Understanding why $\int_{T_1}^{T_2}C_Vd\ln(T) + \int_{V_1}^{V_2} nRd\ln(V) = 0$ is violated when $T_2$ and $V_2$ are picked

Question

In Chapter 4 of Chemical Thermodynamics, by Peter A. Rock published by Oxford University Press, 1983, the author is trying to explain why a new state function is needed, specifically a state function relating entropy. He starts off with an equation from the first law for an adiabatic, reversible expansion of an ideal gas.

$C_V\ln(\frac{T_2}{T_1}) + nR\ln(\frac{V_2}{V_1}) = 0$

He then goes onto explain the equation above by saying:

Equation (1) has an interesting consequence, namely, that given $T_1$ and $V_1$ we are free to pick $T_2$ or $V_2$ but not both; otherwise eq(1), which relates $T_1$, $V_1$ and $T_2$, $V_2$, would be violated.

I don't understand why picking two variables violates the equation.

Answer 1

Let's say $T_1$ and $V_1$ are given. Without loss of generality, let us pick $T_2$ (the argument for choosing the value of $V_2$ is practically the same). Now due to $$C_V\ln(\frac{T_2}{T_1}) + nR\ln(\frac{V_2}{V_1}) = 0$$ the following holds for $V_2$ $$ V_2 = V_1 \left( \frac{T_2}{T_1} \right)^{-\frac{C_V}{nR}} $$ If $V_1$, $T_2$, $T_1$ are known, we can calculate the value of $V_2$. Choosing an arbitrary value of $V_2$ violates above equation quite clearly. Let's say we choose $V_2 = 5\cdot V_1 \left( \frac{T_2}{T_1} \right)^{-\frac{C_V}{nR}}$. Then combining both expressions for $V_2$, we obtain $1=5$ which is wrong ...