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arsen [322]
3 years ago
11

A wire having resistance 20 ohm is bent to make a closed square.What is the resistance across the diagonal of square?

Physics
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

5 Ω

Explanation:

When it is bent into a square, each side of the square will get 5 Ω resistance. As considered diagonally ,path of the wire between to diagonal points is the two sides of the square (that is half the length of the wire connected in parallel position). Which gives 10 ohm parallel to 10 ohm wire. So it finally read as 5 Ω wire

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For the cracking reaction, C3H8(g) ® C2H4(g) + CH4(g) the equilibrium conversions negligible at 300K, but becomes appreciable at
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Answer:

a. Fraction conversion is 0.778,  b. Temperature = 645.22 K

Explanation:

Write the reaction as follows

C₃H₈(g) > C₂H₄(g) + CH₄(g)

Take the basis as 1 mole of feed C₃H₈

Let ε be the reaction coordinate.

Calculate the final moles of each component as follows

             C₃H₈(g)       ⇒       C₂H₄(g) +    CH₄(g)

n₀ (mol)     1                  0           0

n (mol) 1 - ε                         ε                 ε

Write the expression for fractional conversion as follows

Fractional conversion of C₃H₈ = (n₀ – n)/n₀ = [1 – (1 - ε)]/1 = ε

Calculate the mole fraction of each component as follows

y(C₃H₈) = n(C₃H₈)/n(T) = (1 – ε)/1 + ε

y(C₂H₄) = n(C₂H₄)/n(T) = ε/1 + ε

y(CH₄) = n(CH₄)/n(T) = ε/1 + ε

From the table, calculate the enthalpy of the reaction and Gibbs free energy as follows

ΔH(298) = ΔHf°(products) – ΔHf°(reactants)

= ΔHf°(C₂H₄) + ΔHf°(CH₄) – ΔHf°(C₃H₈)

= 52510 - 74520 – (-104680) = 82670 J/mol

ΔG(298) =   ΔGf°(products) – ΔGf°(reactants)

= ΔGf°(C₂H₄) + ΔGf°(CH₄) – ΔGf°(C₃H₈)

= 68460 – 50460 – (-24290) = 42290 J/mol

Expression for heat capacity is as follow

<u>Cp(298)/R = A + BT + CT² + DT⁻²</u>

Write the heat capacity coefficients for the components as follows

Components A              B         C  D

C₃H₈  1.213 28.785 x 10⁻³ -8.824 x 10⁻⁶ 0

C₂H₄  1.424 14.394 x 10⁻³ -4.392 x 10⁻⁶ 0

CH₄  1.702 9.081 x 10⁻³ -2.164 x 10⁻⁶ 0

ΔA = A(CH₄) + A(C₂H₄) – A(C₃H₈)

<u>= 1.702 + 1.424 – 1.213 = </u><u>1.913</u>

ΔB = B(CH₄) + B(C₂H₄) – B(C₃H₈)

<u>= (9.081 + 14.394 – 28.785) x 10⁻³ = </u><u>-5.31 x 10⁻³</u>

ΔC = C(CH₄) + C(C₂H₄) – C(C₃H₈)

<u>= (-2.164 – 4.392 – (-8.824)) x 10⁻⁶ = </u><u>2.268 x 10⁻⁶</u>

a. Calculate the Gibbs free energy at 625 K as follows

ΔG = ΔH – TΔS

= ΔH(298) + ΔH(298 K to 625 K) – T(ΔS(298) + ΔS(298 K to 625 K))

= ΔH(298) – TΔS(298) + ΔH(298 K to 625 K) – TΔS(298 K to 625 K)

= ΔH(298) – T[{ΔH(298) – ΔG(298)}/T₀] + ΔH(298K to 625K) – TΔS(298K to 625K)

<u>Calculate the enthalpy change from 298 K to 625 K as follows</u>

ΔH(298 K to 625 K) = ∫(298,625) (Cp(298))dT

= R∫(298,625) (ΔA + ΔBT + ΔCT² + ΔDT⁻²)dT

= R[1.913T – 5.31 x 10⁻³T²/2 + 2.268 x 10⁻⁶T³/3](298,625)

= 8.314 [1.913 (625 - 298) – 2.655 x 10⁻³(625² - 298²) + 0.756 x 10⁻⁶(625³ - 2983³)]

= 8.314 (625.55 – 801.33 + 164.56)

= -93.283 J/mol

<u>Calculate the entropy change from 298 K to 625 K as follows</u>

ΔS(298 K to 625 K) = ∫(298,625) (Cp(298))dT/T

= R∫(298,625) (ΔA + ΔBT + ΔCT² + ΔDT⁻²)dT/T

= R[1.913/T – 5.31 x 10⁻³ + 2.268 x 10⁻⁶T + (0)T⁻²](298,625)

= R[1.913lnT – 5.31 x 10⁻³T + 2.268 x 10⁻⁶T²/2](298,625)

= 8.314[1.913ln(625/298) – 5.31 x 10⁻³(625 - 298) + 1.134 x 10⁻⁶(625² - 298²)]

= 8.314 (1..417 – 1.736 + 0.342) = 0.191 J/molK

<u>Calculate the Gibbs free energy as follows</u>

= ΔH(298) – T[{ΔH(298) – ΔG(298)}/T₀] + ΔH(298K to 625K) – TΔS(298K to 625K)

= 82670 – (625/298)(82670/-42290) – 93.283 – 119.375

= -2232.26 J/mol

Calculate the equilibrium constant as follows

ΔG = -RTlnK

k = exp(-ΔG/RT)

k = exp(2232.26/8.314 x 625)

k = 1.537

write the expression for the equilibrium for the reaction as follows

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

Substitute the corresponding values in the above equation to calculate the value of ε

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

1.537 = {ε/(1 + ε)}{ε/(1 + ε)}/{(1 - ε)/(1 + ε)}

ε²/(1 – ε²) = 1.537

ε = 0.778

Hence the fraction conversion is 0.778

b. Fractional conversion ε is 0.85

Calculate the equilibrium constant as follows

K = y(C₂H₄) x y(CH₄)/y(C₃H₈)

= ε²/(1 – ε²)

= 0.852/(1 – (0.85)²)

= 2.603

Now calculate the Gibbs free energy as follows

ΔG = -RTlnK

= -8.314 x 625 x ln2.603

= -4971.1 J/mol

By trial and error method calculate the temperature at which the Gibbs free energy is -4971.1 J/mol

T₀          T          ΔH     ΔS     ΔG

298        643   -174.37  0.061461 -4670.39

298        644   -179.066  0.054163 -4805.95

298        645   -183.782  0.046846 -4941.50

298        645.22 -184.822  0.045234 -4971.32

298        645.5   -186.147   0.043181 -5009.27

Therefore, the required temperature is 645.22 K

5 0
3 years ago
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