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Nitrogen in a cylinder - temperature, work and volume

  1. Sep 3, 2015 #1

    Rectifier

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    This problem was translated from Swedish, sorry for any grammatical errors.

    The problem

    A cylinder with nitrogen gas inside is closed by a tight piston (that can slide without friction against the walls of the cylinder). The piston weighs ##s=12.5 kg## and holds the ##m=28## grams of nitrogen gas in constant pressure. The base of the cylinder has the area of ##A=250 cm^2##. In the beginning the temperature is ##T_1 = 25 degrees C ##. The nitrogen gets ##Q=6kJ## of heat. Calculate:

    a) Temperature of the nitrogen after the heating process
    b) The work, W, that nitrogen does during the heating process
    c) Difference of the internal energy of the nitrogen during the heating process
    c) Difference in the volume of nitrogen
    Atmospheric pressure is ## P_0 = 100kPa ## ; the specific heat capacity at constant pressure and volume is ## C_p = 1.04 ## and ##C_v = 0.743 \ kJ / (kg * K) ##

    How I picture the problem:

    ektschsm.png


    ## Pv= nRT##
    ## P = \frac{F}{A} ##
    ## Q = \Delta U + W ##
    ## W = p \Delta V ##
    ## \Delta U = nc_v \Delta T ##

    ##P_0 = 100kPa##
    ##PN_1## = nitrogen pressure
    ##PN_2## = nitrogen pressure after the heat is applied
    ## T_1 = 25C = 25 + 273.15 = 298K ##
    ##A = 250 cm ^2 = 0.025 m^2##
    ## C_p = 1.04 kJ/(kg*K)##
    ##C_v = 0.743 kJ(kg*K) ##
    ##m=28g = 0.028kg ##
    ##m_p=12.5kg ##
    ##R = 8.3145 \ J \ mol ^{-1} K ^{-1} ##
    ##M_N = 28##

    The attempt

    I am a bit unsure about b and I cant complete d. Please help.

    a)

    It seems like this is an isobaric process since pressure is constant.

    ## E = m \cdot c_p \cdot \Delta T \\ 6000 = 0.028 \cdot 1.04 \cdot 10^3 \cdot \Delta T \\ \Delta T = \frac{6000}{0.028 \cdot 1,04 \cdot 10^3} = 206K##

    ##T_2 = T_1 + \Delta T = 298 + 206 = 504K##

    b)
    I am not sure how to solve this. I guess that you can use with
    ## W = p \Delta V ## is also ## W = n \cdot R \Delta T ##

    ## n=\frac{n}{M} =\frac{28}{28} = 1 ##

    ## W = n \cdot R \Delta T \\ W = 1 \cdot 8.3145 \cdot 206 = 1712.787 J = 1.7kJ##

    c)
    ## Q = \Delta U + W \\ \Delta U = Q - W = 6kj - 1.7kj = 4.3kJ ##

    d) ## \Delta V ##

    Before heat:
    ## pV_1= nRT_1 \Leftrightarrow \frac{V_1}{T_1} = \frac{nR}{p}##

    After heat:
    ## pV_2= nRT_2 \Leftrightarrow \frac{V_2}{T_2} = \frac{nR}{p}##

    Together it gives:
    ## \frac{V_2}{T_2} = \frac{V_1}{T_1} \\ V_2 = \frac{V_1 \cdot T_2}{T_1} \\ V_1 = \frac{V_2 \cdot T_1}{T_2}##

    ##\Delta V = \frac{V_1 \cdot T_2}{T_1} - \frac{V_2 \cdot T_1}{T_2} =\frac{V_1 \cdot 504}{298} - \frac{V_2 \cdot 298}{504} = ##

    aaaand I am stuck.

    I guess I could somehow calculate V with the help of height the beginning of the piston in the cylinder. If I had the height I would calculate the volume of wither the before or after case. I just need one.

    That force up from on the piston = the force down from the piston on the gas (gravitational)

    I tried this too but I failed:

    The cylinder that forms when nitrogen lifts the piston has the following proportions:
    ## \Delta V = \Delta h \cdot A ##

    The work done by the gas to lift the piston ## \Delta h ## is W. It is what I calculated in b) earlier. The potential energy of the piston if E_p = mgh. this means that the energy at the new position is
    ## E_2 - E_1 = m g h_2 - m g h_1 ## or ## \Delta E = m g \Delta h##

    ## \Delta E = W = 1712.787 \\ 1712.787 = m g \Delta h \\ 1712.787 = 0.028 \cdot 9.82 \Delta h \\ \Delta h = \frac{1712.787}{0.028 \cdot 9.82} = 6229.2##
    WTF? so it moved 6229 meters? Nah I dont think so...

    Can someone please help me out here?
     
    Last edited: Sep 3, 2015
  2. jcsd
  3. Sep 3, 2015 #2
    You did very well on this. Parts a - c are correct. To complete part d, you need to get V1 from the ideal gas law. Do you know how to determine the starting pressure P1?

    Chet
     
  4. Sep 3, 2015 #3
    Correct.
    it is a constant pressure process (isobaric)
    Formula for A is correct, but we use Q instead of E.
    Did not check your math.
     
  5. Sep 3, 2015 #4
    But you have it. Let's just trim your equations back a bit to where you went astray.

    1: did you try picking a nice, easy number for V1 (such as 1)?
    2: if you plug that in, what is V2?
    3: now what is your delta?

     
  6. Sep 3, 2015 #5

    Rectifier

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    pV=nRT

    Is it that ## p = \frac{F}{A} ## ?
     
  7. Sep 3, 2015 #6

    Rectifier

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    Then ## V_2 = \frac{T_2}{T_1} ##. ## \Delta V ## is therefore ## V_2 - V_1 = V_2 - 1 = \frac{T_2}{T_1} - \frac{T_1}{T_1} = \frac{T_2 -T_1}{T_1} ##
     
  8. Sep 3, 2015 #7
    Not delta V, that's V1.
    Can you calculate T2/T1 for me?
    That is V2/V1, yes?
     
  9. Sep 3, 2015 #8
    No. It's atmospheric pressure plus the weight of the piston divided by the area. That's why they gave you the atmospheric pressure, the mass of the piston, and the area. Once you know the initial pressure, you can calculate the initial volume using the ideal gas law. Then you can calculate the final volume from the temperature ratio. Then you can get the change in volume.

    Chet

    MICHIGAN: GO BLUE
     
    Last edited: Sep 3, 2015
  10. Sep 3, 2015 #9
    Close
    As Chet said, you forgot the Atm pressure pushing down on the weight.

    ## p = \frac{F}{A} + P_0 ##
     
  11. Sep 3, 2015 #10
    Hint: what are the units on R?

    If you are going to apply the ideal gas law, V= P/nRT, you had best remember to select an R with good units and check that the final answer us in the correct units...
     
  12. Sep 4, 2015 #11

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    I am not sure what you mean. Is that the wrong R ? Or should I pick one with more digits (more precise)?
     
  13. Sep 4, 2015 #12

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    Attempt 2:

    ## p_a = 100 \cdot 10^3 \ Pa \\ F_p = 12.5 \cdot 9.82 = 122.75 \\ A= 250 cm ^2 = 250 \ cm^2 = 0.025 \ m^2 ##

    d) ## \Delta V = V_2 - V_1 \\ \frac{V_2}{T_2} = \frac{V_1}{T_1} \\ p_1V_1= nRT \\ p = \frac{F_p}{A} + p_0 = \frac{122.75}{0.025} + 100 \cdot 10^3 =104910 \\ 104910 \cdot V_1= 1 \cdot 8.3145 \cdot 298 \\ V_1 = \frac{8.3145 \cdot 298}{104910} = 0.0236... \\ V_2 = \frac{V_1 \cdot T_2}{T_1} = \frac{0.0236 \cdot 504}{298} = 0.03994 \\ \Delta V = V_2 - V_1 = 0.03994 - 0.0236 = 0.01633 m^3 ##

    Does this look right?
     
  14. Sep 4, 2015 #13
    You can check your answer by dividing the work by the pressure.

    Chet
     
  15. Sep 4, 2015 #14

    Rectifier

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    I guess its right then :D

    Thank you both for your help!!!
     
    Last edited: Sep 4, 2015
  16. Sep 4, 2015 #15
    You are welcome. Very good job.

    Hint: Step your game up by performing a 'unit analysis'.

    So what is a Pascal (Pa)? A Newton per square meter... ##\frac{N}{m^2}##
    And a Newton is ? ##\frac{kg-m}{s^2}##
    Plug that into the Pascal units and you have what? ##\frac{kg-m}{m^2-s^2}##
    And one of the meters in the numerator and denominator cancel to give...##\frac{kg}{m-s^2}##

     
    Last edited: Sep 4, 2015
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