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Question:Two different vessels are filled with 100g of water at 20 C and 50g of ethyl alcohol at 80 C respectively. The specific heat of water and ethyl alcohol are 4.0 and 2.0 J/g C respectively. Answer: 1.86 J/K

Answers:Initially, we have two different solutions at different temperatures. Once they are mixed and equilibrated, the final state of the system is a homogeneous solution at some temperature intermediate between the initial temperatures of the two solutions. (I assume that the solutions are mixed in an insulated container, so that no heat is lost from the system during this process.) I will also assume that water and ethanol mix ideally, so that there is no net change in total volume of the solutions when they are mixed, and that the enthalpy of mixing is zero. There are two things going on here. First, there is an entropy change associated with equilibrating the temperatures of the two solutions. Second, there is an entropy change associated with the mixing of the two solutions. Because entropy is a function of state, we can choose to break apart the problem into tractable processes, so long as the initial and final states of the system correspond to the initial and final states noted above. First, calculate the equilibrium temperature of the combined system. Because this process occurs in a thermally isolated system, energy is conserved, and the heat energy gained by the cold water is equal to the heat lost by the hot ethanol. The heat capacity of a system is defined as C = dq/dT, T is the temperature, and q is the heat energy, so dq = C*dT If we assume the heat capacity is constant (doesn't depend on temperature), this integrates to: q = C* T In this question, the two solutions have different heat capacities, so we have: q_total = 0 = q_w + q_e = m_w*C_w*(T_eq - T_w) + m*e*C_e*(T_eq - T_e)) where q_w, m_w, C_w, and T_w are the change in heat, mass, specific heat capacity, and initial temperature of the water, and q_e, m_e, C_e, and T_e are the corresponding parameters for the ethanol. T_eq is the equilibrium temperature. Solving for T_eq gives: T_eq = (m_w*C_w*T_w + m_e*C_e*T_e)/(m_w*C_w + m_e*C_e) The equilibrium temperature is just the heat-capacity-weighted average of the temperatures of the two solutions. Plugging in the numbers for this problem gives: T_eq = (100gm*4J/(gm* C)*20 C + 50gm*2J/(gm* C)*80 C)J)/(100gm*4J/(gm* C) + 50gm*2J/(gm* C)) T_eq = 32 C This thermal equilibration is an example of an irreversible process, but because the entropy is a state function, it doesn't matter how we get to the final state. We can therefore use a reversible path to calculate the entropy changes, which simplifies things. We can calculate the entropy change involved in reversibly taking the water from T_w to T_eq, and add that to the entropy change involved in taking the ethanol from T_e to T_eq, and we will end up with the desired answer. For a reversible process: dS = dq/T But from above, we know that dq = C*dT, so: dS = C*dT/T If C is constant, S = C*ln(T_final/T_initial) In this case, the total entropy change is: S_thermal = S_e + S_w = m_w*C_w*ln(T_eq/T_w) + m_e*C_e*ln(T_eq/T_e) Note that in this case, we must express the temperatures in absolute terms (in kelvins) because we are dealing with actual temperatures, not just temperature changes. S_thermal = (100gm*4J/(gm* C))*ln(305/293) + (50gm*2J/(gm* C))*ln(305/353) S_thermal = 1.44 J/K. This would be the entropy change if the two solutions were brought into thermal contact in an insulated container, and allowed to thermally equilibrate, but not physically mix. The next step is to calculate the entropy change due to mixing. Assuming ideal mixing, for this 2-component system, the entropy of mixing is given by: S_mix = -n*R*[x_w*ln(x_w) + x_e*ln(x_e)] where n is the total number of moles of water and ethanol, R is the universal gas constant x_w is the mole fraction of water in the final mixture x_e is the mole fraction of ethanol in the final mixture Because this is just a 2-component system, x_w + x_e = 1, so x_e = 1 - x_w The molecular mass of water is 18.02 gm/mol and the molecular mass of ethanol (C2H5OH) is 46.07 gm/mol. The mass of water and ethanol in this problem therefore correspond to: n_w = 100gm/(18.02 gm/mol) = 5.55 mol n_e = 50gm/(46.07 gm/mol) = 1.08 mol n = n_w + n_e = 6.635 mol x_w = 5.55mol/(5.55 + 1.08)mol = 0.836 x_e = 1 - x_w = 0.164 We know that R = 8.314 J/(mol*K). Plugging all this into the expression for S_mix gives: S_mix = -(6.635 mol)*(8.314 J/(mol*K))*[0.836*ln(0.836) + 0.164*ln(0.164)] S_mix = 24.62 J/K The total entropy change upon mixing these solutions is given by the sum of the thermal entropy change and the entropy of mixing: S_tot = S_thermal + S_mix = 1.44 J/K + 24.62 J/K = 26.06 J/K. The answer you were given is incorrect.

Question:i need help! if you can please help me with any of these three please. just put the number you are working on, on the side THANX ALOT :) 1. how do you calculate the entropy change during a change of state equilibrium. 2. why is the entropy of a substance higher in the liquid state than in the solid state? 3. the molar enthalpy of fusion of water is 6.009 kj/mol at 0 degrees Celsius. explain what this statement means.

Answers:1. Equilbrium G = H - T S, equlibrium gibbs free energy, G = 0 0 = H - T S S = H/T 2. Molecules movement are more order in solid state compared to liquid state, thus entropy of solid is lower. 3. Enthalpy of fusion of water is 6.009 kj/mol at 0 degrees Celsius It means for every 1 mol of water at 0 deg. C, water will release 6.009 kJ to change its phase to solid.

Question:I am trying to find the change in entropy if 5 moles of a gas with C_v= 5/2 R is heated from 350 K to 550 K adiabatically. I'm confused about what I'm supposed to integrate to find the change in entropy, since for an adiabatic process there is no Q and I have been using dS= dQ/T Do I calculate the volume based on the change in temperature and somehow use this in another equation (the pressure is 1 atm)? Which equation do I integrate? Sorry, to clarify the pressure is initially at 1 atm

Answers:There is something that puzzles me about this question. if the process is adiabatic, the gas is being heated by compression (DeltaE = W since q = 0) and, as your reasoning correctly implies, the process is reversible and there is no change in entropy. You cannot heat a gas adiabatically at constant pressure, so I don't understand your reference to pressure of 1 atm.

Question:An irreversible adiabatic expansion of one mole of an ideal gas occurs. The initial volume and temperature are (V1; T1) and the final volumes and temperatures are (V4; T4). Compute the entropy change dS for this process by using a path that consists of three steps: (a) a reversible adiabatic compression that decreases the volume to V2 and increases the temperature to T2, (b) a reversible isothermal process that takes the system from (V2; T2) to (V3; T3 = T2) and (c) a reversible adiabatic expansion that increases the volume and decreases the temperature and takes the system from (V3; T2) to the final state (V4; T4). Show your results for the intermediate steps in this path. A complete answer for the overall entropy change should express dS in terms of the properties of the initial and final states.

Answers:a) For a reversible process, dS = q/T where q is the infinitesimal amount of thermal energy added to the system at temperature T from the surroundings. For an adiabatic process, q = 0, so dS = 0 and S(1->2) = 0 Nevertheless, we will need to know how the volume changed in this process so that we can express the total entropy change in terms of the initial and final states. Let the entropy be a function of volume and temperature, so S = S(V,T) Take the total differential of the entropy: dS = ( S/ V)_T dV + ( S/ T)_V dT Identify that ( S/ T)_V = Cv/T dS = ( S/ V)_T dV + (Cv/T) dT Using a Maxwell's Relation (see source), we can write ( S/ V)_T = ( P/ T)_V, so: dS = ( P/ T)_V dV + (Cv/T) dT For 1 mole of an ideal gas, we know that P = RT/V ( P/ T)_V = R/V So: dS = (R/V) dV + (Cv/T) dT Integrate: S = R*ln(V_final/V_initial) + Cv*ln(T_final/T_initial) We already know that S(1 -> 2) = 0, so: 0 = R*ln(V2/V1) + Cv*ln(T2/T1) R*ln(V2/V1) = Cv*ln(T1/T2) R*ln(V2) = Cv*ln(T1/T2) + R*ln(V1) ----------------- b) Go back to the expression we derived above: S = R*ln(V_final/V_initial) + Cv*ln(T_final/T_initial) This next path is specified to be isothermal, so dT = 0, and T_final = T_initial, (T2 = T3) and the temperature term is zero. S(2->3) = R*ln(V3/V2) = R*ln(V3) - R*ln(V2) --------------- c) Step 3 is another adiabatic process, so S(3->4) = 0 and 0 = R*ln(V4/V3) + Cv*ln(T4/T2) R*ln(V4/V3) = Cv*ln(T2/T4) R*ln(V3) = R*ln(V4) - Cv*ln(T2/T4) The total entropy change is S(1->4) = S(1->2) + S(2->3) + S(3->4) = 0 + R*ln(V3/V2) + 0 S(1->4) = R*ln(V3) - R*ln(V2) Plugging in the expressions we had above for ln(V2) and ln(V3) S(1->4) = R*ln(V4) - Cv*ln(T2/T4) - Cv*ln(T1/T2) - R*ln(V1) S(1->4) = R*ln(V4/V1) + Cv*ln(T4/T1) Note that because entropy is a state function, it doesn't matter what path we take to get from an initial state to a final state. We could have gotten to this result by taking a reversible path directly from state 1 to state 4 (T1,V1) to (T4, V4). Going back to our general expression for the entropy change for a reversible process: S = R*ln(V_final/V_initial) + Cv*ln(T_final/T_initial) Plug in state 4 for the final state and state 1 for the initial state, and we get: S(1 ->4) = R*ln(V4/V1) + Cv*ln(T4/T1)

From Youtube

Entropy and Second Law of Thermodynamics (3) :Physics: Entropy and the Second Law of Thermodynamics. Entropy as a state function. General formulas for calculating entropy change. How to calculate entropy change for phase change or temperature change problems. The Second Law of Thermodynamics; reversible vs. irreversible processes. Entropy change in isothermal processes, adiabatic processes, and adiabatic free expansions This is a recording of a tutoring session, posted with the student's permission. These videos are offered on a "pay-what-you-like" basis. You can pay for the use of the videos at my website: www.freelance-teacher.com For printable documents containing the problem and handout discussed in this video series, go to my website. For a list of all the available video series, arranged in suggested viewing order, go to my website. For a playlist containing all the videos in this series, click here: www.youtube.com (1) Entropy as a state function. General formulas for entropy change (2) Continued. How to calculate entropy change for phase change or temperature change problems (3) Continued. The Second Law of Thermodynamics; reversible vs. irreversible processes (4) Continued. An entropy problem (5) Continued (6) Continued (7) Continued. Entropy change in isothermal processes (8) Continued. Entropy change in adiabatic processes. Entropy change in adiabatic free expansions (9) Continued

Entropy and Second Law of Thermodynamics (1) :Physics: Entropy and the Second Law of Thermodynamics. Entropy as a state function. General formulas for calculating entropy change. How to calculate entropy change for phase change or temperature change problems. The Second Law of Thermodynamics; reversible vs. irreversible processes. Entropy change in isothermal processes, adiabatic processes, and adiabatic free expansions This is a recording of a tutoring session, posted with the student's permission. These videos are offered on a "pay-what-you-like" basis. You can pay for the use of the videos at my website: www.freelance-teacher.com For printable documents containing the problem and handout discussed in this video series, go to my website. For a list of all the available video series, arranged in suggested viewing order, go to my website. For a playlist containing all the videos in this series, click here: www.youtube.com (1) Entropy as a state function. General formulas for entropy change (2) Continued. How to calculate entropy change for phase change or temperature change problems (3) Continued. The Second Law of Thermodynamics; reversible vs. irreversible processes (4) Continued. An entropy problem (5) Continued (6) Continued (7) Continued. Entropy change in isothermal processes (8) Continued. Entropy change in adiabatic processes. Entropy change in adiabatic free expansions (9) Continued