Based on the phase diagram illustated below, what is the phase condition of the given two-component system (w and b)? 5 0.8 8.0 1.0 0.2 0.4 0.6 0.8 0.6 0.2 0.0 1.0 0.8 0.4 Z5 = 1 Z, = 0 Z6 = 0 mole fraction Zw = 1 41 3) pressure (bar)
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- You add 28 g of ice (H2O) at 0°C to 179 g of H20 (I) at 63°C at a constant pressure of 1.01 kPa. Calculate the final temperature in K. [AfusH°(H20) = 6.01 kJ mol""; Tfus=273.15 K; Cp.m(H20,)=75.3 J mol Kl; m(H20)=18.02 u].Suppose that a 1.50 mole sample of naphthalene (C10H10, M= 128.17 g/mol) is initially at a temperature of 300.0 K. Calculate the final temperature the sample if 42.2 kJ of heat are added, assuming a constant pressure of exactly 1 atm. Remember to account for phase Cs naphthalene-1.293 I @=1.293 git док changes. Some values from Table I will be needed. CS ATWhat is AG (in kJ) for 2 SO2(g) + O2(g) 2 SO3(g) at 700 K, under standard conditions of 1 bar partial pressure for all gases, given that K = 3.0x104 at 700 K? %3D -36 0.0 O +60. -60. +36
- Based on the phase diagram illustated below, what is the phase condition of the given two-component system (w and b)? 5 5 4 3 2 1F 0.8 8.0 0.2 0.4 0.6 0.8 1.0 1.0 0.8 0.6 0.4 0.2 0.0 Zb = 0 mole fraction Zb = 1 Zw = 1 Zw = 0 3.5 bar at 0.32 of b A. All vapor 4.5 bar at 0.55 of b B. liquid w + vapor 1.2 bar at 0.23 of w C. All liquid 3.5 bar at 0.8 of w pressure (bar)(2) 1 kPa (4) 273 kPa E. At 110°C, which compound has a vapor pressure of 80 kilopascals? (1) ethanoic acid (2) ethanol (3) propanone (4) water 3. Under which conditions of temperature and pressure would a real gas behave most like an ideal gas? (3) 600. K and 50.0 kPa (4) 600. K and 200.0 kPa (1) 200. K and 50.0 kPa (2) 200. K and 200.0 kPa Dilation cent at A with Scale factor greater than 1. What imilar ransform aps he maller riangle nto he larFor liquid with the typical values a= 0.04 K, K= 0.002 atm1, Vm= 73 cm/mol, C, m= 103 J/mol.K. Calculate %3D P,m Cy.m= (J/mol.K) at 241 K and 1 atm. %3D v,m O a. -72410.16 O b. 71.59 C. -144893.32 O d. -1318.51
- Water exists at T= 300'C and P=400 kPa. What is the phase of water at this state? Determine the specific internal energy, u, of water at this stateIn a laboratory experiment, molecular mass of a diethyl ether is determined by using vapour density (Victor Meyer) method at 85°C. When 110 mg diethyl ether is given to the system, 44 mL volume change is observed. Find Mw according to ideal gas equation and Vander Waals Equation respectively constant: R: 0.08206 L-atm, P: 1 atm, a= 17.17 atm.I^2.mol^-2, b= 0.133 I.mol^-1.A 0.5530-kilogram ice cube at −12.40∘C is placed inside a chamber of steam at 365.0∘C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 5.950 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.
- The following data was collected for n-pentane and isopentane: Compound P, (kPa) n-pentane 282.9 at 70°C 895.5 at 120°C Isopentane 955.9 at 115°C 1152.4 at 125°C Develop a correlation for the vapor pressure of Isopentane as a function of temperature. Clearly state the units.Płots of AG" of vaporization vs the pressure(gas) for butane and benzene are shown below. At 1 atm, the boling point of butane is 272 K and the boling point of benzene is 353 K. Assign the plots to the corresponding gas and temperature system. A) A = P(butane) at 300 K; B = P(benzene) at 275 K; C = P(benzene) at 300 K 10- B)A = P(benzene) at 275 K; B = P(benzene) at 300 K; C = P(butane) at 300 K C)A = P(butane) at 300 K; B = P(benzene) at 275 K; C = P(benzene) at 275 K -10- D)A = P(benzene) at 275 K: B = P(butane) at 300 K; C = P(benzene) at 300 K 15- Pressure(gas) in atm E)A = P(benzene) at 300 K; B = P(benzene) at 275 K; C = P(butane) at 300 KA can of hair spray with a pressure of 3.0 atm at 25.0°F is exposed to heat. Should the new temperature become 35.0°C, which of the following solutions computes for the new pressure in kPa? (3.0 atm)(269.26 К) (308.15 K) 101.325 kPa ... a. P2 = = 270 kPa 1 atm (3.0 atm) (298.15 К) 101.325 kPa b. P2 = = 290 kPa (308.15 K) 1 atm (3.0 atm)(308.15 К) (269.26 K) 101.325 kPa c. P2 = = 350 kPa 1 atm ((3.0 atm)(308.15 K)' 101.325 kPay d. P2 = 310 kPa (298.15 K) 1 atm