M Inbox (40) ktomo00 x y BMasteringphysics: Ch x (G A heat engine using x Ci fi

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M Inbox (40) ktomo00 x y BMasteringphysics: Ch x (G A heat engine using x Ci fi https: masteringphysics.com/myct/itemView? assignment ProblemID 5796809 Apps M UCR M PCC f Facebook o My subscription Twitter Do welcome, Kurt B LancerPoint Log... NBA.com E LeBrown James D WeBWork MA... Slicethepie Time Warner Ca f Electronics, Car.. PHYS040B(020) General Physics, Winter 2016 ITsai) Ch Problem 19.53 Problem 19.53 Part A Determine W,, Q. and AEM for prooess 1-v2 A heat engine using a diatomic gas follows the cycle shown in the figure. Its temperature at point 1 is 30.0 C. Enter your answers numerically separated by commas. w.. Q. AE 52, 5.32 3.80 incorrect Try Again; 9 attempts remaining Part B Determine W, Q. and AEER for prooess 2 3 Enter your answers numerically separated by commas. w.. q AEA 50,382,584 submit up incorrect Try Again; 11 attempts remaining Part C Determine W., Q, and AEA for prooess 3 Enter your answers numerically separated by commas W Q. AEA 515 26450.99 26449.476 J MLAnswer: Give Up incorrect Try Again; 11 attempts remaining Part D What is the thenmal efficiency of this heat engine? Express your answer using two significant figures. 2.186 10 PSI Pr Kurt G1 Other bookmarks Signed in as Kurt Tomongn l Hada 3:08 PM 2/14/2016

Explanation / Answer

Diatomic Gas Constants
Cv= 5/2*R , Gamma = 7/5 (or 1.40)
Cp= 7/2*R , Eth = 5/2*nRT

Wout = 0.5 x 101325Pa x 0.00003 m^3 = 1.52J (Area of the Triangle)
QH = The Sum of all Positive Q Values from A to C
Use PV=nRT T2 = 360 C, T3 = 120 C, n= 2*10^-4

Start with Part B (Points 2-3) (Constant Volume)
Ws = 0
Q = n x CV x (T3 - T2) = 2*10^-4*2.5*8.314*(120-360) = -0.9977 J
Eth = Q =-0.9977 J

Part C (Points 3-1) (Constant Pressure)
Ws= P x (V1-V3) = 0.5 x 101325Pa*(10-40)*10^-6 = -1.52 J
Q = n x Cv x (T1-T3) = 2*10^-4*2.5*8.314*(30-120)= -0.374 J
Eth = Q - Ws = -0.374 + 1.52 = 1.146 J

Part A (1-2)
Ws = Wout - Ws(2-3) - Ws(3-1) = 1.52 - 0 +1.52 = 3.04 J
(delta) Eth = n x CV x (delta T)(1-2) = 2*10^-4*2.5*8.314*(30-360) = -1.37 J
Q = Eth + Ws = -1.37+3.04 = 1.67 J

Part D
Thermal Efficiency = Wout / QH x 100 =(1.52/1.67)*100 =91.02 %

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