# Thermodynamic, ideal gas problem

I already know solution, I just want to understand something in part b.

A vertical cylinder fitted with a piston contains air initially at 120kPa and 100 celcius. The piston is restrained by a set of stops. The piston diameter is 0.1m, its mass is 5kg, and the ambient pressure and temperature are 100kPa and 20 celcius. The cylinder is then cooled as heat is transferred to the surroundings.

a) At what temperature does the piston fall below the stops?

b) How far down (as a fraction of its initial height) has the piston moved when the contents of the cylinder has cooled to the ambient temperature?

a) So this is assumed to be a 3-step process: $P_2 = P_{external} + \frac{mg}{A}$

Volume is supposedly constant from 1 to 2:

$\frac{P_1 V_1}{T_1} = m R = \frac{P_2 V_2}{T_2}$

So $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

$T_2 = T_1 \cdot \frac{P_2}{P_1}$

b)

Since 2 -> 3 is supposedly a constant pressure process:

$\frac{P_3 V_3}{T_3} = \frac{P_2 V_2}{T_2}$ or $\frac{V_3}{V_2} = \frac{T_3}{T_2}$

So $\frac{V_2 - V_3}{V_2} = \frac{\delta h}{h_2}$

So my question is, if the ideal gas equation is:

$P V = mR \cdot T$

$m$, $R$ are constants

and if the temperature changes, then on the left hand side either pressure of volume can change, how is it known that from process 1 to 2, volume stays constant?

The problem postulates that heat of system will be cooled down by the surrounding temperature (20 C), nothing's said about neither pressure nor volume. So either of them can change... how does one know that volume stays the same from process going from 1 to 2?