KEY W5 - Thermo 1st Law S14(1)

# KEY W5 - Thermo 1st Law S14(1) - Names 20 points total CHEM...

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Unformatted text preview: Names:__.__ 20 points total__ CHEM 104- Workshop 5: 1st Law of thermodynamics Thermodynamics is all about the study of energy transfers. The first law of thermodynamics states that AE = q + w. This means that all energy changes can be computed in terms of heat and work. In todas workshop you will compute energy changes for a gaseous system and then consider calorimetry and bond energy. To complete the workshop you need to remember some key concepts about gases. Specifically you will need to remember the ideal gas equation: PV = nRT and how to compute the energy of a gaseous system: E = (3/2) nRT You will also need to know the gas constant: R = 0.0820 L-atm /mole-K = 8.315 I/mole-K and remember that there are 1000 liters in a cubic meter. Lets try and solve some problems: [20 points] 1. Consider a system that contains two [2] moles of an ideal gas in a cylindrical syringe. Initially the temperature of the system is T1 = 300 K and the pressure is 1 atm. Now, the gas is slowly heated (at a constant pressure of latm) until the gas reaches a temperature of T2 = 350 K. During this constant pressure process, the volume expands to a higher value V2 and the energy of the gaseous system changes from E1 to E2. a) Compute the change in energy [AB = E2 - E1] for this gas in units of loules. [5 pts) E1 is the energy at 300K and E2 is the energy at 350 K E1 = [3/2) nRT1 = (3/2)[2 mol)(8.315]/m01-K](300 K) = 7480] E2 = [3/2] nRT2 = [3/2)[2 mol)[8.315 I/mol-K)[350 K) = 8730] AB: E2—E1 = 87301—7480121250] b) Find the change in volume (AV) for this gas in units of liters (5 pts) PV = nRT -) V1 is the volume at 300K and V2 is the volume at 350 K V1 = nRT1/P1 = 2 mol(0.0820 L-atm/mol-K][300 K)/[1 atm) = 49.2 L V2 = nRT2/P2 = 2 mol(0.0820 L-atm/mol-K)[350 K)/(1 atm) = 57.4 L AV = V2 —V1= 57.4 L - 49.2 L = 8.2 L c) Now find the work [W = -PAV) of the system, in loules. Is the gas doing work, or is work being done on the gas? Does the sign ofW in your answer make sense to you? Note: 101 Jis equivalent to 1 L-atm. (4 pts) va = -PAV = -1 atm [8.2 L) = -8.2 L-atm You must now convert L-atm to Joules using the conversion factor -8.2 L-atm [1011/ 1 L-atm) = = -830] Work is being done BY the gas so the sign should be negative _ _ 8.315J/mole-K It IS fun to see where the COIlVeI‘SlOl'l factor comes from: — = 101 I/L-atm 0.08201-atm/m01- K Names:_____ 20 points total__ d) Using your answers in [a] and [c], and the First Law of Thermodynamics, compute the heat, in Joules, that must have been added to the gas in the process described. (1 pt) AE=q+w q=AE-w=1250]—[-830])=2080] 2. I add a 50 g piece ofAl [C = 0.88 I/g-deg) that is at 225°C to 100 ml ofwater at 20°C. What is the final temperature ofthe water? The density ofwater is approximately 1g/ml. [5 pts] We know that q = chT Heat lost = heat gained -q [aluminum] = + q (water) -m1C1[Tf-T1i) = m2C2[Tf-T21) -[50 g)*(0.88 I/g-deg)*[Tf- 225 °C] = [100 g)*[4.18 I/g-deg]*[Tr— 20 °C) -44Tr+ 9900 = 418Tf— 8360 -44Tf' 418Tf= - 8360 -9900 -462Tf= -18260462 Tf= 39.5 °C And here is one for extra credit: [5 points] Consider the formation of ammonia: 3H2[g) + 2N2[g) 9 2NH3[g] AH = -46.11 kI/mol NH3 a) Is the energy of the product higher or lower than that of the reactants? [1 pt) Lower 9 the reaction is exothermic so heat has been lost by the system to the surroundings b) As the reaction proceeds, is the mixture getting hotter or colder? (1 pt) Hotter 9 Energy is released by the system as the reaction proceeds energetically downhill toward the lower energy products. That released energy heats the surroundings — the solvent. c) Given that the H2 bond energy is 435 kI/mol, and the N2 bond energy is 946 kJ/mol What is the strength of the N-H bond in NH3? (3 pts] Let the NH3 bond strength = x Bonds broken — bonds formed = total energy change for the reaction. %[946 kI/mol] + 2(435 kI/mol) - 3[x kI/mol) = -46.11 kJ/mol , x = -391 kI/mol ...
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