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Entropy: gas heated by resistor

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A current ## I=0.2 A ## flows in a resistor ##R = 50 Ω## immersed in a rigid adiabatic vessel that contains ##n=3## moles of Helium. The initial temperature of the system is ##T_0 = 27 °C##. The resistor has a mass ## m = 10 g## and specific heat ## c = 0.2 (cal/K)/g ## and runs for ##t= 10 min##. Find:
    1. the internal energy change of He.
    2. the entropy change of He and the entropy change of the Universe.

    2. Relevant equations
    ##P = i^2R##
    ##dS = \left( \frac {\delta Q} T \right)##

    ...

    3. The attempt at a solution
    I think I solved the problem, but I'm not completely sure of my reasoning, especially on the last question. Could you tell me what you think?

    1) Helium is a noble gas, so I assumed it's an ideal gas. Therefore ## U = U(T) ##.
    Since the volume is kept constant: ## ΔU = n c_V ΔT = n \frac 3 2 RΔT##. (Helium is monoatomic)
    I calculated the temperature at equilibrium (##T_{eq}##) by knowing that the electrical energy is converted into heat that heats up both the resistor and the gas:
    $$E_{in} = (mc + n c_v) (T_{eq} - T_0) $$
    $$ I^2 R t + T_0 (mc + n c_v) = T_{eq} (mc + n c_v)$$
    and hoping I converted all units accordingly:
    $$T_{eq} = \frac {I^2 R t + T_0 (mc + n c_v)} {(mc + n c_v)} ≈ 431 K $$
    Therefore: $$ΔU = n \frac 3 2 RΔT ≈ 4.89 kJ$$

    2) Helium absorbs heat at constant volume, so ##TdS =\delta Q = nc_v dT##
    So: $$ ΔS_{He} = \int_{T_0}^{T_{eq}} \frac {nc_v} T \, dT = n c_v \ln{\frac {T_{eq}} {T_0}} ≈13.5 J/K $$
    As for the entropy change of the universe, I calculated the entropy change due to the heating of the resistor and added it to that of the gas:
    $$ ΔS_R = \int_{T_0}^{T_{eq}} \frac {mc} T \, dT = mc \ln{\frac {T_{eq}} {T_0}} ≈3.02 J/K $$
    $$ ΔS_{univ} = ΔS_{He} + ΔS_R ≈ 16.52 J/K$$

    Since the energy flowing into the system comes only through electrical work, and the vessel is adiabatic, there are no other heat exchanges that could cause a change in entropy of the universe, right?
     
  2. jcsd
  3. Sep 3, 2016 #2

    TSny

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    Gold Member

    Hello and welcome to PF!

    Your method and resoning look very good to me. However, I'm getting different numbers.

    For example, you have
    The formula looks correct, but I don't get 431 K. Can you tell us what numerical values you have for ##mc## for the resistor and ##nc_v## for the He?
     
  4. Sep 3, 2016 #3
    Sure. I might have been a bit sloppy with numbers. So using the given data:
    ##mc = 10 g ~ 0.2~ \frac {cal} {K~g} = 2~ \frac {cal} K ##
    ##nc_v = 3~ mol~ \frac {3}{2}~ R = 3~mol~ \frac {3}{2}~ 1.987~ \frac {cal}{K ~mol} ≈ 8.94 \frac {cal} K##
    Therefore, since ## 1 ~cal = 4.18 ~J##:
    $$T_{eq} = \frac {6000 ~J + 300.15~ K (2+8.94)~ 4.18 \frac J K} {(2+8.94)~ 4.18 \frac J K} ≈ \frac {6000 ~J + 13727.5 ~J } {45.74 \frac J K} ≈ 431.3 K$$
     
  5. Sep 3, 2016 #4

    TSny

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    OK. The problem appears to be with your calculation of ##I^2Rt##. Did you square the current?
     
  6. Sep 4, 2016 #5
    Ahaha. Yes, you're right. Apparently I didn't. So ##I^2Rt = 1200 ~J## and therefore:
    ##T_{eq} ≈ 326.4~ K##
    ##ΔU ≈ 982.1 ~J##
    ##ΔS_{He} ≈ 3.14 ~J/K##
    ##ΔS_R ≈ 0.7 ~J/K##
     
  7. Sep 4, 2016 #6

    TSny

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    OK. I get about 987 J for ΔUHe, but that's close to what you get.
     
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