a. Find an appropriate expression for the change in entropy in the followingtwo cases:1) S=S(T, V)2) S= S(T, P)Where: S is entropy. T is temperature. V is volume. P ispressureb. Prove the following two thermodynamic property relationshipsor).OTT²(S)イ),oPWhere: T. P. V are temperature. pressure and volume, respectively.C. and C, are specific heats at constant volume and constant pressure, respectively.

We are suppose to solve only one question. Please post other question as a separate question.

Answered: a. Find an appropriate expression for…

a. Find an appropriate expression for the change in entropy in the following two cases: 1) S=S(T, V) 2) S= S(T, P) Where: S is entropy. T is temperature. V is volume. P is pressure

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: The process below shows a reversible two-step path for expansion of 0.050 mole of monatomic ideal…

A: Given: Process 1-a is constant process. Process a-2 is constant pressure.

Q: (1) Steam at a temperature of 250°C has a specific volume of 0.15436m/kg. What are the pressure,…

Q: A- Define the following with mathematical expression if needed: 1- Closed system, 2- Cycle, 4-…

A: The given question is

Q: a. Find an appropriate expression for the change in entropy in the following two cases: 1) S=S(TV)…

Q: 1 x 106 cm3 of Nitrogen at a pressure of 1 N/mm2 and temperature 300°C, expands 10 times its…

Q: 10. Find the change in entropy upon the isothermal (T = const) expansion of 40 grammes of hydrogen…

Q: 1.2 m' of air is heated reversibly at constant pressure from 300 K to 600 K, and is then cooled…

A: Given:V=1.2 m3T1=300 KT2=600 KP1=1 barFor Process 1-2:Constant pressure…

Q: In a reversible process the rate of heat transfer to the system per unit temperature rise is given…

A: Write the given data with suitable variables. dQdT=0.5 kJ/°CT1=500 KT2=800 K

Q: Ten kmol per hour of air is throttled from upstream conditions at 25°C and 10 bar to a downstream…

Q: 4. Steam at a temperature of 210°C has a specific entropy of 7.1167 kJ/kg.K. Determine the pressure,…

A: According to the problem, the steam is at a temperature of 210oC. From the saturated steam tables,…

Q: 12.2: m, kg of water at absolute temperature T, is isobarically and adiabatically mixed with m2 kg…

Q: (a) How much heat flows from 1.00 kg of water at 38.2°C when it is placed in contact with 1.00 kg of…

Q: An ideal gas undergoes a process from state 1 (T, = 300o K,P, = 100 kPa) to state 2(T, = 600 K, P, =…

Q: 6. Identify the process/es in which the entropy change is positive among the following. i, Increase…

Q: If 2 pmole of a hydrogen (H2) are throttled adiabatically (Q = 0) from an initial volume of 4 ft³…

A: Since the process is an adiabatic throttling process, therefore

Q: a) Starting with AS-dQr/T, clearly explain all steps - evaluating the integral and finding the…

Q: ygotins lo 6.69) By injecting liquid water into superheated steam. the desuperheater shown in Fig.…

A: given data; mass enter at superheated vpor(m2)=0.03kg/sentropy at superheated vapor at pressure…

Q: Compute for (a) Entropy, (b) Internal Energy, (c) Enthalpy of compressed water. a. Compressed water…

Q: the process? Calculate the amount of entropy produced, in kJ/K per kg of steam. What is the maximum…

Q: (B) 0.56 m3 of air at a temperature of 250°C and a pressure of 6.5 bar expand according to the law…

A: Thermodynamics : It is a branch of science which deals with work, heat, and energy. Also, it studies…

Q: 4.0 Four pounds of air gain 0.491 Btu/"R of entropy during a nonflow isothermal process. If p; = 120…

Q: QUESTION 7 Calcluate the entropy that is created by mixing Vh = 52 litres of hot water at Th = 51°C…

A: Let s=entropy created by mixing xh=Mole fraction of hot water n1=no. of moles of hot water=(mass of…

Q: A rigid, insulated vessel is divided into two compartments connected by a valve. Initially, one…

Q: One-quarter Ibmol of axygen gas (02) undergoes a process from p; = 20 lb/in?. T, = 500°R to p2 = 150…

Q: Problem #3: Assume air is an ideal gas with variable specific heat, and that the reference…

A: Given data: To = 0 °C = 273 KP0 = 101.325 kPaandu0 = h0 = s0 = 0 from ideal gas equation P0v0 =…

Q: One-quarter Ibmol of oxygen gas (O2) undergoes a process from p1 = 20 lbę/in?, T1 = 500°R to p2 =…

Q: Q1: A rigid cylinder containing o.006mt at nitrogen at 1-04 bar, 15c , is heated reversibly until…

A: Ideal gas equation relates the pressure, temperature, and volume of ideal gas. Ideal gas equation is…

Q: An ideal gas undergoes a process from state 1 (T, = 300 K,p, = 100 kPa) to state 2(T, = 600 K, p, =…

A: Given:- State 1: T1=300 K P1= 100 kPa State 2: T2=600 K P2=500kPa Cp= 1 kJ/kg-K Cv= 0.7 kJ/kg-K To…

Q: 1 m3 of carbon monoxide at a pressure of 1 N/mm2 and temperature 300°C, expands 10 times its…

Q: 2 kg of N2 initially at P1 = 2 bars and T1 350 K and nally at P2 = 1.5 bars and T2 = 1400 K. For…

A: Given data: The mass of nitrogen gas is m=2 kg The pressure of the nitrogen gas at the state 1 is…

Q: _3.0 A vessel of 0.058 m3 capacity is well insulated and is divided equally by a rigid conducting…

A: Entropy defines the randomness of the system. It measures the thermal energy of a system per unit…

Q: To obtain the change in entropy in an irreversible process, we have to calculate S along a…

A: Given statement is FALSE.

Q: An ideal gas initially has a macrostate (temperature, volume, pressure) that corresponds to W…

A: Initial microstates ri=WTemperature T=300 Krf=.2W (a The change in entropy We know that entropy…

Q: uestion: 10 cubic meters of air at 300 psia and 400 deg F is cooled to 140 deg F at constant volume.…

A: Hi and thanks for the questions. Since you have asked question with multiple sub-parts, as per our…

Q: 2-An automobile wheel contains air with a pressure of 3x10 'Pa at 25 ° C. The sipop cover was…

A: Equilibrium is a state of a body at which there is no disturbance in a system neither inside nor…

Q: The most central principle is the second law: Entropy tends to increase. Explain this ?

A: The most central principle is the second law because the second law of thermodynamics quantifies the…

Q: Example (8.1): Volume of 0.14 m³ of air at 0.2 bar and 90°C is compressed to 0.014 m³ according to…

A: Given Data V1=0.14m3 P1=0.2bar T1=90C=363K PV1.3=constant At state 2…

Q: Question attached

A: Given: State1: P1=4 MPaT1=400 °C State 2: P2=0.1 MPa

Q: Define a new property called entropy to quantify the second-law effects.

A: Chemical and physical changes in a system may be accompanied by either an increase or a decrease in…

Q: Calculate the change in entropy when 0.2 m3 of air at 1 bar and 59.9 °C is compressed to 0.051 m3…

Q: Argon gas enters an adiabatic compressor at 100 kPa, 25 ∘C, 27 m/s and exits at 1 MPa, 550 ∘C, 100…

Q: Show the whole solution and conversion. At 1.3 MPa, a mixture of steam and water has an entropy of…

Q: Calculate the change in entropy of air, if it is throttled from 5 bar, 27°C to 2 bar adiabatically

A: Data given- P1 = 5 bar T1 = 27°C = 300 K P2 = 2 bar Throttling process

Q: One-quarter Ibmol of oxygen gas (O2) undergoes a process from p1 = 20 lb/in?, T1= 500°R to p2 = 150…

Q: 2B-1. Get the volume, enthalpy, entropy, and internal energy of 10 lbm psia and 500 F. [V = 162.07…

A: Given:mass of water, m=10 lbmPressure of water, P=35 psiaTemperature of water, T=500 F

Q: A. (5) determine the power output of the turbine in kW. B. (5) determine the rate of entropy…

Q: C. Steam at 173.46 °C with a specific entropy of 6 kJ/kg-K undergoes a polytropic process with n=…

A: Given Data The initial temperature is: T1=173.46°C=446.46 K The polytropic constant is: n=1.18 The…

Q: If 2 pmole of a hydrogen (H2) are throttled adiabatically (Q = 0) from an initial volume of 4 ft³…

A: Given Data: The number of moles of the hydrogen is n = 2 pmol. The initial volume of the gas is Vi…

Concept explainers

Entropy

In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its molecules are in random motion, vibrating and colliding with each other. During this scenario, the probability of finding a molecule at a particular place twice, is low. Under this circumstance, the system can be said to be in a high entropy.

Question

a. Find an appropriate expression for the change in entropy in the following
two cases:
1) S=S(T, V)
2) S= S(T, P)
Where: S is entropy. T is temperature. V is volume. P is
pressure
b. Prove the following two thermodynamic property relationships
or).
OT
T²
(S)イ),
oP
Where: T. P. V are temperature. pressure and volume, respectively.
C. and C, are specific heats at constant volume and constant pressure, respectively.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

Can someone explaine this problem please? A closed system contains an ideal gas, which molecular weight is W = 63.47 kg/kmol, and its standard state entropy is S0 = 0. The system undergoes the following cycle: at state 1 the temperature is 295.15 K, the pressure is 95 kPa (absolute), and the entropy is 46.450 J/(kg/kmol). The gas is compressed polytropically at n = 1.46 until the specific volume is 10 times lower than that at state 1 ( state 2 ). Then 85651.7 J/kg of heat is added at constant specific volume ( state 3). After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the next process the system undergoes isentropic expansion (and reaches state 5 ). Finally there is a constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each cycle point; b) the p-V and T-s diagrams of cycle; c) the heat added to the system; d) the heat rejected by the system; e) the cycle net work; f) the mean effective pressure; g)…
When a system undergoes a Isochoric process (P = 5 bar), the change in internal internal is 5 kJ and temperature during the process is 15°C. Then its change in entropy would be approximately 1.736 kJ/kgK Select one: True False
Water of m= 5 kgm. P 1 MPa, H 10,280 kJ. Determine temperature, total volume, totalinternal energy and total entropy.
Select the best answer and explain why you chose that answer. Entropy as a state function can be proven, by showing that the integral of the change in entropy for a cyclic process is zero. Considering the Carnot cycle, step 1 is a reversible isothermal expansion from A to B at Th. The entropy change can be actually calculated using: (A) qh/Th (B) zero (C) nR ln (Vb/Va) (D) nRTh ln (Vb/Va)
0.25kg/s of Carbon Dioxide (R=0.1889 kJ/kg-K, k=1.289) undergo a certain process in a close system from initial volume and temperature of 268cm^3 and 134K to a final volume and temperature of 804cm^3 and 402K respectively. a. Work done by/on the system (kJ/s) b. Change in Entropy (kW/K) c. Heat added/rejected by the system (kJ/s) Please include the figure, clearly state the formulas used and the complete solution. Thank you!
An oxygen gas R = 0.2598 KJ/kg°k and k = 1.395. If 4 kg of oxygen undergo a reversible non flow constant pressure process from initial volume =1.2 cubic meter and initial pressure = 690 kPa to a state where final temperature = 600°C. What is the initial temperature ? Determine the constant volume specific heat. need a complete solution, symbol, and cancellation:)
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.
Q1 Explain the following sub steps breifly. (a) Explain The Following Terms: Isolated System, Open System And Closed System And Give Example Where Ever Possible (b)Explain the following with the help of diagram: Extensive properties, Intensive properties and specific properties. Enthalpy, Entropy and Internal energy. (c) What Happens If Petrol (Gasoline) Is Used In A Diesel Engine, Siesel Engine Will Work?
1 x 10^6 cm³ of Nitrogen at a pressure of 1 N/mm² and temperature 300°C, expands 10 times its original volume and its final temperature will be 25°C. Calculate (i) mass (ii) change in entropy by considering temperature, volume, entropy relationship (iii) change in entropy by. considering temperature, pressure, entropy relationship (iii) change in entropy by considering pressure, volume, entropy relationship.
A reversible non-flow, constant volume process decreases the internal energy by 400 kJ for 3 kilograms of gas. For the process, determine (a) the work, (b) the heat, (c) the change in entropy if the initial temperature is 250 degrees Celsius. Use R = 450 J/kg-K and k = 1.35.
Two vessels A and B of different sizes are connected by a pipe with a valve. Vessel A contains 142 L of air at 2,767.92 kPa, 93.330C. Vessel B, of unknown volume, contains air at 68.95 kPa, 4.440C. The valve is opened and, when the properties have been determined, it is found that pm = 1378.96 kPa, tm = 43.330C. What is the volume of vessel B?
The entropy of an ideal gas depends on both T and P. The function s° represents only the temperature-dependent part of entropy.