **Total Pressure:** - t. \( P_{\text{total}} = P_1 + P_2 + \ldots \) - **Ideal Gas Law:** - u. PV = nRT ### Finally: Select the correct answer for the new volume: - v. \[4.000 L\] - w. \[2.250 L\] - x. \[0.7500 L\] - y. **1.333 L**
Transcribed Image Text:### Problem: A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon? ### Step-by-Step Solution: #### First: Identify the type of problem: **This is a:** - [ ] a. missing variable - [x] b. change in conditions - [ ] c. 22.414 L/1 mol - [ ] d. density = molar mass/22.414 L - [ ] e. partial P #### Second: Determine what needs solving: **We need to solve for:** - [x] k. V₂ (not V) **Using the equation:** - v. \[ \frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2} \] ### Information and Formulas: - **Symbols and values:** - f. P - g. P₁ - h. P₂ - i. V - j. V₁ (not V) - l. n - m. n₁ (not n) - n. n₂ (not n) - o. T - p. T₁ (not T) - q. T₂ (not T) - r. R = 8.205 x 10⁻² L atm/K mol - s. 1 atm = 760 torr - **Total Pressure:** - t. \( P_{\text{total}} = P_1 + P_2 + \ldots \) - **Ideal Gas Law:** - u. PV = nRT ### Finally: Select the correct answer for the new volume: - v. \[4.000 L\] - w. \[2.250 L\] - x. \[0.7500 L\] - y. **1.333 L**
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A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a a. missing variable d. density = molar mass/22.414 L second: we need to solve for o. T 760 torr f. P g. P₁ h. P₂ i. V j. V₁ (not V) n. n₂ (not n) p. T₁ (not T) t. Ptotal = P₁ + P₂ + u. PV = nRT v. P₁V₁ P2V₂ n₁T₁ n₂T2 finally: The answer is w. 0.7500 L problem b. change in conditions x. 1.333 L q. T₂ (not T) e. partial P y. 2.250 L c. 22.414 L/1 mol using the equation k. V₂ (not V) z. 4.000 L I. n r. R = 8.205 x 10-² L atm/K mol m. n₁ (not n) s. 1 atm =
Transcribed Image Text:A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a a. missing variable d. density = molar mass/22.414 L second: we need to solve for o. T 760 torr f. P g. P₁ h. P₂ i. V j. V₁ (not V) n. n₂ (not n) p. T₁ (not T) t. Ptotal = P₁ + P₂ + u. PV = nRT v. P₁V₁ P2V₂ n₁T₁ n₂T2 finally: The answer is w. 0.7500 L problem b. change in conditions x. 1.333 L q. T₂ (not T) e. partial P y. 2.250 L c. 22.414 L/1 mol using the equation k. V₂ (not V) z. 4.000 L I. n r. R = 8.205 x 10-² L atm/K mol m. n₁ (not n) s. 1 atm =
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