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Solutions Thermodynamics Test Term 2, 2015 Define: i. System: the object(s) or physical situation that is being studied Boundary: is what separates the system from the environment Environment: everything external to the system that has bearing on the system ii. Thermodynamic process A process that changes from one state (p,v, and t) to another state examples include: isobaric, isovolumetric, isothermal, adiabatic, and polytrophic. iii. Adiabatic A thermodynamic process where no heat is exchanged between the system and the environment, ΔQ = 0 iv. Polytrophic Thermodynamic process where the relationship between pressure, volume is 𝑝𝑣 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 v. The four laws of thermodynamics 0th Law of thermodynamics: There exists a useful scalar quantity called temperature that is a property of all thermodynamics systems. When 2 bodies (A and B) are in thermal equilibrium (same temperature) with a third body (C), then all bodies (AC, BC, AB) are in equilibrium with each other. 1st Law of thermodynamics: Energy is neither created or destroyed, energy can only change form. 𝑑𝑈 𝑑𝑄 There is exists a useful quantity called internal energy. 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 𝑜𝑟 ∫ 𝑑𝑡 = ∫ 𝑑𝑡 − ∫ 𝑑𝑊 𝑑𝑡 2nd Law of thermodynamics: Heat must flow from hot to cold. The second law deals with entropy or the measure of the quality of energy in natural processes and states if a process can happen in nature. 3rd Law of thermodynamics: it is impossible by any procedure, no matter how idealized, to reduce any system to absolute zero temperature in finite number of operations. List as many as you know: i. System variables Pressure, Volume, Temperature, Mass/Moles and Internal Energy ii. Equations of state ∆𝑈 = ∆𝑄 − ∆𝑊 𝑝𝑣 = 𝑛𝑅𝑇 𝑝𝑣 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎 ) (𝑣 − 𝑏) = 𝑅𝑇 𝑣2 𝑝𝑉𝑚 𝑍= 𝑅𝑇 (𝑝 + iii. Thermodynamic processes A process that involves heat transfer, work or a change in internal energy of a system. 1. Sketch the following processes on a pV diagram, Isobaric, isovolumetric, isothermal and adiabatic Pressure A Process A to B: Isothermal Process A to C: Adiabatic Process A to D: Isovolumetric B D Process B to D: Isobaric C Volume 2. A gas expands from an initial state where p1 = 400 kPa and V1 = 0.2 m3 to a final state where p2 = 200 kPa. The relationship between pressure and volume during the process is pV = constant. Sketch the process on a p-V diagram and determine the work, in kJ. Pressure p1V1 p2V2 Volume 𝑝𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝1 𝑉1 = 𝑝2 𝑉2 𝑝1 𝑉1 𝑉2 = 𝑝2 400 𝑘𝑃𝑎 × 0.2 𝑚3 = 𝑚3 200 𝑘𝑃𝑎 𝑽𝟐 400 × 0.2 𝑚3 = 𝟎. 𝟒 𝒎𝟑 200 𝑉 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉 1 𝑊 = ∫ 𝑝 𝑑𝑉 = ∫𝑉 2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ln = 𝑝1 𝑉1 ln 𝑉2 𝑉1 𝑑𝑉 𝑉2 𝑉1 𝑾 = 400(0.02) ln 0.4 = 𝟓𝟓. 𝟓 𝑱𝒐𝒖𝒍𝒆𝒔 0.2 3. One thousand kg of natural gas at volume of 10 m3 and 255 K is stored in a tank. If the pressure, p, specific volume, v and temperature, T, of the gas are related by the following expression 𝑝 = [(5.18 𝑥 10−3 ) { 𝑘𝑃𝑎 𝑚3 } 𝑘𝑔 𝐾 𝑇 ]− 𝑚3 (𝑣 − 0.002668 ) 𝑘𝑔 (8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9 { } 𝑣2 𝑘𝑔2 where v is in m3/kg, T is in K, and p is in bar, determine the pressure of the tank. 𝑝 = [(5.18 𝑥 10 −3 ) 𝑘𝑃𝑎 𝑚3 { } 𝑘𝑔 𝐾 𝑝 = [(5.18 𝑥 10−3 ) 𝑝 = [(5.18 𝑥 10−3 ) 𝑉 𝑚 𝑣= 10 𝑚3 𝑚3 = 0.01 1000 𝑘𝑔 𝑘𝑔 (8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9 ]− 2 { 𝑘𝑔2 } 𝑚3 𝑚3 𝑚3 (0.01 − 0.002668 ) (0.01 ) 𝑘𝑔 𝑘𝑔 𝑘𝑔 255 𝐾 (8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9 255 𝑘𝑃𝑎 ]{ }− 2 { 𝑘𝑔2 } (0.01 − 0.002668 ) 𝑚3 (0.01 ) 𝑘𝑔 (8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9 255 𝑘𝑃𝑎 ]{ }− { } 𝑚9 (0.01 − 0.002668 ) 𝑘𝑔2 (0.0001 ) 𝑘𝑔2 𝑝 = [(5.18 𝑥 10−3 ) (8.91 𝑥 10−3 ) 𝑘𝑃𝑎 255 𝑘𝑃𝑎 ]{ }− { } (0.01 − 0.002668 ) (0.0001 ) 𝑝= [ 𝒑= [ 𝑣= (89.1 ) 𝑘𝑃𝑎 1.32 𝑘𝑃𝑎 ]{ }− { } (0.007332 ) 180.0 ]{ 𝑘𝑃𝑎 }− (89.1 ) 𝑘𝑃𝑎 { } = 𝟗𝟏. 𝟏 𝒌𝑷𝒂 4. One mole of a monatomic gas goes in a cycle ABCA as in figure 3. Let (P0, V0, T0) be the parameters at A and 2T0 the temperature at B. Solve algebraically: i. The values of TC and PB ii. The work done in each part: AB, BC and CA iii. The heat input in each part: AB, BC and CA Given: State A B C Pressure P0 P0 Volume V0 V0 2V0 Temperature T0 2T0 Find: PB 𝑝𝑉 = 𝑛𝑅𝑇 Ideal Gas Law allows us to write 𝑃0 𝑉0 = 𝑃𝐵 𝑉𝐵 (1) Also, at A 𝑃0 𝑉0 = 𝑛𝑅𝑇0 (2) From the table above at B 𝑃𝐵 𝑉0 = 𝑛𝑅(2𝑇0 ) (3) Thus, (2) ÷ (3) yields 𝑃0 𝑉0 𝑛𝑅𝑇0 = 𝑃𝐵 𝑉0 𝑛𝑅(2𝑇0 ) Canceling 𝑃0 1 = 𝑜𝑟 𝑷𝑩 = 𝟐𝑷𝟎 𝑃𝐵 2 Given State A B C Pressure P0 𝟐𝑷𝟎 P0 Volume V0 V0 2V0 Temperature T0 2T0 Find: TC 𝑝𝑉 = 𝑛𝑅𝑇 Ideal Gas Law allows us to write 𝑃0 𝑉0 = 𝑃𝐶 𝑉𝐶 (1) Also, at A 𝑃0 𝑉0 = 𝑛𝑅𝑇0 (2) From the table above at C 𝑃0 (2𝑉0 ) = 𝑛𝑅𝑇𝐶 (3) Thus, (2) ÷ (3) yields 𝑃0 𝑉0 𝑛𝑅𝑇0 = 𝑃0 2𝑉0 𝑛𝑅𝑇𝐶 Canceling 1 𝑇0 = 𝑜𝑟 𝑻𝑪 = 𝟐𝑻𝟎 2 𝑇𝐶 State A B C Pressure P0 𝟐𝑷𝟎 P0 Volume V0 V0 2V0 Temperature T0 2T0 𝟐𝑻𝟎 𝟐𝑷𝟎 𝟐𝑻𝟎 𝟐𝑻𝟎 Find the work done and heat input in each process Process A -> B ∆𝑈 = ∆𝑄 − ∆𝑊 𝑉𝐵 𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑝 𝑑𝑉 𝑉𝐴 Inputting values 𝑽𝟎 𝑾 = ∫ 𝒑 𝒅𝑽 = 𝟎 𝑽𝟎 ∆𝑈 = 3𝑛𝑅(𝑇𝐵 − 𝑇𝐴 ) 3(1)𝑅(2𝑇0 − 𝑇0 ) = 2 2 3𝑅𝑇0 ∆𝑈 = 2 Plugging into Energy equation 3𝑅𝑇0 = ∆𝑄 − 0 2 or ∆𝑸 = 𝟑𝑹𝑻𝟎 𝟐 State A B C Pressure P0 𝟐𝑷𝟎 P0 Volume V0 V0 2V0 Temperature T0 2T0 𝟐𝑻𝟎 𝟐𝑷𝟎 𝟐𝑻𝟎 𝟐𝑻𝟎 Process B -> C ∆𝑈 = ∆𝑄 − ∆𝑊 𝑉𝐶 𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑝 𝑑𝑉 𝑉𝐵 Because pV = constant 𝑉𝐶 𝑊= ∫ 𝑉𝐵 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑉 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ln 𝑉𝐶 𝑉𝐵 Because 𝑝𝐵 𝑉𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉𝐶 = 𝑝𝐵 𝑉𝐵 ln 𝑉𝐵 Putting in values for pB, VB and VC 2𝑉0 = 2𝑝0 𝑉0 ln 𝑉0 𝑾 = 𝟐𝒑𝟎 𝑽𝟎 𝐥𝐧 𝟐 ∆𝑈 = 3𝑛𝑅(𝑇𝐶 − 𝑇𝐵 ) 3(1)𝑅(2𝑇0 − 2𝑇0 ) = 2 2 ∆𝑼 = 𝟎 Plugging into Energy equation 0 = ∆𝑄 − 2𝑝0 𝑉0 ln 2 or ∆𝑸 = 𝟐𝒑𝟎 𝑽𝟎 𝐥𝐧 𝟐 State A B C Pressure P0 𝟐𝑷𝟎 P0 Volume V0 V0 2V0 Temperature T0 2T0 𝟐𝑻𝟎 𝟐𝑷𝟎 𝟐𝑻𝟎 𝟐𝑻𝟎 Process C -> A ∆𝑈 = ∆𝑄 − ∆𝑊 𝑊 = ∫ 𝑝 𝑑𝑉 Because pressure is constant C -> A 𝑉𝐴 𝑊 = 𝑝 ∫ 𝑑𝑉 = ∫ 𝑑𝑉 𝑉𝐶 = 𝑝(𝑉𝐴 − 𝑉𝐶 ) = 𝑝0 (𝑉0 − 2𝑉0 ) 𝑾 = −𝒑𝟎 𝑽𝟎 ∆𝑈 = 3𝑛𝑅(𝑇𝐴 − 𝑇𝐶 ) 3(1)𝑅(𝑇0 − 2𝑇0 ) = 2 2 3𝑅𝑇0 ∆𝑈 = − 2 Plugging into Energy equation − 3𝑅𝑇0 = ∆𝑄 − (−)𝑝0 𝑉0 2 or ∆𝑸 = − 𝟑𝑹𝑻𝟎 − 𝒑𝟎 𝑽𝟎 𝟐