T _c = 273.15 + 320 = 593.15 K T/T _c = 0.7

Therefore, T = 0.7(693.15) = 415.2 k = 142 C

Vapor pressure of toluene at 142 C by interpolation,

P _v = (3-2) + 2 = 2.1325 atm.

Answer is (D)

Watson's equation: ? H _{v 2} = ? H _{v 1}

? H _{v 2} = 164.36 = 157.5 cal/g

= 157.5(4.1868)(60.09)

= 39625 J/gmol

Answer is (D)

From steam tables, at 70 F, h _L = 38.04 Btu/lb

Again from steam tables, at 260 psia and 1500 F,

Enthalpy change = 10 (1605.8 - 38.04) = 15677.6 Btu

Answer is (B)

Concentration of NaOH: Basis of calculation 100 g of solution gmol of NaOH = 10/40 = 0.25 gmol

90 g of water = 90 cc with sp gr = 1.0

C _i = (0.25/90)(1000) = 2.7778 gmol/Liter

( assumes no volume change on dissolving NaOH in water)

k of water at 20 C = 0.340 Btu/(h.ft. F) =

k _mix = 0.0014055 + ?(2.7778 0 + 2.7778x20.934x10 ^-5)

= 0.0015444 cal/(cm.s.K) = 0.3736 Btu/(h.ft. F)

Answer is (A)

Cascade temperature control alone is good to overcome the time lags inherent in a reactor system. However, when a batch is required to attain a certain temperature to initiate the reaction, the heating and cooling medium control valves are split range controlled. The heating medium...