If pressure and temperature depend on volume, then why are pressure and temperature intensive variables and volume an extensive variable? Pressure and temperature of a system are intensive variables, and volume is an extensive variable. If pressure and temperature change with volume, won't they affect the pressure and temperature? If a tank with volume V, pressure P, and temperature T is cut in half, the volume becomes (V/2); then doesn't the pressure and temperature change accordingly?
 A: In the ideal gas equation,
$$ pV = NkT, $$
the pressure $p$ and temperature $T$ are intensive, and the volume $V$ and number of gas molecules $N$ are extensive. So each side of the equation is the product of an intensive term and an extensive term. The remaining term, $k \approx \frac{25\rm\,meV}{300\rm\,K}$, is a unit-conversion constant attributed to Boltzmann.
(Chemists deal with molecules in large numbers instead of one at a time, and use $N$ to mean the number of moles instead of the number of molecules. In that case you have a different unit-conversion constant, usually called $R$.)
It is certainly possible to subdivide a volume of gas without changing the pressure and temperature in the halves; I do this in my house, where I have a sliding door between two large rooms.
A: For a system in equilibrium pressure and temperature are the same regardless of the mass, and are therefore intensive properties.  Volume depends on the mass enclosed and is an extensive property.  Specific volume (volume per unit mass) is an intensive property.
For a tank with an ideal gas in equilibrium $PV = nRT$ where $P$ is pressure, $V$ is volume, $n$ is moles of gas, $R$ is the universal gas constant, and $T$ is temperature.  If you cut the tank in half, both $V$ and $n$ decrease by half: $P$ and $T$ do not change.
