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Author Question: For a particular process that is carried out at constant pressure, q = 140 kJ and w = -30 kJ. Therefore, (Read 106 times)

mspears3

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on: Feb 15, 2020

Question 1

Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 35 L against a constant external pressure of 1.5 atm. 1 L ∙ atm = 101 J
◦ -3.0 kJ
◦ +5.3 kJ
◦ -5.3 kJ
◦ +3.0 kJ

Question 2

For a particular process that is carried out at constant pressure, q = 140 kJ and w = -30 kJ. Therefore,
◦ ΔE = 110 kJ and ΔH = 140 kJ.
◦ ΔE = 140 kJ and ΔH = 110 kJ.
◦ ΔE = 170 kJ and ΔH = 140 kJ.
◦ ΔE = 140 kJ and ΔH = 170 kJ.
Chemistry ¦ McMurry, Fay, Robinson ¦ 7th Edition
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Marked as best answer by mspears3 on Feb 15, 2020

ryansturges

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Reply #1 on: Feb 15, 2020
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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mspears3

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Reply 2 on: Feb 15, 2020
Wow, this really help

dantucker

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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