Thermodynamics: ideal gas undergoing an isothermal process

[atlantic]

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?

 

[atlantic]

Or does these equations not apply for isothermal processes?

 

[rude man]

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.

 

[atlantic]

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?

 

[vela]

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.