# 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?

## 2 Answers

The solution for this problem relies upon the fact that the process is reversible, that is every state undergone by the system is in equilibrium. Consider the initial state. Now the pressure inside the cylinder is 120 kPa. If we consider the equilibrium of the piston, 4 forces are acting upon it. They are: gravitational force(weight of piston) acting downward, atmospheric pressure acting downward,reaction force exerted by the stops acting downward andpressure inside cylinder acting upward. By doing the calculations we can see that the net downward force exerted by the atmospheric pressure and the weight of piston is 834 N. But the upward force exerted by the gas in the cylinder is 942 N. Hence a downward force of 108 N is required to be applied by the stops as a reaction force. Since reaction force is a surface force this force cannot be applied unless the piston and the stops remain in contact, ie the volume of the gas inside the cylinder remains constant. The need for such a reaction force to maintain equilibrium will exist until the pressure inside the cylinder reduces to about 106.24 kPa . This is the process 1-2. After reaching the state point 2 , the force exerted by the stops will be zero. So further reduction in pressure inside cylinder is impossible since it would disrupt the equilibrium. Hence, if temperature is again reduced the volume will decrease, as seen in process 2-3

The process 1>2 is constrained to a change in temperature as the volume is held constant by the stops as in "The piston is restrained by a set of stops."