**Question:**A sample of an ideal gas has a volume of 3.40 L at 14.60 °C and 1.90 atm. What is the volume of the gas at 23.80 °C and 0.990 atm?**Solution:**_V =_ [Box for answer] _L_**Explanation:**To determine the final volume of the gas under new conditions, you should use the Combined Gas Law:\[\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\]Where:- \( P_1 = 1.90 \, \text{atm} \)- \( V_1 = 3.40 \, \text{L} \)- \( T_1 = 14.60^\circ C + 273.15 = 287.75 \, \text{K} \)- \( P_2 = 0.990 \, \text{atm} \)- \( T_2 = 23.80^\circ C + 273.15 = 296.95 \, \text{K} \)Rearrange the equation to solve for the new volume \( V_2 \):\[V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}\]Plug in the known values and calculate \( V_2 \).

Given data is as follows: The initial volume of sample V1= 3.40 L The initial temperature of the sample T1= 14.60 °C=14.60+273.17 =287.75 K The initial pressure of the sample P1= 1.90 atm The final temperature of the sample T2= 23.80 °C=23.80+273.15=296.95 K The final pressure of the sample P2= 0.990 atm The final volume of the sample V2= ? According to the ideal gas equation: P1V1T1=P2V2T21.90 atm×3.40 L287.75 K=0.990 atm×V2296.95 KV2=1.90 atm×3.40 L×296.95 K287.75 K×0.990 atm=1918.297284.8725 L=6.73 L Hence, the final volume of the gas sample is 6.73 L.

Answered: A sample of an ideal gas has a volume…

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A sample of an ideal gas has a volume of 3.40 L at 14.60 °C and 1.90 atm. What is the volume of the gas at 23.80 °C and 0.990 atm? V =

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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