$P$, $V$, $T$ in case of pressure cookers?

Question

In a pressure cooker, when T increases -P increases which means that the K.E of molecules increases. But According to equation PV/Tinc so T inc P inc and V dec. If they less volume I.e gas molecules does it means they occupy less space in the cooker.Does that mean that molecules don’t reach or go to some areas of cooker. If it is that,Then how can there any pressure. Since they are not trying to come out.

Mark H · Accepted Answer

One way to describe the behavior of a gas is by an equation of state that relates the measurable properties of a gas. One of these is
$$frac{PV}{T} = textrm{constant}.$$
Equivalently, this can also be written as
$$frac{P_1 V_1}{T_1} = frac{P_2 V_2}{T_2}.$$
What these equations mean is that, if one quantity changes--whether pressure, volume, or temperature--then one or more other variables will change to keep the quantity $PV/T$ constant. It is not necessary for all of the quantities to change, thought at least one must.
So, in the case of the pressure cooker, the temperature increases due to the heater. This means that, in response, the pressure of the pressure cooker must increase or the volume of the cooker must increase. Or, the volume could decrease and the pressure increases even more. There are many things that can happen, as long as the original value of $PV/T$ is conserved. In the case of the pressure cooker, since it is constructed of metal that keeps its shape, the volume will remain constant. This means that the only option is that the pressure in the cooker will increase in response to the increase in temperature.
In math,
$$frac{P_1 V_1}{T_1} = frac{P_2 V_2}{T_2}.$$
Since $V_1 = V_2,$
$$frac{P_1}{T_1} = frac{P_2}{T_2}.$$
THus,
$$P_2 = P_1left(frac{T_2}{T_1}right).$$
So, if $T_2 > T_1,$ then $T_2/T_1 > 1$ and $P_2 > P_1.$

Cort Ammon · Answer

$frac{PV}{T}$ does not mean that the volume has to decrease in this situation.  There are other situations where it means volume has to decrease, but not this one.  The rule, as Mark H pointed out, is that $frac{PV}{T}$ must not change.  So if the temperature goes up, P and V must somehow change to increase $PV$ so that $frac{PV}{T}$ remains constant.
It doesn't say anything about how the variables change to accomplish this.  That has to be gathered from the problem.  In this case, the pressure cooker is a closed volume that isn't changing.  Volume is constant.  So this means that if $T$ increases, the only way for $PV$ to increase accordingly is for $P$ to increase.  This makes intuitive sense, because that's what a pressure cooker does.
The case where $V$ must decrease to allow $P$ to increase is a situation where the temperature is held constant.  In this case, to increase the pressure, you must squeeze the gas, decreasing volume.  This is probably what you were thinking of, but it doesn't apply here.  We don't have a fixed temperature.  Our temperature is increasing... quite a lot!

$P$, $V$, $T$ in case of pressure cookers?

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