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Physics / Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 3 / Chapter 17 / Problem 77CP

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics | 3rd Edition | ISBN: 9780321740908 | Authors: Randall D. Knight

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Chapter 17, Problem 77CP

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QUESTION:

Figure CP17.77 shows a thermodynamic process followed by \(120 \mathrm{mg}\) of helium.

a. Determine the pressure (in atm), temperature (in \(\left.{ }^{\circ} \mathrm{C}\right)\), and volume (in \(\mathrm{cm}^{3}\) ) of the gas at points 1,2 , and 3 . Put your results in a table for easy reading.

Equation Transcription:

$120\ mg$

${\ }^{\circ{}}C$

$c{m}^{3}$

Text Transcription:

120 mg

^circC

cm^3

Questions & Answers

QUESTION:

Figure CP17.77 shows a thermodynamic process followed by \(120 \mathrm{mg}\) of helium.

a. Determine the pressure (in atm), temperature (in \(\left.{ }^{\circ} \mathrm{C}\right)\), and volume (in \(\mathrm{cm}^{3}\) ) of the gas at points 1,2 , and 3 . Put your results in a table for easy reading.

Equation Transcription:

$120\ mg$

${\ }^{\circ{}}C$

$c{m}^{3}$

Text Transcription:

120 mg

^circC

cm^3

ANSWER:

Step 1 of 3

Part a

We are required to calculate the pressure, temperature and volume at the given points 1,2 and 3.

At the point 1, the pressure is 3 atm. The volume is 1000 $c{m}^{3}$ .

Pressure $P=3$ atm $=3\times{}1.01\times{}1{0}^{5}$ Pa

Volume $V=1000$ $c{m}^{3}$ = 0.001 ${m}^{3}$

The mass of helium is $m=120$ mg = 120 $\times{}1{0}^{-3}$ g

The molar mass of helium is 4.0 g.

Therefore, the number of moles of helium is $n=120\times{}1{0}^{-3}/4$ = 30 $\times{}1{0}^{-3}$ = 0.03 mole

Therefore, the temperature of helium at the point 1 is,

$T{\ }_{1}=P{\ }_{1}V{\ }_{1}/nR$

$T{\ }_{1}=3.0\times{}1.01\times{}1{0}^{5}\times{}0.001/(0.03\times{}8.314)$ K

$T{\ }_{1}=1215$ K

$T{\ }_{1}=1215-273$ °C

$T{\ }_{1}=942$ °C

Therefore, at the point 1,


3.0 atm	1000 ${cm}^{3}$	942 °C

The system reaches point 2 after an isothermal expansion.

At this point, the volume is $V{\ }_{2}=3000$ ${cm}^{3}$ = 0.003 ${m}^{3}$

The temperature is 942 °C as the process is isothermal.

The process is isothermal and the volume increases by three times. So, the pressure at this point will be reduced by three times. So, the pressure at the point 2 is 1 atm.

Therefore, at the point 2,


1.0 atm	3000 ${cm}^{3}$	942 °C

The system reaches point 3 at constant volume. This is followed by an adiabatic compression to state 1. Helium has $\gamma{}=1.67$ .

Therefore, to calculate the pressure the point 3, we have to consider the following equation,

$P{\ }_{1}V{\ }_{1}{\ }^{\gamma{}}=P{\ }_{3}V{\ }_{3}{\ }^{\gamma{}}$

$3\times{}0.00{{1}^{\ }}^{1.67}=P{\ }_{3}\times{}0.00{3}^{1.67}$

$P{\ }_{3}=3\times{}(\frac{0.001}{0.003}{)}^{1.67}$ atm

$P{\ }_{3}=0.48$ atm

Therefore, the pressure at the point 3 is 0.48 atm.

For an adiabatic process, the temperature of the point 3 can be calculated from the following equation,

$(\frac{T{\ }_{3}}{T{\ }_{1}}{)}^{\gamma{}}=(\frac{P{\ }_{3}}{P{\ }_{1}}{)}^{\gamma{}-1}$

$T{\ }_{3}=T{\ }_{1}\times{}(\frac{P{\ }_{3}}{P{\ }_{1}}{)}^{(\gamma{}-1)/\gamma{}}$

$T{\ }_{3}=1215\times{}(\frac{0.48}{3.0}{)}^{(1.67-1)/1.67}$ K

$T{\ }_{3}=1215\times{}(0.16{)}^{0.40}$ K

$T{\ }_{3}=583$ K

$T{\ }_{3}=583-273$ °C

$T{\ }_{3}=310$ °C

Therefore, the temperature at the point 3 is 310 °C.

Therefore, at the point 3,


0.48 atm	3000 ${cm}^{3}$	310 °C

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