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Isobaric reversible ideal gas system

Isobaric reversible ideal gas system. The concepts deals with system in which the pressure of the system is constant. With the help of this animation user will be able to do expansion and contraction of the system at the constant pressure and check how does the graph of P v/s V is plotted.

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Isobaric reversible ideal gas system

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  1. Isobaric reversible ideal gas system The concepts deals with system in which the pressure of the system is constant With the help of this animation user will be able to do expansion and contraction of the system at the constant pressure and check how does the graph of P v/s V is plotted By Rachit Agarwal Department of chemistry IIT Bombay

  2. thermometer System piston weight Master layout Pressure tab graph volume tab Up and down arrow key Instruction for user Start and end button start Temperature tab Change in Internal energy temperature Change in enthalpy volume work Heat pressure end Heat, change in enthalpy, work and change in internal energy displays Instruction box Q W V P T

  3. Instructions for the animator: Working of the animation • A constant pressure will be maintained during the process. • User can choose a pressure from the list given (this list will be shown in the instruction box) • User can change either temperature or the volume and other parameter volume or temperature respectively will be changed by following the law PV= NRT (N=1) • The parameter can by changed by using the down and up arrow tabs • Both the volume and temperature have their limit. Volume 20 to 100 lts, temperature from 500K to 1000K , these limits are used only when the parameter is changed • Start button – used to start the process • End button – used to stop the process • The graph should be plotted simultaneously once the user click the start button till the user presses the end button, • for volume change :- for the change in 10 units the graph should be plotted • for temperature change :- for the change in 50 units the graph should be plotted select the pressure from the following 1atm 2atm 3atm 4atm start end

  4. Instructions for the animator: Plotting of the graph The x axis of the graph is the volume and the y axis of the graph is the pressure Scale X axis 1 cm = 20 lts of volume Y axis 1cm = 1atms of pressure 1 2 3 4 20 40 60 80 100 Eg P= 3 For T= 500 and V = 25.6 The red dot shows the mark For T = 900 and V = 46.1 The blue dot shows the mark And then both the points are joint by straight line, shown by black line

  5. Definition and rules for the users An isobaric is a process in which the pressure is maintained constant during the process, The work done by the process is –P(Vf-Vi) using the first law of thermodynamics the Q (heat) can be calculate as dU = dq + dw U= Cv∆T = Cv/r*P(Vf-Vi) here r= .0821 Q= P(Vf-Vi)(Cv/r + 1 ) Cv is taken as 2.5R (diatomic gas), R= 8.314 (S.I units) the enthalpy can be calculated as H = U + PV , therefore the change in enthalpy is equal to heat User will read the instruction in the instruction box and then follow the steps as said It will choose a pressure at which the system process has to be carried Then he will choose any of the two parameter i.e. volume or temperature and enter the initial value of one of the parameter, the value of the other parameter will come automatically by using the equation PV=nRT , here n =1 , both volume and temperature have limits (used only when the parameter is changed) Then the user will click the start button and can change the selected parameter using the arrows tab and reach his final destination and then click the end button to stop the process and to get the values of work, heat, change in internal energy and change in enthalpy

  6. Analogy / Scenario / Action / Example Initial stage Final stage 5 cm 2 cm Temperature gradient: the color code (R:G:B 0:0:0) signifies the temperature equal to 500 K and color code (R:G:B 0:255:255) signifies the temperature equal to 5000 K e.g. initially white thermometer (0:0:0) signifies the temperature = 500K and finally the blue thermometer (0:150:150) signifies temperature = 3000 K volume gradient: the height of the yellow box 5 cm signifies the volume equal to 100 lts and height = 0 cm signifies the volume equal to 0 lts e.g. initially height = 5 cm  100 lts and finally the height = 2 cm  40 lts

  7. Analogy / Scenario / Action / Example P = 2 atm P = 1 atm P = 3 atm P = 4 atm There are four figures each corresponds to a particular pressure the user will choose one the pressure from the list and the animator has to show the respective system e.g. the user choses the pressure as 4 atm, then animator has to choose the first fig

  8. Step 1 initial Instruction for user click the pressure tab start pressure end P

  9. Step 1 final Instruction for user click the pressure tab start pressure end P

  10. Step 2 initial Instruction for user select the pressure from the following 1atm 2atm 3atm 4atm start pressure end P

  11. Step 2 final 2 atm Instruction for user select the pressure from the following 1atm 2atm 3atm 4atm start pressure end P

  12. Step 3 initial 2 atm temperature volume Instruction for user click on the one of the tab from volume or temperature that you want to change start pressure end P

  13. Step 3.1 final 2 atm temperature volume Instruction for user click on the one of the tab from volume or temperature that you want to change start temperature pressure volume end V P T

  14. Step 4.1 (volume) initial 2 atm Instruction for user set the initial volume from where you want to start the process using the down and up arrow key, range 20lts to 100 lts with 10 lts as least count then click the start tab to start the process start temperature pressure volume end V P T

  15. Step 4.1 (volume) final 2 atm 20 lts Instruction for user set the initial volume from where you want to start the process using the down and up arrow key, range 20lts to 100 lts with 10 lts as least count then click the start tab to start the process start temperature pressure volume end V P T

  16. Step 5.1 (volume) initial 2 atm 20 lts Instruction for user set the initial volume from where you want to start the process using the down and up arrow key, range 20lts to 100 lts with 10 lts as least count then click the start tab to start the process start temperature pressure volume end V P T

  17. Step 5.1 (volume) final 2 atm 50 lts start ` end Instruction for user set the initial volume from where you want to start the process using the down and up arrow key, range 20lts to 100 lts with 10 lts as least count then click the start tab to start the process start temperature pressure volume end V P T

  18. Step 6.1 (volume) initial 2 atm 50 lts Instruction for user The temperature is calculated using the equation PV= NRT start temperature pressure volume end V P T

  19. Step 6.1 (volume) final 2 atm 50 lts Instruction for user 1218.02 K The temperature is calculated using the equation PV= NRT start temperature pressure volume end V P T

  20. Step 7.1 (volume) initial 2 atm 50 lts Instruction for user 1218.02 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  21. Step 7.1 (volume) final 2 atm 50 lts 1 2 3 4 20 40 60 80 100 Instruction for user 1218.02 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  22. Step 8.1 (volume) initial 2 atm 50 lts 1 2 3 4 20 40 60 80 100 Instruction for user 1218.02 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  23. Step 8.1 (volume) final 2 atm 60 lts 1 2 3 4 20 40 60 80 100 Instruction for user 1461.63 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  24. Step 9.1 (volume) initial 2 atm 60 lts 1 2 3 4 20 40 60 80 100 Instruction for user 1461.63 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  25. Step 9.1 (volume) final 2 atm 100 lts 1 2 3 4 20 40 60 80 100 Instruction for user 2436.05 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  26. Step 10.1 (volume) initial 2 atm 100 lts 1 2 3 4 20 40 60 80 100 Instruction for user 2436.05 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  27. Step 10.1 (volume) final 2 atm 100 lts 1 2 3 4 25316.68 J - 10126.67 J 35443.36 J 35443.36 J end 20 40 60 80 100 Instruction for user 2436.05 K Your process has started, please change the value of volume to change the state of the system and then click the end button once the process is over start Change in Internal energy temperature pressure Change in enthalpy volume work Heat end Q W V P T

  28. Step 3.2 final 2 atm temperature volume Instruction for user click on the one of the tab from pressure or temperature that you want to change start temperature pressure volume end V P T

  29. Step 4.2 (temperature) initial 2 atm Instruction for user set the initial temperature from where you want to start the process using the down and up arrow key, range 500K to 1000 K with 50 K as least count then click the start tab to start the process start temperature pressure volume end V P T

  30. Step 4.2 (temperature) final 2 atm 500 K Instruction for user set the initial temperature from where you want to start the process using the down and up arrow key, range 500K to 1000 K with 50 K as least count then click the start tab to start the process start temperature pressure volume end V P T

  31. Step 5.2 (temperature) initial 2 atm 500 K Instruction for user set the initial temperature from where you want to start the process using the down and up arrow key, range 500K to 1000 K with 50 K as least count then click the start tab to start the process start temperature pressure volume end V P T

  32. Step 5.2 (temperature) final 2 atm 600 K start ` end Instruction for user set the initial temperature from where you want to start the process using the down and up arrow key, range 500K to 1000 K with 50 K as least count then click the start tab to start the process start temperature pressure volume end V P T

  33. Step 6.2 (temperature) initial 2 atm 600 K start end Instruction for user The volume is calculated using the equation PV= NRT start temperature pressure volume end V P T

  34. Step 6.2 (temperature) final 2 atm 600 K start end Instruction for user 24.6 lts The volume is calculated using the equation PV= NRT start temperature pressure volume end V P T

  35. Step 7.2 (temperature) initial 2 atm 600 K start end Instruction for user 24.6 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  36. Step 7.2 (temperature) final 2 atm 600 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 24.6 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  37. Step 8.2 (temperature) initial 2 atm 600 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 24.6 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  38. Step 8.2 (temperature) final 2 atm 650 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 26.6 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  39. Step 9.2 (temperature) initial 2 atm 650 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 26.6 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  40. Step 9.2 (temperature) final 2 atm 1000 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 41 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  41. Step 10.2 (temperature) initial 2 atm 1000 K 1 2 3 4 start end 20 40 60 80 100 Instruction for user 41 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start temperature pressure volume end V P T

  42. Step 10.2 (temperature) final 2 atm 1000 K 1 2 3 4 start 10392.5 J - 4157 J 14549.5 J 14549.5 J end end 20 40 60 80 100 Instruction for user 41 lts Your process has started, please change the value of temperature to change the state of the system and then click the end button once the process is over start Change in Internal energy temperature pressure Change in enthalpy volume work Heat end Q W V P T

  43. Links for further studies

  44. Questionnaire A system started at P=2.5 atm and volume equal to 10 lts , then the following steps were performed on the system P constant , calculate the heat, work, change in internal energy, change in enthalpy in each case Q1) temperature is reduced to triced Q2) volume is doubled Q3) temperature is 500K Q4) volume is taken to 50 lts `

  45. Answers A2) a 8860.6,2531.6,6329,8860.6 b 8860.6,-2531.6,6329,8860.6 c 6329,-2531.6,6329,8860.6 d 6329.6,2531.6,6329,8860.6 A1) a 17721.3,-5063.3,12658,17721.3 b 17721.3,5063.3,12658,17721.3 c 12658,-5063.3,12658,17721.3 d 12658.3,5063.3,12658,17721.3 A4) a 25316,-10126.5,25316,35442.5 b 25316,10126.5,25316,35442.5 c 35442.5,10126.5,25316,35442.5 d 35442.5,-10126.5,25316,35442.5 A3) a 4061.4,-1624.5,4061.4,5685.9 B 4061.4,1624.5,4061.4,5685.9 c 5685.9,-1624.5,4061.4,5685.9 d 5685.9,1624.5,4061.4,5685.9 The order of the answer is as follows heat, work, change in internal energy, change in enthalpy in each case Answers A1) a A2) b A3) c A4) d

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