Thermodynamics: ideal gas undergoing an isothermal process

atlantic
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Homework Statement



For an ideal gas, undergoing a quasistatic process, the equations below are correct. Evaluate them given that we have an isothermal process

Homework Equations



PV^\alpha=K where K is a constant and \alpha=C-C_P/C-C_V

W = \frac{K}{\alpha -1} (\frac{1}{V_f^{\alpha-1}}-\frac{1}{V_i^{\alpha-1}})
Q = C(T_f -T_i)
\Delta S= Cln\frac{T_f}{T_i}



The Attempt at a Solution


For an isothermal process, ΔT = 0, but what does that mean for the equations given? First I though it would mean that C→∞, but that would mean that Q=0 and W→∞ (because \alpha→1), which clearly is not correct.

How should I argue?
 
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Or does these equations not apply for isothermal processes?
 
atlantic said:

Homework Statement



For an ideal gas, undergoing a quasistatic process, the equations below are correct. Evaluate them given that we have an isothermal process

Homework Equations



PV^\alpha=K where K is a constant and \alpha=C-C_P/C-C_V
This looks wrong. For example, if it's an isothermal process, α = 1 but then (C-Cp)/(C-Cv) = 1 or Cp = Cv which is definitely not true for an ideal gas.
 
rude man said:
This looks wrong. For example, if it's an isothermal process, α = 1 but then (C-Cp)/(C-Cv) = 1 or Cp = Cv which is definitely not true for an ideal gas.

I though α=1 because C→∞ (C=Q/dT, where dT→0)?

Anyways, I'm thinking that these equations are not good to use when the process is isothermal, as the equations for the work, heat and entropy becomes of the type: ∞ muliplied with 0. Do you think this is a good conclusion?
 
The point of the problem might be to figure out how these indeterminate forms can be evaluated and to show the result is what you'd expect for an isothermal process.
 
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