Please go over my answers, help me find the mistake(s), and showed the step(s) t

Question

Please go over my answers, help me find the mistake(s), and showed the step(s) to fix it if there is any. Thank you!

PV= nRT A Problems P: NR/ Circle the problems you want me to grade. _u2k the steps of a reversible Stirling engine (A. Isothermal expansion at 148°C from 0.20m to 0.60mº. e as follows. For this problem we will use B. Isochoric cooling at 38°C=3UK 6.0025 mole of a monatomic gas that starts at a temperature of 148°C and a volume of GC, Isothermal compression at 38°C from 0.60m to 0.20m. 0.20m3, which will be called point A. Then, it SD. Isochoric heating back to 148°C and 0.20m. goes through the following steps Pleague (PA) dd i 5o ..... Complete the graph. Fill in the table below. How does the efficiency of the Stirling engine compare to the Carnot engine working within the same two heat reservoirs? 4S.. 40.- 3 .. 3.. 13:56 0.2O to.2 ZL12K O Pa- (0.002 NX *52%alm Xu 2 Ik)/=ly3.2rPa Jan - Xarid (3) to.6 Volkmars ()] x(0.002s nis/x*. Xsn v21. ln("%) , -11.12.: V. os41% = 19.J Qoy xnr 1. xc0.002r mol )( *. nuste )( 311K) = 9.1032 ) Ooscº 4: 9.70usun 0.02 % Ower: Xnrik ()= Y(0.0025 mol )(**Yasin X31K ) n (2)=-10.665 Ooser: 4 - - 10.6%ux - - 0.0 14 % O au-XnRT > 300. vv2 sm ol )( *. sayesinX4 zik) = 13.135 ) Dos - Xardn (%) + X(0.002s (r.Aprn) 1-(2x): 0.001919/ On: nny : tovorf mol (v.2 pon )( su%.com - 10. 2* P Ore : nny: (0.001 i mal)(n. 12 %s11. SK Yo.1o- pr. 344 P:/ DP, - vry: (0.0025 mol ( *. 32 Zin ( 12 in U. , 2. 744 PA Path Process asuing | ay Process AS(J/K) wa Aun. Alapostela e .. W (J) AU (J) Sare spt es Prerre CA Isclose jo.03y2 % 9.70] Something wrons ? -0.01434-10.667 Tcliorie 0.0094EX /4 J Auso Path . Q (J) Ishorne lo.034) Jul bu.43J 14. 43 J 9. 70 J w: au: W: 0 Co share -10. 66] a A pso O 13.14] Aus klo

Explanation / Answer

You missed the step of finding internal energy change in isochoric cooling when gas is cooled from 138 degrees c to 38 degree c. That path should be shown in the graph as B to C with a straight line. As pressure drops with cooling, it should be a vertical line directly below B.

The change in internal energy is given by dE = mCvdT where m is the mass of the gas.

What you did there is dE = 3/2nRT and it should be dE = mCvdT

Where Cv for diatomic molecules is 3/2R. Once you incorporate this step, in other isochoric process as well, other part is completely fine.

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