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Q1.Prove that PV^Y=C for reversible adiabatic process.

 Q1.Prove that PV^Y=C for reversible adiabatic process. Solution:        δQ = dU + PdV Reversible Adiabatic: δQ=0              (:: There is no heat interaction between                                                                   system and surroundings = adiabatic  dU + dV = 0 dU = -PdV m(Cv)dT = -PdV  ----------------------------------------------(1) Now we want 'Y' , so, We know      H = U + PV                  dH = dU + PdV + VdP                  dH =   δQ + VdP (:: There is no heat interaction δQ =0)              dH = VdP       m(Cp)dT = VdP       ------------------------------------------(2) Dividing (2) by (1)     Y = -V/P(dP/dV)                              (:: Y= Gama=Cp/Cv)     Y = -V/dV(dP/P) Integral both side     ∫YdV/V  +  ∫dP/P  = ∫0 Y(lnV)  + lnP = lnC ln PV^Y = lnC PV^Y = C  ------------Since, this has been found for  δQ =0 so, only                                                     apply it for reversible adiabatic 
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Q. A 1 gm nitrogen undergoes a following sequence of process in a piston cylinder arrangement.(1) An adiabatic expansion in which volume double. (2) A constant pressure process in which volume is reduced to its initial state. Represent the cycle on p -v diagram, calculate net work done by the gas.

 Q. A 1 gm nitrogen undergoes a following sequence of process in a piston cylinder arrangement.(1) An adiabatic expansion in which volume double. (2) A constant pressure process in which volume is reduced to its initial state. Represent the cycle on p -v diagram, calculate net work done by the gas. Assuming  t1 = 150°C , P1 = 5 atm, Y = 1.4, R = 297 J/kg K. Solution: t1 = 150°C                   T1 = 150 + 273 = 423 K                   P1 = 5 × 1.01325 = 5.06625 bar Process (1-2): T2/T1 = (V1/V2)^Y-1 T2/423 = (V1/2V1)^1.4-1 T2 = 320.5K Process(2-3): T3/T2 = V1/V2 T3/T2 = V1/2V1 T3 = T2/2 T3=320.5/2=160.25K Now : W(net) = W(1-2) + W(2-3) + W(3-1) W(net) = P1V1 -P2V2/Y-1  + P3V3 -P2V2 + 0 W(net) = mR(T1-T2)/Y-1   + mR(T3-T2) W(net) = 10^-3 × 297(423 - T2)/1.4 -1  + 10^-3×297(T3 - T2) W(net) = 10^-3 × 297(423 - 320.5)/0.4 + 10^-3×297(160.25 - 320.5) W(net) = 28.86 J
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Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find u and the max height stone will attain.

Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find 'u' and the max height stone will attain. Solution:                              s = ut + 1/2 at^2                    0 = u(5) + 1/2(-9.81)(5)^2                                       (9.81/2)(5×5/5) = u                    u = 4.905×5                    u= 24.55 m/sec Now we find maxi height                       V^2 = u^2 + 2as                        0   = (24.55)^2 + 2(-9.81)s                       24.55×24.55/2(9.81) = s                         s = 30.72 m
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Question 2. An insulated pressure vessel divided into two parts. One part of vessel is occupiedby an ideal gas at a pressure P1, volume V1 & temp T1.The other part occupied by the same gas at a pressure P2, volume V2, & temp T2. The partition is removed and the two parts are mixed. Then show that the final pressure P3 & final T3 is

Thermodynamics  Chapter: First law of Thermodynamics  Question 2. An insulated pressure vessel divided into two parts. One part of vessel is occupiedby an ideal gas at a pressure P1, volume V1 & temp T1.The other part occupied by the same gas at a pressure P2, volume V2, & temp T2. The partition is removed and the two parts are mixed. Then show that the final pressure P3 & final T3 is   Solution:  Same ideal gas  PV = mRT   ...................................................................(1) Mass Conservation : m1+m2 = m3      ......................................(2) Energy conservation: m1 u1 + m2 u2 = m3 u3                                       m1 Cv T1 + m2 Cv T2 = m3 Cv T3                                 m1 T1 + m2 T2 = m3 T3 Use PV =  mRT,                 P1 V1/ R + P2 V2/R = P3 V3/R "R" Is canceled out with each other, so P1 V1 + P2 V2 + = P3 V3 P3 = P1 V1 + P2 V2/V3                                                            (:: V3= V1+V2 ) P3 = P1 V1+ P2
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Question 1. Air at initial temp of 300K & initial volume of 0.002 m^3 is contain in a piston cylinder arrangement. Initially a spring is attached to the piston having a spring constant of 10 k N/m is touching the piston but no force exerting on it. Now heat is added to the air & it expended slowly to occupy a final volume of 0.03 m^3 and the area of piston is 0.02 m^2. The atmospheric pressure assume to be 100KPa. Then determine the final pressure inside the cylinder.

THERMODYNAMICS CHAPTER: FIRST LAW OF THERMODYNAMICS Question 1. Air at initial temp of 300K & initial volume of 0.002 m^3 is contain in a piston cylinder arrangement. Initially a spring is attached to the piston having a spring constant of 10 kN/m is touching the piston but no force exerting on it. Now heat is added to the air & it expended slowly to occupy a final volume of 0.03 m^3 and the area of piston is 0.02 m^2. The atmospheric pressure assume to be 100KPa. Then determine the final pressure inside the cylinder. Solution  T1 = 300K V1 = 0.002 m^3  V2  = 0.03 m^3 Spring constant (K) = 10 KN/m A = 0.02 m^2 Patm = 100 KPa  Psp = pressure applied by spring Patm = pressure applied by atmosphere P1 = inside pressure of piston at position one  P2 = inside pressure of piston at position two   P1 =  Patm + Psp1   ……………..............................(0)                                                   ( Psp1 = pressure of  spring at position one  is zero =0) P1 = Patm = 100 KPa ………
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Question 3. What does a temperature difference of 10 on °C scale correspond to on a Fahrenheit scale.

THERMODYNAMICS  CHAPTER: ZEROTH LAW OF THERMODYNAMICS Question 3. What does a temperature difference of 10 on °C scale corresponding toon a Fahrenheit scale. Solution:-                            Δtc = 10              (:: Δ= d/dt) We know     tf = 9/5tc + 32   ....................(0) Applying Δ on both sides of equation (0),                            Δtf  = 9/5Δtc   ....................(1)                                                                      (:: here 32 is zero because it is constant   Put Δtc = 10  in equation (1)                     (:: Δ this use for differentiation  Δtf  = 9/5(10) Δtf = 18°F *at any temperature difference in °C find by Δtf = 1.8(Δtc) equation 
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Question 1. A spherical balloon of 1m diameter contains a gas at 150kpa. The gas inside balloon is heated until pressure reaches 459kpa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas.

THERMODYNAMICS QUESTION SOLUTION, CHAPTER. ENERGY INTERACTION. Question 1. A spherical balloon of 1m diameter contains a gas at 150 KPa. The gas inside balloon is heated until pressure reaches 450 KPa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas. Solution: D1= 1 m    P1= 150 KPa                                           V = (4/3)  π R^3 =    πD^3/6        P2 = 450 KPa                                             dV = ( π/6)  (3D^2)    δD P directly proportional D^3                         dV = ( π/2) (D^2)   δD P = k(D^3)                                                   W =  ∫ P dV  --------------(0) P1=k(D1^3)                                                  150 = k(1)^3                                              k = 150 KN/m^3 ----------------(1)                P2 = k(D2^3) 450 = 150(D2)^3 D2 = 1.44 m -----------------------(2) W =  ∫ P dV  W =  ∫ (kD^3)   ( π/2) (D^2)   δ
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