FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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Using steam table data, determine the indicated property data for a process in which there is no change in specific entropy between state 1 and state 2. In each case, locate states on a sketch of the T-s diagram.
a. Find an appropriate expression for the change in entropy in the following
two cases:
1) S=S(TV) 2) S=S(TP)
Where: S is entropy. T is temperature. V is volume, P is pressure
b. Prove the following two thermodynamic property relationships
ac
Where: T. P. V are temperature, pressure and volume, respectively. C. and C, are specific heats at constant volume and constant pressure, respectively.
Identify the Enthalpy of State 1 to 4 (h1, h2, h3 and h4)
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- Two moles of a diatomic ideal gas such as oxygen are compressed adiabatically and reversibly from an initial state (1.96 atm, 19.11 L) to a final state with a pressure of 7.5 atm. Assume that Cv=5R Determine the change in the internal energy of the gas in this process (in J). 4,756.2334 margin of error +/- 1%arrow_forwardUsing the rogers and Mayhew's table can you solve this as I'm struggling.arrow_forwardAnswer on a piece of paper WITH COMPLETE AND CLEAR UNITS/CONVERSION FACTORS.arrow_forward
- i) Use the thermodynamic identity for a P-V-T system and the equation of state to show that the entropy change of one mole of an ideal gas of vibrating diatomic molecules, as volume and temperature are changed, is given by; 7 AS = S(V;,T;) – S(V;, T;) = ;Rln + Rln T ii) The gas undergoes an isobaric compression from V; to V;/2. Evaluate the change in the number of microstates of the system that occurs as a result. ii) Using the equation derived in i), demonstrate that for an ideal gas undergoing an adiabatic expansion from initial volume V; to final volume V; the change in entropy is zero.arrow_forwardTopic: Ideal gas process Instructions: Answer the following review questions and show complete solutions. *please write legibly. Thank youarrow_forwardQ.1 A insulated container is divided into two parts ,one containing the oxygen and other nitrogen at same temperature and pressure. The number of kilo moles of each is 2 and 4 respectively. The partition is removed and the gas is allowed to mix. What is the entropy generated in the process? A 31.751 kJ/K В -37.751 kJ/K C C 24.751 kJ/K D -24.751 kJ/Karrow_forward
- 4. Choose the correct statement/s with respect to entropy change during a process a. Entropy increases with increase in pressure at constant temperatureb. Entropy increases with increase in temperature at constant pressurec. Entropy can be kept constant by systematically increase both pressure and temperatured. Entropy can not be changedarrow_forwardIf you're not sure ,Skip it . Do not give Wrong answer. [Explain properly, you can Include Equations for Better understanding]arrow_forwardQuestion #2 For the following processes, find the changes in h as appropriate. The initial state pressure is p1 = 0.5 MPa. the final state is 2. a. constant volume : v1 = 0.3 m3/kg, p2 = 0.3 MPa; b. constant entropy : s1 = 6.3 kJ/kg K, p2 = 0.15 MPa; c. constant volume : h1 = 2500 kJ/kg, p2 = 0.2 MPa; d. constant enthalpy : s1 = 6.4 kJ/kg K, p2 = 0.2 MPa;arrow_forward
- C6 2.arrow_forward5) A fluid fits the following equation of state. a= 10-3 m3K/(bar*mol) ve b = 8*10-5 m3/molFor this fluid, the average Cp value between 0-300 0C at 1 bar is 33.5 J / mol.K.a) Calculate the average Cp value in the 0-300 0C temperature range for 12 barpressure.b) Calculate the enthalpy and entropy change when the fluid is brought from 4 bar and300 K to 12 bar and 400 K.arrow_forwardStep by step question please I only have 1 attempt thank you.arrow_forward
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