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Thermodynamics - Temperature change of Argon

  1. Dec 3, 2017 #1
    1. The problem statement, all variables and given/known data
    The temperature of n = 19 mol of argon gas is increased from T1 = 21 oC by Q = 4.4 kJ heat transfer, while the gas pressure is kept constant. What is the new gas temperature in Celsius degrees?

    2. Relevant equations
    cp1.gif and as its a monoatomic gas I think this means that the equation is just Q=Cp*change in temp. But I'm not really sure. The Cv of Argon is 0.3122, with R being 0.2081 and Cp being 0.5203.

    3. The attempt at a solution
    My last attempts have come up with the final temp as 44.9 - where I had used the mass of the object as 749. I also have got 30.88 by subbing the values into the unknown. I keep getting the answer marked as wrong and I'm not sure where I'm going wrong
     
  2. jcsd
  3. Dec 3, 2017 #2

    BvU

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    HI,

    Perhaps you should add dimensions to your quantities and check that the answer has the dimension you want
     
  4. Dec 3, 2017 #3
    Do you mean - Cp-0.5203[kJ/kg.K] Cv-0.3122Cp[kJ/kg.K] and then R=0.2081Cp[kJ/kg.K]?
     
  5. Dec 3, 2017 #4
    What is ##\Delta T##?
     
  6. Dec 3, 2017 #5
    That's what I want in celcius
     
  7. Dec 3, 2017 #6
    Yep. Me too.

    But what is it?

    Can you elaborate it?
     
  8. Dec 3, 2017 #7
    It's the temperature change in celsius.
     
  9. Dec 3, 2017 #8
    cool.

    Write that in symbols.
     
  10. Dec 3, 2017 #9
    What is the value of the molar heat capacity at constant volume of a mono atomic gas (in terms of R)? What is the molar heat capacity at constant pressure?
     
  11. Dec 3, 2017 #10
    CV = (3/2)R and CP = (5/2)R

    Or Q = (3/2)nRΔT + nRΔT = (5/2)nRΔT for Cp

    and Q = (3/2)nRΔT for Cv

    Is that right?
     
  12. Dec 3, 2017 #11
    I have 32.14 when I do - (4400/((5/2)*(759)*0.2081))+21 thats when everything is in grams? Is that correct
     
  13. Dec 3, 2017 #12
    So, ##\Delta T## is change in temperature, right?

    Now, say, initially the temperature is ##T_i## and finally ##T_f##.
    What is the change in temperature?
     
  14. Dec 3, 2017 #13
    Yes so the final Temperature was 32.14 - I submitted it and it was correct - with the change in temp being 11.14. This was found but using the molar mass - The molar mass of Argon is 39.95 which is multiplied by the 19mol so 759.05. Then sub into the equation above and then add the initial value to get 32.14
     
  15. Dec 3, 2017 #14
    How did you obtain the value ##32.14##?

    So your initial equation was wrong!

    It should be $$Q=m\cdot {C_p} \cdot {(T_f \sim T_i)}$$
     
  16. Dec 3, 2017 #15
    The initial equation is correct I have just subbed in the equation for Cv and the value for n in the initial equation is 1 or excluded as Argon is monoatomic and I got 32.14 by adding the initial value of 21 to the change in Temp which was found using the equation. To give the final Temperature
     
  17. Dec 3, 2017 #16
    The question clearly says "the temperature of ##n=19~\rm {mol}## of argon gas ..." why did you use ##n=1##?

    How did you find the change in temperature to add with?
     
  18. Dec 3, 2017 #17
    You didn't even need to know the molar mass of argon if you just used the universal gas constant R = 8.314 J/mole.K:
    $$(19)(2.5)(8.314)\Delta T = 4400$$
    So, $$\Delta T=11.14\ C$$
     
  19. Dec 3, 2017 #18
    Isn't your unit for ##T## - ##K##?
     
  20. Dec 3, 2017 #19
    For a temperature change, K and C give the same number.
     
  21. Dec 3, 2017 #20
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