Question

Part A:Refer to diagram 2. A flask contains 85.2 moles of a monatomic ideal gas at pressure 6.9 atm and volume 13.4 liters (point A on the graph. Now, the gas undergoes a cycle of three steps: - First there is an isothermal expansion to pressure 3.65 atm (point B on the graph). - Next, there is an isochoric process in which the pressure is raised to P1 (point C on the graph). - Finally, there is an isobaric compression back to the original state (point A on the graph). Find the total work done, in kJ, over the entire cycle (from A back to A). A positive value means work was done by the gas; a negative value means work is done on the gas. Why is this a poor sort of engine?

Part B:

Refer to diagram 2.

A flask contains 60.4 moles of a monatomic ideal gas at pressure 9.21 atm and volume 15.2 liters (point A on the graph. Now, the gas undergoes a cycle of three steps:

- First there is an isothermal expansion to pressure 4.2 atm (point B on the graph).

- Next, there is an isochoric process in which the pressure is raised to P₁ (point C on the graph).

- Finally, there is an isobaric compression back to the original state (point A on the graph).

Find the heat added or removed, in KJ, over the isobaric process (from C to A). A positive answer means heat is added; a negative answer means heat was removed.  

HW Diagra X s Microsoft X Take Test: X C Get Homex C A Flask Holx C Get Home X C Reter To Dix a Atm to Pa X O Significant X + х O File C:/Users/andre/Downloads/physics%20121%20diagramsM2012.pdf 0 Microsoft Word - physics 121 diagrams 12 1/1 DIAGRAMS FOR ASSIGNMENT 12 Diagram 1 Diagram 2 F P: 1 71 V: Type here to search O 9 D (9) ENG 5:13 PM 5/15/2020

Answer 1

Diagram: 9 A C Pi B P Part -A No. of moles (monoatomic gas) = 85.2 Pressure Pi 6.9 atm Vj = 13.4 lit Step-I Isothermal expansion to pressure 3.65 atm AB Since work done in isothermal exponsion Wi= ART In ( Step-II Isochoric process. BC Since volume is constant in this process. Hence worke doneo Wzro

Step- I sobranc Compression CA Since in compression, work is done on the gas. Hence W: -Pop (Vg-Vo) + 3 atm f Now at point A- PV = nRT P= 6.9 atm = 6.98 105 N/m2 v= 13.4 lit = 13.4 x 103 m3 85.2 R = 8.31 J/molo-K niR T- 6.9x105 x 13.48153 NRT = 92.46 x 10² Ws 92.46 8102 ln 6.9 3.65 Wi- 98,46 x 10? x 0.6366 Ws = 5.88 KJ W3 - -P2 (V-V.)

She at Point B - Pi Vi= Pa Va V = Pus P2 Pi = = 6.9 atm P2 = 3.65 atm Vj = 13.4 le Va= legges 6.9x13.4 3.65 = 25.33 lit Vo 25.33 XI 03 m3 Work dine in process TA Wz: - Pj (V-V.) - 6.9 x 105 x 11.93 x 103 Wz = 8.23 KI Hence total werk dine- W = WitwatW3 W= 5.88 to- 8.23 W = -2.35 KI This is a poor sort of Engine because total work done is negative that means we have to do work on gas.

Part B No. of moles (nl= 60.4 Ps = 9.21 atm = 9, 2dxtos N/m2 V = 15.2 lt = 15.2 X10 m3 First Step Isothermal expansion AB P₂ = 4.2 atm W, ART Dn 16 MRT = PV (cat point A) MRT = 9.2J X 105 x 15-28103 MRT = 139.99 X10? W, = 139.99 %10². In 9.9] 4.2 Wi= 10.97 KI Step I geocharic process Btc sino olime is constant hence We =0

Thind Step :- Isobaric process CHA (compression) at P = 9.21 XIOS N/m2 point c Work clone W3 = -Pj (Vg-us) ( work done on the Gas ) W3 = -9. dd 105 (V-VI) Pj Vi= Pava Ve Prvi P, P = 9.2 l ahm P2 = 4.2 atm Vj = 15.2 lt 9.21 X 15.2 lit 4.2 Ve= 3333 X163 . 3 Wz 9.21X10SX 18.13 x103 Wz=-16.7 KI Hence Net work done W = Witw, tW3 W = -5.73 KJ

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