How to derive this temperature-pressure-specific volume relationship?
For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:

$$\mathrm dU = \delta Q + \delta W$$

$$\mathrm dU = - p\,\mathrm dV$$

This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...

$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$

$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$

$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$

$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$

From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.