Answer / kangchuentat

QUANTUM COMPUTING - ANSWER 32.5 : (a) Ba + Be = (00 + 11) + (00 - 11) = 2 (00), 00 = (Ba + Be) / 2. Bc - Bm = (10 + 01) - (10 - 01) = 2 (01), 01 = (Bc - Bm) / 2. Bc + Bm = (10 + 01) + (10 - 01) = 2 (10), 10 = (Bc + Bm) / 2. Ba - Be = (00 + 11) - (00 - 11) = 2 (11), 11 = (Ba - Be) / 2. (b) (C0 + D1) (00 + 11) = C000 + C011 + D100 + D111 = (00) C0 + (01) C1 + (10) D0 + (11) D1 = (1/2) (Ba + Be) C0 + (1/2) (Bc - Bm) C1 + (1/2) (Bc + Bm) D0 + (1/2) (Ba - Be) D1 = (1/2) [ (Ba + Be) C0 + (Bc - Bm) C1 + (Bc + Bm) D0 + (Ba - Be) D1 ] = 0.5 ( Ba C0 + Be C0 + Bc C1 - Bm C1 + Bc D0 + Bm D0 + Ba D1 - Be D1) = 0.5 [ Ba C0 + Ba D1 + Bc C1 + Bc D0 + Be C0 - Be D1 + Bm (-C1) + Bm D0 ] = 0.5 [ Ba (C0 + D1) + Bc (C1 + D0) + Be (C0 - D1) + Bm (-C1 + D0) ] (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 31 – 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.

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