p(o ₁19 (pv)T93. At the inversion temperature of a gas, the value of€8a) Positive and equal to the magnitude of (2), (),Tb) Negative onlyTc) Positive and is greater than the magnitude of (ov), (or),d) Negative and is less than the magnitude of (3), (),Tis

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Answered: т 93. At the inversion temperature of a…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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In the relation, W= JxQ, 1 calorie = 4.18400 joules false true At isobaric process, W=AU-q false true The relation, AU=Q+W means the process is expansion false true The relation PV=nRT is for real gas false true
3. Using the definition of isothermal compressibility (K): 1 (av -= (+)₂, ӘР T Solve for K for an ideal gas. K
For the reaction H2S(g) + 2H20((1)3H2(g) + SO2(g) AH° = 295 kJ and AS° = 295 J/K %3D At standard conditions, this reaction would be product favored O at relatively high temperatures. at relatively low temperatures. at no temperature. O at all temperatures.
The N2O molecule has the structure N-N-O. In anordered crystal of N2O, the molecules are lined up in aregular fashion, with the orientation of each determinedby its position in the crystal. In a random crystal (formedon rapid freezing), each molecule has two equally likelyorientations.(a) Calculate the number of microstates available to a random crystal of NA (Avogadro’s number) of molecules.(b) Calculate the entropy change when 1.00 mol of a random crystal is converted to an ordered crystal.
A sample of an ideal gas in a cylinder of volume 3.83 L3.83 L at 298 K298 K and 2.26 atm2.26 atm expands to 8.10 L8.10 L by two different pathways. Path A is an isothermal, reversible expansion. Path B has two steps. In the first step, the gas is cooled at constant volume to 1.45 atm1.45 atm . In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.45 atm1.45 atm until the final volume is 8.10 L8.10 L . Calculate the work for path B.
26. A mole of perfect gas at 27°C is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm in two ways (a) reversibly, (b) against a constant external pressure of 1.00 atm. Cal- culate q, w, AU, AH, AS, ASsurr, AStot for both. (+2.74 kJ, -2.74 kJ, 0, 0, 9.13 J/K, -9.13 J/K, 0 ||1.66 kJ, -1.66 kJ, 0, 0, 9.13 J/K, -5.53 J/K, 3.60 J/K) T=27°C 300 K b) w - - Peut av = -(1 atm) (ner - her NRT ) Pf P₁ IsoT w AU÷0 q = -((atm) 1 mol. 8.314 /mol.K - 300K I atm W = -16.6 kJ. 9 = 16.6 kJ AS: grev= UT Imol. 8.314 /mol.K·3.00K 3 atm 27. A block of copper of mass 500 g and initially at 293 K is in thermal contact with an electric heater of resistance 1 kOhm and negligible
A Al(s) + 12(s) = All3(s) В CH4(g) + 4CI2(g) = C(4(1) + 4HCI(g) C Fe,O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g) Which of these processes is AH < 0 OB and C only OA and B only O A, B and OC only
Calculate S ( in J/K) for the following changes in state of 2.50 mol of a perfect monatomic gas with CV,m = 1.5R. (1.50 atm, 400 K) → (3.00 atm, 600 K)
1.94 x 1024 molecules of an ideal gas with energy E = NKBT is subjected to a cyclic, quasi- static (i.e., reversible) process shown in the figure. What is the change in the entropy (in J/K) of the gas as it moves from state A to state D? B A D. 2 3 Volume (m³) Type your answer. Pressure (Pa) 2.
Q1: Cp of gases H2 , 02 and H20 are 29 JK-1 mol-1 , 29.5 JK-1 mol-1 and 33.6 JK-1 mol-1. The enthalpy of formation of gases of H20 at 25 oC is -241.82 kJ. mol-1. What is its value at 100 oC.
Derive an expression for (∂S/∂P)T of an ideal gas. Then determine (∂S/∂P)T o for a sample containing 2.00 mol Ne (ideal gas) at 1.50 atm and 298 K. (answers: (∂S/∂P)T = -nR/P; -11.1 J/atmK)
4- An ideal gas is taken through a closed path 1→2→3→4→1. the net work done by the gas is (A) 6PV (В) 9PV (C) 4PV (D) -6PV (E) -9PV ЗР 2 3 1 4 V 4V V P.

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т 93. At the inversion temperature of a gas, the value of (GV) is T T a) Positive and equal to the magnitude of (V), (2), b) Negative only T c) Positive and is greater than the magnitude of (27), (37), d) Negative and is less than the magnitude of (3V) (37) T ^