When a gas is taken from a to c along the curved path in the figure (Figure 1) , the work done by the gas is W = -40 J and the heat added to the gas i

Nannie Mack

Nannie Mack

Answered question

2021-05-04

 
When a gas is taken from a to c along the curved path in the figure (Figure 1) , the work done by the gas is W = -40 J and the heat added to the gas is Q = -140 J . Along path abc, the work done by the gas is W = -50 J . (That is, 50 J of work is done on the gas.) 
I keep on missing Part D. The answer for part D is not -150,150,-155,108,105( was close but it said not quite check calculations) 
Part A 
What is Q for path abc? 
Express your answer to two significant figures and include the appropriate units. 
Part B 
f Pc=1/2Pb, what is W for path cda? 
Express your answer to two significant figures and include the appropriate units. 
Part C 
What is Q for path cda? 
Express your answer to two significant figures and include the appropriate units. 
Part D 
What is Ua?Uc? 
Express your answer to two significant figures and include the appropriate units. 
Part E 
If Ud?Uc=42J, what is Q for path da? 
Express your answer to two significant figures and include the appropriate units.

Answer & Explanation

Anonym

Anonym

Skilled2021-05-06Added 108 answers

Part D:
For the variation of the internal energy in the path cda we have that
Uca=Ucd+Uda
Uca=(QcdWcd)+(QdaWda)
the path da is a constant volume, so the work in this path is zero
Uca=(QcdWcd)+(Qda0)
Uca=Qcd+QdaWcd
Uca=QcaWcd
Here we have, form the parts B and C, the Qca=130J and the Wcd=25 J, so
Uca=105J
An Answer that you already said is incorrect. But there is a "trick", the significant figures. You need an answer with two significant figures but that has three. So, rounding, the answer is
Uca=110J
Part E:
Since the path dc is at constant volume
Ucd=Qcd
And like we have Qcda
Qcda=Qcd+Qda
Qda=QcdaQcd
Qda=130J42J
Qda=88J

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