# Thermodynamics: ideal gas undergoing an isothermal process

1. Oct 11, 2012

### atlantic

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

3. 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?

2. Oct 12, 2012

### atlantic

Or does these equations not apply for isothermal processes?

3. Oct 12, 2012

### rude man

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.

4. Oct 12, 2012

### atlantic

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?

5. Oct 12, 2012

### vela

Staff Emeritus
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.