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Airida [17]
2 years ago
15

s) Use Cramer’s rule to solve the system below, and state the condition at which solution exists. ax+by = 1 cx+dy =−1

Engineering
1 answer:
Zanzabum2 years ago
4 0

(a) The solution to the system of equation is, x = (d + b)/(ad - cb) and y = (-a - c)/(ad - cb).

(b) The condition at which the solution exists is, ad - cb ≠ 0.

<h3>Solving the system of equation with Cramer's rule</h3>

ax + by = 1

cx + dy = -1

D = [a   b]

      [c    d]

D = ad - cb

Dx = [1   b]

       [-1   d]

Dx = d + b

Dy = [a   1]

       [c  - 1]

Dy = -a - c

x = Dx/D

x = (d + b)/(ad - cb)

y = Dy/D

y = (-a - c)/(ad - cb)

Cramer's rule applies to the case where the coefficient determinant is nonzero.

Thus, D ≠ 0 (ad - cb ≠ 0).

Learn more about Cramer's rule here: brainly.com/question/10445102

#SPJ1

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An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
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Ts =Ta E)- 300(

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A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

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10

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Due to step 12 is isothermal: T1 = T2= 300 K and

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Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

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And Ts =Ta E)- 300(

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5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

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Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

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Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

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For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

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AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

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