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owlet

  • one year ago

I need a step-by-step guidance on how to solve this question. I keep getting it wrong. Question below with my solution. I don't think my solution is right though

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  1. owlet
    • one year ago
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    A gas is confined to a container with a massless piston at the top. A massless wire is attached to the piston. When an external pressure of 2.03 bar is applied to the wire, the gas compresses from 6.40 to 3.20 L . When the external pressure is increased to 2.53 bar, the gas further compresses from 3.20 to 2.56 L . In a separate experiment with the same initial conditions, a pressure of 2.53 bar was applied to the gas, decreasing its volume from 6.40 to 2.56 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

  2. owlet
    • one year ago
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    @Photon336

  3. Photon336
    • one year ago
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    @Cuanchi @Rushwr

  4. Photon336
    • one year ago
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    @abb0t

  5. Rushwr
    • one year ago
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    Q = U + W right?

  6. owlet
    • one year ago
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    no U is the energy, q is the heat..so it should be U=q+w I'm not really sure what to do with the one-step process

  7. Rushwr
    • one year ago
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    Bubs what I meant was that is the standard equation right?

  8. Photon336
    • one year ago
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    @owlet but if U = Q+W then the change in U deltaU should be equal to the change in Q + change in W. based on the way you've done it implies that the change in work is equal to the -of the Q so that means that the change in internal energy would be zero.

  9. Photon336
    • one year ago
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    I think @Woodward @Empty would be able to help solve this question

  10. Rushwr
    • one year ago
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    delta Q = delta u + delta W delta Q= heat supplied delta U = increase in internal energy Delta W = work done That is a standard equation. Infact that is the first the law of thermodynamics.

  11. Rushwr
    • one year ago
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    This is an isothermal process right? So delta U = 0 Then delta Q = delta W right?

  12. Rushwr
    • one year ago
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    that means the only thing we have to do is finding delta W right?

  13. owlet
    • one year ago
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    yeah, difference between work done of the one-step process and the two-step process

  14. Rushwr
    • one year ago
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    yp

  15. Photon336
    • one year ago
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    iSO = SAME thermal temperature. so the temperature is constant.

  16. owlet
    • one year ago
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    wait.. so it means.. delta w= w3- (w1+w2) ? w3 is the one step process and (w1 + w2) is the two step process

  17. Photon336
    • one year ago
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    Maybe this could help us understand this: the derivation of work at constant pressure. work = pdv isothermal. pV = nRT \[\frac{ nRT }{ V } = P \] substitute this into P \[p \int\limits_{v_f}^{v_i} dv \] \[\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV \] \[\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV = nRTLn \frac{ v_{2} }{ v_1 }\]

  18. Photon336
    • one year ago
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    this must be an isothermal process Work = nRT*Ln(v2)-Ln(v1) for isothermal expansion if this is an isothermal process the the internal energy change is 0 \[\Delta U = 0 \] \[\Delta U = 0 = q + w ; \] that would mean that q = -w \[q = -w \]

  19. Rushwr
    • one year ago
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    Wait I'm tryna attach my work can u please check on that @Photon336 @owlet

  20. owlet
    • one year ago
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    i already got 160J which is the same as the answer on the book. i just did what i understand awhile ago: delta w= w3- (w1+w2)

  21. Rushwr
    • one year ago
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    Hold on I can't attach it

  22. Rushwr
    • one year ago
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    Oh then that's fine ! I think I might have gone wrong somewhere I got 1600J LOL

  23. owlet
    • one year ago
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    maybe you got wrong with converting 1 Lbar : 100J thanks btw for both of your help :)

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