mid thermo

# mid thermo - CR 223 i 1 Materials Thermodynamics CR 223...

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Unformatted text preview: CR 223 i 1 Materials Thermodynamics CR 223 Mid-Semester Examination ~ Autumn 2012 National Institute of Technology Rourkela Maximum Time Allowed: 2 Hrs; Maximum Credit: 60; Bold numbers at the end of the questions indicate credits. Answer all of the questions. 1. Write the equation, in the line of the ﬁrst law of thermodynamics, for the following events (please state all your assumptions and mention the parameters): (a) A person steps on to the weighing machine. 4 (b) A child blows a soap bubble. 4 (C), A cubic Ceramic, such asBaTiOg, transforms to tetragonal. 4 2. Calculate the change in enthalpy and the change in entropy when 1 mole of 810 is heated from 25 °C to 1000 °C. The variation of the molar heat capacity of SiC can be expressed as follows: 12 0,, = 5.79 + 1.97 * 10‘3T — 4.92 *106T‘2 + 8.2 *108T“3J.m0le‘1.K"1 3. (a) If the undergraduate programme in NIT Rourkela were to be considered a ther- modynamic function, would you call it a state function or a path function. Justify your answer shortly. 3 (b) Sort out the intensive and extensive variables: Pressure, Volume, Gibbs Free Energy, Molar Speciﬁc Heat, Temperature. 3 (c) A baseball player hits a 5 cm diameter spherical water ball (instead of the usual baseball) into uniform spherical droplets of 5 u in radius. If the surface tension of water is 0.075 N.m"1, how much of work the captain must have done at least? 6 4. A mole of N2 gas is contained at 273 K and 1 atm. presure. Addition of 3 kJoules of heat into the gas at constant pressure causes 1000 joules of work done during the expansion. Assuming reversible process and that the gas behaves ideally, (a) Calculate the ﬁnal temperature of the gas. 4 (b) Calculate the values of AU and AH for the process. 4 (0) Calculate the values of CI, and CU for the gas. 4 5. Write down the two oxidation reactions for the two forms of carbon; Cgmphite and Cdtamond' Which of the two reaction is more exothermic at 1000 K based on the speciﬁc heats of the two forms over the designated temperature ranges as follows: 12 0298—1101“ = 0.11 + 38.94 * 10-37“ — 1.48 * 105T‘2 —— 17.38 * 10‘GT2J.K‘1 p,g'raphite Owes—120% ~ 9.12 + 13.22 * 1073’!" — 6.19 * 105T’2J.K‘1 p,diamond H ...
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