Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 years ago

An isobaric process where heat Q is absorbed by a diatomic gas with 5 degrees of freedom, then undergoes an isometric process where the...?

same Q is released. Find an equation for the change in temperature across the whole process?

Please help, I'm not sure how to approach this. Does Qp=-Qv? I'm sure this will come out really simply but I just don't know. Thank you in advance!

Update:

Isometric and Isochoric are interchangeable I believe. I got to where you did but just became confused. If you do remember to check back please do :) thanks!

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  • 6 years ago
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    Isometric? A 2-sides-are-equal-triangular process?

    Isobaric - constant pressure

    Isothermal - constant temp

    Isentropic - constant entropy

    Isochoric - constant volume

    Adiabatic - No heat added

    I'm pretty sure you ment Isochoric.

    Define the specific heats

    Cv = degrees or freedom / 2 * R = 5/2*R

    Cp = Cv + R = 7/2*R

    I got a doc's appointment I have to run to, but I hope I helped you just a little bit. It will be tomorrow befoe I can revie this again.

    1st Law of Thermo

    dU = Q - W

    Q = ΔU + W

    For the isobaric process:

    W = integral (P dV)

    Since pressure is constant

    W = P*ΔV

    From the ideal gas law

    P*ΔV = n*R*ΔT

    Also,

    ΔU = n*Cv*ΔT

    Plug those into the Q equation

    Q = ΔU+W

    Q = n*Cv*ΔT + n*R*ΔT

    Q = n*ΔT*(Cv + R) <---- This is the equation we want

    Q = n*ΔT*Cp

    For the isochoric process

    Q = ΔU + W

    Well, since volume isn't changing there is no work being done, W = 0

    Q = ΔU

    Sub in ΔU

    Q = n*ΔT*Cv

    So, the change in temps are different though. Call the ΔT for the isbaric process ΔT1 and ΔT2 for the isochoric process. The heat gained during the isobaric process is lost in the isochoric process. So the change in heat is 0.

    Qp + Qv = 0

    n*ΔT1*(Cv+R) + n*ΔT2*Cv = 0

    ΔT1*(Cv+R) + ΔT2*Cv = 0

    Even though I left it Qp+Qv, the change in temps will take care of making the + sign a - sign.

    During the isobaric process the temperature starts at T1i and ends up at T1f. During the isochoric process the temperature starts at T2i and ends at T2f.

    However, T1f = T2i = T

    ΔT1*(Cv +R) + ΔT2*Cv = 0

    (T - T1i)*(Cv + R) + (T2f - T)*Cv = 0

    T*Cv - T1i*Cv + R*T - R*T1i + T2f*Cv - T*Cv = 0

    Cv*(T2f - T1i) + T*(Cv+R) = 0

    (T2f-T1i) = -T*Cp/Cv

    T1i - T2f = T*Cp/Cv

    With Cp = 7/2*R and Cv = 5/2*R

    T1i - T2f = 7/5*T

    Now, this is as far as I got. I'm not sure if they want you to use a relationship to find what T is, the final temp of Qp/initial temp of Qv. Hope I helped you this far.

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