Question { 1644 }

Question 100 - (a) Time evolution in Heisenberg picture, according to Ehrenfest theorem is m (d / dt) < r > = < p >, where m = mass, r = position, p = momentum of a particle. If v = velocity, prove that m < v > = < p >. (b) Lande g-factor is given by Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. If Gl = 1 and under approximation of Gs = 2, prove by calculation that Gj = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ].

Answer

Answer 100 - (a) Let (d / dt) < r > = < v >. Substitute it into m (d / dt) < r > = < p > so that m < v > = < p >, proven by momentum = mass x velocity. (b) Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. When Gl = 1 and Gs = 2, Gj = [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + 2 [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] = [ J (J + 1) + 2J (J + 1) ] / [ 2J (J + 1) ] + [ 2S (S + 1) - 2L (L + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] = [ 3J (J + 1) ] / [ 2J (J + 1) ] + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ] (proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1505 }

ENGINEERING MATHEMATICS - EXAMPLE 8.4 : Let 1 ^ 1 = 1, 2 ^ 2 = 4, 3 ^ 3 = 27. By using the Excel computer programming - either by Solver or Goal Seek, find the value of v for the Van der Waals equation (P - a / v ^2) (v - b) = RT where a = 18.82, b = 0.1193, P = 2, R = 0.082, T = 5000 for benzene. Describe briefly how to use Solver and Goal Seek in Excel program of computer to find the solution quickly.

Answer

ENGINEERING MATHEMATICS - ANSWER 8.4 : Rearranging the equation gives P v^3 - (Pb + RT) v^2 - av + ab = 0. By using Excel program of computer v = 0.19322. Solver : FILE -> Options -> Add-Ins -> Solver Add-in. Goal seek : DATA -> What-If-Analysis -> Goal Seek. In both functions, Set cell / Objective, To value / Value Of, By changing cell / Changing Variable Cells should be used where To value / Value Of = 0, By changing cell / Change Variable Cells is the assumed v value, Set cell / Objective is the equation of function. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1574 }

UNIT OPERATION - EXAMPLE 9.1 : Which of the sequence below represent a feasible flows of ethanol processing plants using cellulose as starting material? A. raw material -> heat exchanger -> distillation column -> reactor. B. reactor -> distillation column -> raw material -> heat exchanger. C. heat exchanger -> raw material -> distillation column -> reactor. D. raw material -> heat exchanger -> reactor -> distillation column. E. distillation column -> raw material -> reactor -> heat exchanger.

Answer

UNIT OPERATION - ANSWER 9.1 : D. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 2384 }

Question 101 - (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.

Answer

Answer 101 - (a) | A > + | C > - | B > = (Aa + Ca - Ba Ab + Cb - Bb Ac + Cc - Bc). (b) Let d = 10, Ea = 6, Eb = 7, Ec = 8. Then 10 | E > = (10 x Ea 10 x Eb 10 x Ec) = (10 x 6 10 x 7 10 x 8) = (60 70 80). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1945 }

UNIT OPERATION - EXAMPLE 9.2 : A distillation column separates 10000 kg / hr of a mixture containing equal mass of benzene and toluene. The product D recovered from the condenser at the top of the column contains 95 % benzene, and the bottom W from the column contains 96 % toluene. The vapor V entering the condenser from the top of the column is 8000 kg / hr. A portion of the product from the condenser is returned to the column as reflux R, and the rest is withdrawn as the final product D. Assume that V, R, and D are identical in composition since V is condensed completely. Find the ratio of the amount refluxed R to the product withdrawn D. Hint : Solve the simultaneous equations as follow in order to find the answer (R / D) : 10000 = D + W; 10000 (0.5) = D (0.95) + W (0.04); 8000 = R + D.

Answer

UNIT OPERATION - ANSWER 9.2 : Overall Process - Total Balance : 10000 = D + W. Benzene Balance : 10000(0.50) = D (0.95) + W (0.04). Solving simultaneously, D = 5050 kg / hr; W = 4950 kg / hr. Total balance around the separator : 8000 = R + D; R = 2950 kg / hr. Answer : Ratio R / D = (2950 / 5050) = 0.58. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1534 }

UNIT OPERATION - EXAMPLE 9.3 : In the distillation of binary systems by Mc Cabe Thiele method, the equation for the line of top section is given by y = [ R / (R + 1) ] x + XD / (R + 1). 2 points on the line are (0.99, 0.99) and (0, 0.36). Find the reflux ratio of R and XD.

Answer

UNIT OPERATION - ANSWER 9.3 : The slope is given by R / (R + 1) = (0.99 - 0.36) / 0.99 = 0.6363. Rearranging the equation gives R = 0.6363 / (1 - 0.6363) = 1.7495. Choose the point (0, 0.36) as reference, 0.36 = XD / (R + 1), XD = 0.36 (R + 1) = 0.36 (1.7495 + 1) = 0.9898. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 2072 }

UNIT OPERATION - EXAMPLE 9.4 : Acetone and ethanol are separated using a distillation column with a partial condenser and partial reboiler. An equimolar, sub-cooled liquid feed enters at 100 kmol / hr and condenses 1 mole of vapor for every 6 moles of feed. The separation requires a distillate vapor that is 95 mol % acetone and bottoms liquid that is 5 mol % acetone. The reflux is returned from the condenser to the column as a saturated liquid and the operation is run at (L / V) = 1.4 * (L / V) min. Assume constant overflow conditions. (a) Feed operating line is y = [ q / (q - 1) ] x - z / (q - 1) where z = 0.5 for equimolar liquid mixture of 2 components, q = (L'- L) / F where L' = L + F + (F / 6) for condensation of 1 mole of vapor / 6 moles of feed. What is y = f(x)? (b) The rectifying operating line is y = (L / V) x + (D / V) (xd) where (L / V) min goes through the points A (0.95, 0.95) and B (0.53, 0.69), V = L + D. What is y = f(x)? Let xd = 0.95. (L / V) min is the slope of the 2 points A and B.

Answer

UNIT OPERATION - ANSWER 9.4 : (a) q = (F + F / 6) / F = 7 / 6, then y = [ (7 / 6) / (1 / 6) ] x - 0.5 / (1 / 6) or y = 7x - 3. (b) (L / V) min as a slope of AB = (0.95 - 0.69) / (0.95 - 0.53) = 0.62. (L / V) = 1.4 * (L / V) min = 1.4 x 0.62 = 0.868. V = L + D, 1 = (L / V) + (D / V) then (D / V) = 1 - (L / V) = 1 - 0.868 = 0.132, y = 0.868 x + (0.132) (0.95) then y = 0.868 x + 0.1254. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 2204 }

Question 102 - (a) As an approximation, let v = Zc / 137 where v is the radial velocity for 1 s electron of an element, c is the speed of light, Z is the atomic number. For gold with Z = 79, find the radial velocity of its 1 s electron, in term of c and percentage of the speed of light. (b) As an approximation, let A x A = 1 - Z x Z / 18769 where A is the ratio of the relativistic and non-relativistic Bohr radius. Find the value of A.

Answer

Answer 102 - (a) Approximately v = Zc / 137 = (79 / 137) c = 0.5766 c or 57.66 % of the speed of light. (b) Approximately A x A = 1 - Z x Z / 18769 = 1 - 79 x 79 / 18769 = 0.667 then A = 0.817. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1596 }

THERMODYNAMIC - EXAMPLE 10.1 : The water is superheated steam at 440 degree Celsius and 17.32 megapascals. Estimate the enthalpy of the steam above. From the steam table for water at 440 degree Celsius, enthalpy of steam, h at 18 megapascals is 3103.7 kilojoules per kilogram and at 16 megapascal is 3062.8 kilojoules per kilogram. Assume that h = mP + c where P is pressure; m and c are constants at fixed temperature with small differences in P.

Answer

THERMODYNAMIC - ANSWER 10.1 : By interpolation method, h = 3107.3 + (17.32 - 16) / (18 - 16) x (3062.8 - 3103.7) = 3076.7 kilojoules per kilogram. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1740 }

THERMODYNAMIC - EXAMPLE 10.2 : A cylinder with a movable piston contains 0.1 mole of a monoatomic ideal gas. The piston moves through state a, b and c. The heat Q, changes from state c to a is + 685 J. The work W, changes from state c to a is - 120 J. The work, W performed from state a to b then to c is 75 J. By using the first law of thermodynamic, U = Q + W where U is the internal energy : (a) Determine the change in internal energy between states a and c. (b) Is heat added or removed from the gas when the gas is taken along the path abc? (c) Calculate the heat added or removed when the gas is taken along the path abc?

Answer

THERMODYNAMIC - ANSWER 10.2 : (a) U = Q + W = 685 J - 120 J = 565 J. (b) For path a to c, U = - 565 J - heat is removed. (c) Q = U - W = - 565 J - 75 J = - 640 J (heat energy is removed when the value is negative). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1528 }

THERMODYNAMIC - EXAMPLE 10.3 : According to Raoult's law, the pressure when vapor is completely condensed, P = x(1) P(1) + x(2) P(2) ... + x(n) P(n) when x(1), x(2) ... x(n) are the mole fractions of component 1, 2 ... n and P(1), P(2) ... P(n) are the vapor pressures of component 1, 2 ... n. A vapor at 74 degree Celsius containing 70 mole % water and 30 mole % ethanol is to be completely condensed. At the temperature of 74 degree Celsius vapor pressure is 0.38 atm for water and 0.97 atm for ethanol. What is the minimum pressure the compressor must be operated?

Answer

THERMODYNAMIC - ANSWER 10.3 : Let component 1 is water, component 2 is ethanol, x(1) = 70 / 100 = 0.7, x(2) = 30 / 100 = 0.3, P(1) = 0.38 atm, P(2) = 0.97 atm, then P = x(1) P(1) + x(2) P(2) = 0.7 (0.38) + 0.3 (0.97) = 0.557 atm. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1463 }

CHEMICAL ENERGY BALANCE - EXAMPLE 11.1 : Please match the term A - E with the stated definition i - v. Terms : A. Yield. B. Selectivity. C. Relative saturation. D. Molal saturation. E. Absolute saturation. Definitions : i. (moles of desired product formed) / (moles that would have been formed if there were no side reactions and the limiting reactant has reacted completely); ii. (moles of desired product formed) / (moles of undesired product formed); iii. (relative humidity 40 % means partial pressure of water vapour equals 4 / 10 of the vapour pressure of water at the system temperature); iv. (moles of vapour) / (moles of vapour dry gas); v. (mass of vapour) / (mass of dry gas).

Answer

CHEMICAL ENERGY BALANCE - ANSWER 11.1 : i. A (moles of desired product formed) / (moles that would have been formed if there were no side reactions and the limiting reactant has reacted completely); ii. B (moles of desired product formed) / (moles of undesired product formed); iii. C (relative humidity 40 % means partial pressure of water vapour equals 4 / 10 of the vapour pressure of water at the system temperature); iv. D (moles of vapour) / (moles of vapour dry gas); v. E (mass of vapour) / (mass of dry gas). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1609 }

CHEMICAL ENERGY BALANCE - EXAMPLE 11.2 : Calculate the cooling duty, H required to condense and cool acetone from 100 degree Celsius to 25 degree Celsius at atmospheric pressure. The heat of vaporization for acetone at its normal boiling point is 30.2 kJ / mol. The boiling point of acetone at atmospheric pressure is 56 degree Celsius. The flowrate of acetone through the condenser is 100 mol / s = N. Value of sensible heat needed to increase the temperature of acetone in liquid form from 25 to 56 degree Celsius is 4.06 kJ / mol. Value of sensible heat needed to increase the temperature of acetone in vapor form from 56 to 100 degree Celsius is 3.82 kJ / mol. Unit of H is kJ / s.

Answer

CHEMICAL ENERGY BALANCE - ANSWER 11.2 : In the same state, total heat changes in heating = total heat changes in cooling for acetone. H = N ( total heat of sensible + heat of vaporization) = 100 (mol / s) [ (30.2) + (4.06 + 3.82) ] (kJ / mol) = 3808 kJ / s. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 2238 }

CHEMICAL ENERGY BALANCE - EXAMPLE 11.3 : For liquid benzene, the CP constants are : a = 129440, b = - 169.5, c = 0.64781. Reference temperature is 298 K. The temperature of benzene is 60 degree Celsius. Calculate the enthalpy of benzene by using the formula H = a (DT) + (b/2) (T^2 - TREF^2) + (c/3) (T^3 - TREF^3) where ^ is power, DT is temperature difference with TREF = 298 K. H is in J / kmol. DT = T - TREF.

Answer

CHEMICAL ENERGY BALANCE - ANSWER 11.3 : T = 60 + 273 = 333 K. H = 129440 (35) + (-169.5 / 2) (333^2 - 298^2) + (0.64781 / 3) (333^3 - 298^3) = 4917920 J / kmol. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1775 }

Question 103 - (a) Let | - > = 1 | x > + 0 | y >, | | > = 0 | x > + 1 | y >. Find the value of 2 | x > + 3 | y > in term of | - > and | | >. (b) Let m to be the reduced mass. Find the value of m in term of Ma and Mb where 1 / m = 1 / Ma + 1 / Mb.

Answer

Answer 103 - (a) For | - > = 1 | x > + 0 | y >, multiply it with 2 to produce 2 | - > = 2 | x > + 0 | y >, then 2 | x > = 2 | - > - 0 | y > as first equation. For | | > = 0 | x > + 1 | y >, multiply it with 3 to produce 3 | | > = 0 | x > + 3 | y >, then 3 | y > = 3 | | > - 0 | x > as second equation. Finally first equation plus second equation to produce 2 | x > + 3 | y > = 2 | - > + 3 | | >. (b) Let 1 / m = 1 / Ma + 1 / Mb = (Ma + Mb) / (Ma x Mb). Then m = (Ma x Mb) / (Ma + Mb). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.