Nannie Mack

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.

Anonym

Skilled2021-05-06Added 108 answers

Part D:

For the variation of the internal energy in the path cda we have that

$\mathrm{\u25b3}{U}_{ca}=\mathrm{\u25b3}{U}_{cd}+\mathrm{\u25b3}{U}_{da}$

$\mathrm{\u25b3}{U}_{ca}=({Q}_{cd}-{W}_{cd})+({Q}_{da}-{W}_{da})$

the path da is a constant volume, so the work in this path is zero

$\mathrm{\u25b3}{U}_{ca}=({Q}_{cd}-{W}_{cd})+({Q}_{da}-0)$

$\mathrm{\u25b3}{U}_{ca}={Q}_{cd}+{Q}_{da}-{W}_{cd}$

$\mathrm{\u25b3}{U}_{ca}={Q}_{ca}-{W}_{cd}$

Here we have, form the parts B and C, the${Q}_{ca}=130J$ and the ${W}_{cd}=25\text{}J$ , so

$\mathrm{\u25b3}{U}_{ca}=105J$

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

$\mathrm{\u25b3}{U}_{ca}=110J$

Part E:

Since the path dc is at constant volume

$\mathrm{\u25b3}{U}_{cd}={Q}_{cd}$

And like we have$Q}_{cda$

$Q}_{cda}={Q}_{cd}+{Q}_{da$

$Q}_{da}={Q}_{cda}-{Q}_{cd$

${Q}_{da}=130J-42J$

${Q}_{da}=88J$

For the variation of the internal energy in the path cda we have that

the path da is a constant volume, so the work in this path is zero

Here we have, form the parts B and C, the

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

Part E:

Since the path dc is at constant volume

And like we have

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