Year 12: Cambridge Physics Problem (Pressure inside a vessel)

In summary, the problem involves a vessel divided into two parts with equal volume and a small hole in the partition. Initially, both parts have the same volume, pressure (p), and temperature (300K). One part is heated to 600K while the other remains at 300K. The goal is to find the resulting pressure difference between the two parts at steady state, where the rate of molecules passing through the hole is the same for both sides. This can be solved using the ideal gas law, assuming no heat transfer between the two parts. However, if there is heat transfer, solving the heat equation would be necessary to determine the temperature distribution inside both parts.
  • #1
johnconnor
62
0
A vessel is divided into two parts of equal volume by a partition in which there is a very small hole. Initially, each part contains gas at 300K and a low pressure, p. One part of the vessel is now heated to 600K while the other is maintained at 300K. If a steady state is established when the rate at which molecules pass through the hole from each side is the same, find the resulting pressure difference between the two parts.

Attempt:

I'm assuming that the number and mass of molecules inside remain the same and that the temperature of the two parts during the steady state is the same.

So we have
[tex]N_1+N_2=2N[/tex], where N is the number of molecules inside each part before heating and N1 and N2 denote the number of molecules inside each part after heating.

Also pressure is proportional to <c>2, implying T is proportional to <c>2, and that p is proportional to T.

We also have [tex]N_1<c>_1=N_2<c>_2[/tex], where N_i<c>_i denotes the rate at which molecules pass through the hole from one side to another.

So now [itex]<c>^2 \propto T

\text{and }N_1<c>_1=N_2<c>_2

\Rightarrow \dfrac{N_1}{N_2}= \dfrac{<c>_2}{<c>_1}

\Rightarrow \dfrac{N_1^2}{N_2^2}= \dfrac{<c>_2^2}{<c>_1^2}

\Rightarrow \dfrac{N_1^2}{N_2^2}= \dfrac{T_2}{T_1}

\Rightarrow \dfrac{N_1}{N_2}= \(\dfrac{T_2}{T_1})^{1/2}[/itex]

And I'm stuck. I'm supposed to find the difference of pressure in terms of p but how do I do that when the terms which I have introduced are nowhere close to p? The closest one I could get are p1 and p2. Help?
 
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  • #2
I believe this is for ideal gases. Let the volume of each compartment be V. Write down the ideal gas equations for the initial and final conditions(separately). The initial condition will give you an equation in p, which you can use to find out difference in pressures.

[tex](2V)p = 2N R T_i[/tex]
 
  • #3
This question is kind of stupid because if there is any kind of mass transfer between the two sections of the vessel, there will also be heat transfer making the problem quite difficult. If there is no heat transfer then as a first approximation you could apply the ideal gas law to each section of the vessel:

PV = nRT;
P = pressure
V = volume
n = number of molecules
R = universal gas constant
T = temperature of the section
 
  • #4
Aero51 said:
This question is kind of stupid because if there is any kind of mass transfer between the two sections of the vessel, there will also be heat transfer making the problem quite difficult. If there is no heat transfer then as a first approximation you could apply the ideal gas law to each section of the vessel:

PV = nRT;
P = pressure
V = volume
n = number of molecules
R = universal gas constant
T = temperature of the section

At the initial situation of the problem, the ideal gas law, as I suggested, can obviously be applied for the whole vessel. For the final situation, at equilibrium, meaning net transfer of heat being zero, the ideal gas law is applicable.
 
  • #5
You can have equilibrium with a temperature gradient inside both the chambers, which again would make the problem much more difficult. If you want to solve the problem at "steady state" you need to solve the heat equation and determine the temperature distribution inside both vessels. It may not vary with time but it certainly won't be an abrupt change at the interface. of the wall.
 

1. What is the definition of pressure?

Pressure is defined as the force exerted on a surface divided by the area of that surface. It is typically measured in units of Pascals (Pa), which is equal to 1 Newton per square meter.

2. How does the pressure inside a vessel change with depth?

The pressure inside a vessel increases with depth, as the weight of the fluid above the point of measurement adds to the force exerted on the surface. This is known as hydrostatic pressure and follows the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

3. What is the difference between gauge pressure and absolute pressure?

Gauge pressure is the pressure measured relative to atmospheric pressure, while absolute pressure is the total pressure including atmospheric pressure. Gauge pressure is typically used in everyday situations, while absolute pressure is used in scientific calculations.

4. How does the ideal gas law apply to this problem?

The ideal gas law, PV = nRT, can be used to relate the pressure, volume, temperature, and number of moles of a gas inside the vessel. This law assumes that the gas behaves according to certain ideal conditions, such as low pressure and high temperature.

5. What are some real-world applications of understanding pressure inside a vessel?

Understanding pressure inside a vessel is important in many industries, including chemical engineering, meteorology, and diving. It is also crucial for designing and maintaining equipment such as boilers, pressure cookers, and hydraulic systems. Additionally, knowledge of pressure is essential for predicting weather patterns and studying the behavior of gases in various environments.

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