Determine the entropy change in each of the following processes:4a. One mole of an ideal gas is allowed to expand from a volume of 1.50 L to 15.0 L at 298 K.4b. 10 grams of methanol vapor, initially at its normal boiling temperature of 337 K, is allowed to condense and cool to a liquid at a final temperature of 298 K. Useful infoNA = 6.022x103 molg =9.807 m/s?P[H>O] =1.00 g/mL1 bar = 10° Pa1 atm =1.013barR = 8.314 J/K-mol= 8.206x10? L-atm/K-mol= 8.206x10° m-atm/K-mol= 62.36 L-Tor/K-mol1J = 10 L-bar = 9.870×10³ L-atm= 0.2390 cal1m = 10° cm = 10° mm = 10° um = 10° nm = 101º A= 760 Torr1 mbar = 10 barRT = 2.478 kJ/mol for T = 298 K. (= 24.46 L-atm/mol)1 L-atm = 101.3J1 m = 10° L = 10° cm1 kg = 1000 g1 Watt (W) = 1 J/sPhysical Properties of Selected SubstancesCp.mJ/(mol-K) J/(mol-K) kJ/mol24.3AH melting temp AHap boiling tempkJ/molmol. wt.fasg/molkJ/molKKAl(s)CH,OH(I)CH;OH(g)H,(g)H;O(1)H,O(g)H,O(s)1,(s)L(g)I(g)N2(g)Na(s)NaN,(s)0:(g)Properties of pure phases are for standard conditions of 298 K and 1 atm.Phase transition enthalpies are valid at the indicated phase transition temperature, at 1 atm.27.028.332.04-238.6681.63.16175.235.29337.232.04-200.6643.92.0131.728.818.0-285.869.975.36.027340.637318.0-241.8188.833.618.038.037.0253.8116.1454.4415.5386.841.8458.4253.8+62.44260.6936.90126.9+106.84180.7920.7928.0191.6129.123.051.228.22.637197.4115665.0+21.7670.576.632.0205.129.4IntegralsMaxwell relationsn# -1n+1avasIn zasPrp-1- In(az – b)asavTTProperties of Some MaterialsDensity (293 K)g/cmKT10-4 K-110-" barC, graphiteC, diamondCu, solidWater, liquid2.263Euler chain rule3.50.0300.187az8.90.5010.735= -1az0.9972.149.6

Answered: Image /qna-images/answer/e2d850f2-9a34-4b39-bf5d-25371b2a259d.jpg

Determine the entropy change in each of the following processes: 4a. One mole of an ideal gas is allowed to expand from a volume of 1.50 L to 15.0 L at 298 K. 4b. 10 grams of methanol vapor, initially at its normal boiling temperature of 337 K, is allowed to condense and cool to a liquid at a final temperature of 298 K.

Determine the entropy change in each of the following processes: 4a. One mole of an ideal gas is allowed to expand from a volume of 1.50 L to 15.0 L at 298 K. 4b. 10 grams of methanol vapor, initially at its normal boiling temperature of 337 K, is allowed to condense and cool to a liquid at a final temperature of 298 K.

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter8: Thermochemistry

Section: Chapter Questions

Problem 12QAP: The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q,...

See similar textbooks

Similar questions

The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q, per mole of acid (or base) neutralized. Hneut for nitric acid is -52 kJ/mol HNO3. At 27.3C, 50.00 mL of 0.743M HNO3 is neutralized by 1.00 M Sr(OH)2 in a coffee-cup calorimeter. (a) How many mL of Sr(OH)2 were used in the neutralization? (b) What is the final temperature of the resulting solution? (Use the assumptions in Question 11.)
Constants: • c = 2.9979 x 108 m/s h = 6.626 x 10-34 J s per one photon • R = 8.314 J/(K mol) = 0.08206 L atm/(K mol) NA = 6.022 x 1023 particles/mol RH = 1.097 x 107 m1 = 2.178 x 10-18 J . . You have an aqueous Glucose solution that is 12.0% Glucose by mass. What is the molality of Glucose in the solution? Molar Mass of Water = 18.015 g/mol Molar Mass of Glucose 180.16 g/mol =
N2(g) + 20,(g)–→2NO2(g) Clear All > 0 AHrxn 0 low T, 0 high T
Mixing 10.0 g of CaC2 and 25.0 mL of H2O (p = 0.997044 g/mL) conforms to the thermochemical equation below: CaC2(6) + 2 H20() → Ca(OH)2(6) + C2H2(g) AH° = -127.2 kJ Molar masses: CaC2: 64.100 g/mol H2O: 18.015 g/mol Ca(OH)2: 74.092 g/mol C2H2: 26.038 g/mol A. Which statement applies? O. 4.06 g C2H2 is formed. O- 51.3 g Ca(OH)2 is formed. O- 1.81 g CaC2 remains. O- H20 is the limiting reagent. B. Determine the amount of heat involved. O- 88.0 kJ O- -88.0 kJ O. 19.8 kJ O- -19.8 kJ
[References] Given the following data СэНа (9) + 302(9) — 2 СО2(9) + 2Н,0() 2C2 H6 (g) + 702(9) → 4 CO2(g) + 6H2O(1) 2H2 (g) + O2 (g) → 2H2 O(1) ΔΗ -1411.0 kJ AH = -3119.8 kJ AH=-571.7 kJ calculate A H for the reaction C2H4 (9) + H2 (g) → C2H6 (9) ΔΗΞ kJ Submit Answer Try Another Version 3 item attempts remaining
AH = ? CIF 3(g) 2.) CIF(9) + F2(g) AH = - 49.4 kJ Ozlg) + 2F2(g) 20F2(9) AH= 205.6 kJ Cl,O (G) + OF2(g) + Ozlg) 2CIF (9) + + O2(g) AH = 266.7 kJ %3D ½ Cl,0 (G) + 3/2 OF2(9) CIF319)
What volume of benzene (C&H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to the following reaction? %3D = -6278 kJ 2 C6H6(1) + 15 O2(g) – 12 CO2(g) + 6 H2O(g) AH°, rxn 75 mL 37 mL 21 mL 19 mL 42 mL
Мо Peer Leader: Date: Name: calculation and follow significant figure rules. _CH18(g) + O2(g) → CO2(g) + _H;O(1) H2O(1) > AH°pxn = -10940 kJ/mol CO:(g) AH°F = -393.5 kJ/mol %3D H¿O(1) AH°f = -285.8 kJ/mol
The dissolution reactions of gold (Au) in 1M HNO3 and 1M aqua regia (1M HNO3 + 1M HCI) are given below. Аu) + NOз (аq) + 4 H* » Au3+ + NO (g) (aq) + 2 H20(1) (1) (ад) Au(s) + NO3 + 4 H* (ag) + 4 Cl¯ (aq) (aq) → AuCl4¯ + NO(9) + 2 H20) (g) (2) (aq) Reduction Half – Reaction E° (V) Au3+ + 3 e- = Au 1.36 AuCl4 + 3 e¯ → Au + 4 Cl- 1.00 NO3- + 4 H* + 3 e¯ → NO + 2 H20 0.96 2 H30+ + 2 e 2 H2 + 2 H20 Calculate E°cel| and Keg for the dissolution of gold in 1M HNO3 under standard conditions. Calculate E°cell and Keg for the dissolution of gold in aqua regia under standard conditions. Using the values calculated in a) and b), determine the formation constant (K;) of AuCl4-. The formation reaction is given below. [Hint, express K; as a function of the Keq values from a) and b)]. Au3+ (аq) + 4 Cl (aq) → AuCl4 (aq)
Given the following data C,H,(9) + 30,(9) → 2 CO,(g) + 2H,0(1) 2C,H6 (g) + 702(g) → 4 CO2(g) + 6H2O(1) AH=-3119.8 kJ 2H,(g) + O2(9) → 2H,O(1) AH= -1411.0 kJ AH=-571.7 kJ calculate A H for the reaction C2H4 (9) + H2 (g) → C,H6 (g) AH = kJ
The British thermal unit (Btu) is commonly used in engineering applications. A Btu is defined as the amount of heat required to raise the temperature of 1 lb of water by 1 °F. There are ________ Btu in one Joule. 1 lb = 453.59 g; °C = (5/9)(°F - 32°); specific heat of H2O (l) = 4.184 J/g-K.
Molar Masses A = 290.41 D = 250.53– 9 %3D E = 75.44 9 %3D mole mole %3D mole G = 117.05 X = 10.93 Z = 50.01 mole mole mole If 18.23 g A, according to the balanced thermochemical equation below, how much heat (in kJ) is absorbed? 5 A+6 D→ 5 Z +5G AH = 515 kJ Please input your answer to the second decimal place Type your answer...