For items 1-4, consider the isothermal expansion of a 1.5 mole of monoatomic gas gas from a pressure of 1 atm to 0.45 atm at 333 K. 1.) What is the work done for the process? A.) -3.32 kJ B.) + 3.32 kJ C.) -2.28 kJ D.) +2.28 kJ E.) 0 kJ 2.) What is the heat transferred for the process? A.) -3.32 kJ B.) +3.32 kJ C.) -2.28 kJ D.) +2.28 kJ E.) 0 kJ 3.) What is the change in internal energy for the process? A.) +7.61 kJ B.) -7.61 kJ C.) +12.6 kJ D.) -12.6 kJ E.) 0 kJ 4.) What is the change in enthalpy for the process? B.) -12.6 kJ C.) +17.8 kJ 'D.) -17.8 kJ E.) 0 kJ A.) +12.6 kJ

For items 1-4, consider the isothermal expansion of a 1.5 mole of monoatomic gas gas from a pressure of 1 atm to 0.45 atm at 333 K. 1.) What is the work done for the process? A.) -3.32 kJ B.) + 3.32 kJ C.) -2.28 kJ D.) +2.28 kJ E.) 0 kJ 2.) What is the heat transferred for the process? A.) -3.32 kJ B.) +3.32 kJ C.) -2.28 kJ D.) +2.28 kJ E.) 0 kJ 3.) What is the change in internal energy for the process? A.) +7.61 kJ B.) -7.61 kJ C.) +12.6 kJ D.) -12.6 kJ E.) 0 kJ 4.) What is the change in enthalpy for the process? B.) -12.6 kJ C.) +17.8 kJ 'D.) -17.8 kJ E.) 0 kJ A.) +12.6 kJ

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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For the reaction 2N₂(g) + O2(g) → 2N₂O(g) AG = 214 kJ and AS = -148.5 J/K at 334 K and 1 atm. This reaction is product favored under standard conditions at 334 K. The standard enthalpy change for the reaction of 1.51 moles of N₂ (9) at this temperature would be For the reaction 2H₂O(1)→ 2H₂(g) + O₂(g) AG=483 kJ and AH = 572 kJ at 273 K and 1 atm. This reaction is reactant favored under standard conditions at 273 K. The entropy change for the reaction of 2.18 moles of H₂O(1) at this temperature would be kl. J/K.
7. (a A 5.00 mol sample of an ideal gas is compressed irreversibly and isothermally at 150.00 K from an initial volume of 30.00 L in two steps: Step #1: Pext = 5.00 bar, Step #2: Pext = 20.00 bar. Determine the value of the total work and heat involved (in J). (b) Describe in words what happens to the total amount of work if the compression occurs in three steps. (c). Assuming the process could be done in one or multiple steps and the process could be reversible or irreversible, determine the minimum and maximum amount of work that could be done to complete the compression.
4b. Combustion of 3.500 g of butane gas (C4H10; Mm = 58.12 g/mol) causes a temperature rise (AT) of 29.09°C in a constant-pressure calorimeter that has a heat capacity (C,) of 5.500 kJ/°C. o Use calorimetry to calculate the heat (q,) released by the combustion of 3.500 g of butane gas in the calorimeter Яр [ Select ] kJ / 3.500 g of C4H10 o Use the heat (qp) released by the combustion of 3.500 g of butane gas, calculated in part (a) to calculate the change in enthalpy (AHrxn) for the following reaction. (for 1 mole of butane gas): C4H10(9) + 13/2 02(9) → 4 CO2(9) + 5 H20(g) AHrxn [ Select ] kJ / mol of C4H10 • Use the heats of formation (AH¢°) to calculate the change in enthalpy (AHrxn) for the same reaction. (for 1 mole of butane gas): Heats of Formation: CO2 (g) - A H°f = -393.5 kJ/mol H2O(g) - A H°f = -241.8 kJ/mol O2 (9) - A H°f 0.0 kJ/mol C4H10 (g) - A H°f = -125.6 kJ/mol C4H10(g) + 13/2 02(g) → 4 CO2(9) + 5 H20(g) ΔΗη [ Select ] kJ / mol of C4H10
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(2B) A heat engine operates between a hot source at 130.°C and a cold sink at 10.°C. How much workcan be done by the engine if 523 kJ of heat are extracted from the hot source?
Find the mass of water that vaporizes when 2.10 kg of mercury at 205 °C is added to 0.10 kg of water at 30°C. Assume the following: Cmercury (specific heat capacity of mercury) - 139 J/(kg.°C): Cwater (specific heat capacity of water) = 4186 J/(kg.°C); and L, (water) = 2.26x106 J/kg
Cl2 (g) + H2 (g) → 2 HCl (g) ΔH0rxn = ‒184.6 kJ 8. What is ΔH0rxn when 4.50 mol HCl (g) forms? (Slides 17 – 18: Stoichiometry of Thermochem. Eq.) Group of answer choices A.‒831 kJ B.‒415 kJ C.‒207 kJ D.831 kJ E.415 kJ
The internal pressure of a gas, π = au dv)T' is a measure of intermolecular forces, because it indicates a change in internal energy as the molecules get closer or further apart, without temperature changing. = a Calculate the change in internal 1 Vm T energy, AU, when 1.5 mols of acetone (a = 16 l² atm/mo1²), expands from 2 liters to 5 liters at a constant temperature. ii) Would this change in internal energy be greater or smaller for an ideal gas? Why? i) Using the van der Waals equation of state, T = au
2) Consider the situation: 2CH,O, + 30) 4H,0 + e) AH --1453.0 kJ 2 CO, ΔΗ, 2(8) Showing your work, how much heat would this represent for the combustion of 12.74 mol of liquid methanol (CH4O0)? Answer to 1 decimal place. [K3] the follouing processes: (2 marks each) [C4]