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Date Posted: 6/25/2008 7:14:23 PM  Status: Live
General Chem 152. heat energy.rate lifesaver
Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
Calculate the energy required to heat 2.00 kg of ethane gas (C2H6) from 25.0°C to 80.0°C first under conditions of constant volume and then at a constant pressure of 2.00 atm. Calculate ΔE, ΔH, and w for these processes. (See Table 9.1 for relevant data.)


Table 9.1

constant V constant P
q WebAssign will check your answer for the correct number of significant figures. kJ WebAssign will check your answer for the correct number of significant figures. kJ
ΔE WebAssign will check your answer for the correct number of significant figures. kJ WebAssign will check your answer for the correct number of significant figures. kJ
ΔH WebAssign will check your answer for the correct number of significant figures. kJ WebAssign will check your answer for the correct number of significant figures. kJ
w WebAssign will check your answer for the correct number of significant figures. kJ WebAssign will check your answer for the correct number of significant figures. kJ

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Sage
Karma Points: 4,187
Date Posted: 6/26/2008 12:48:03 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
m= 2kg, T1= 250C= 25+273=298K
T2= 800C=80+273=353K, P= 2atm
molar mass of ethane = 30g/mol
moles,n= 2000g/30g/mol= 66.67mol
constant volume process:
for a constant volume process, the equations are
Q=n ∫ CvdT= nCv(T2-T1) =66.67mol*44.60J/mol.K*(353-298)K= 163.53KJ
ΔE=n ∫ CvdT =66.67mol*44.60J/mol.K*(353-298)K=163.53KJ
ΔH= n ∫ CpdT = 66.67mol*52.92J/mol.K*(353-298)K= 194.05KJ 
W=- PΔV= 0
constant pressure process:
Q=n ∫ CpdT = 66.67mol*52.92J/mol.K*(353-298)K= 194.05KJ 
 ΔE=n ∫ CvdT =66.67mol*44.60J/mol.K*(353-298)K=163.53KJ
ΔH= n ∫ CpdT = 66.67mol*52.92J/mol.K*(353-298)K= 194.05KJ 
W= -PdV= -nR(T2-T1) = -(66.67mol)* (8.32J/mol.K)*(353-298)K= 30.51KJ
gvirk18's Comment:
thank you



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