Introduction
In this article, we will delve into the fascinating phenomenon of free expansion of an ideal gas. Imagine a scenario where an ideal gas is trapped in a box, which is situated in a vacuum. When a puncture occurs in the wall of the box, what happens to the temperature and pressure of the gas? This exploration will shed light on the thermodynamic principles at play during this process.
Initial Conditions
Pressure and Temperature
We start with an ideal gas characterized by:
- Initial Pressure (P1)
- Initial Temperature (T1)
- The gas is confined in a box with vacuum surrounding it.
Understanding Free Expansion
Once the wall of the box gets punctured, the gas molecules rush out into the larger vacuum container. This rapid expansion is an example of free expansion, and it occurs without external pressure acting against the gas, which leads us to explore the implications of this process.
The Thermodynamics of Free Expansion
Adiabatic Process
The rapid nature of the gas's expansion means that there is insufficient time for any heat transfer to occur between the gas and its surroundings. As a result, the process can be classified as adiabatic. In an adiabatic process:
- Heat Transfer (Q) = 0
Work Done by the System
Similarly, since the gas expands into a vacuum (where the external pressure is zero), it does not perform work against an opposing pressure. Therefore:
- Work Done (W) = 0
This leads to the conclusion that during free expansion, both heat transfer and work done by the system are zero.
Change in Internal Energy
Internal Energy Equation
The change in internal energy ( ( \Delta U )) can be calculated with the formula:
[ \Delta U = Q + W ]
Since both Q and W are zero:
- ΔU = 0
This indicates that the internal energy of the gas remains constant during the free expansion process.
Relating Internal Energy to Temperature
The change in internal energy for an ideal gas can also be expressed as:
[ \Delta U = n C_V \Delta T ]
Where:
- n = number of moles of the gas
- C_V = molar heat capacity at constant volume
- ΔT = change in temperature
Given that ( \Delta U = 0 ), it implies:
- ΔT = 0
Thus, the final temperature equals the initial temperature, confirming that the temperature remains constant throughout free expansion.
Final Conditions
Changes in Pressure and Volume
Following the expansion, the conditions of the gas change:
- Final Volume (V_f) > Initial Volume (V_i)
- This increased volume means that the gas now occupies a much larger space.
Impact on Pressure
From the ideal gas law, we can assert that as volume increases (while the amount of gas and temperature remains constant), pressure decreases. Thus:
- Final Pressure (P_f) < Initial Pressure (P1)
Summary
To summarize, when an ideal gas undergoes free expansion into a vacuum, several key thermodynamic principles emerge:
- The process is adiabatic with zero heat transfer and work done.
- The internal energy remains constant, leading to no change in temperature.
- Although the temperature remains at the initial condition (T1), the pressure significantly drops as the gas occupies a greater volume.
Understanding these concepts helps illuminate the intriguing behavior of gases under unrestrained conditions, showcasing the underlying principles of thermodynamics.
in this video we're going to discuss a problem on free expansion of ideal gas now the question says an ideal gas at
pressure P1 and temperature T1 in a box is kept in vacuum within a large container now the wall of the Box gets
punctured what happens to the temperature as the gas occupies the entire container okay so let's look at
the initial conditions it says that the gas is initially at pressure P1 and temperature T1 and this gas inside the
box is confined within another container and it has only vacuum outside here correct now what do you think happens
when the wall of this box gets punctured obviously all the gas molecules Inside the Box will rush outside into the
Container correct exactly in other words what is happening here is free expansion of gas because this process happen so
quickly or this free expansion is very rapid there is no sufficient time for heat transfer to take place between the
gas molecules inside and the environment outside and that is why this process is adiabetic which is why Q or the heat
transferred will become zero all right what else is happening here we can see that in this free expansion the work
done by the system or the gas particles is also zero and why is that because we know that work done by the system is
equal to minus P external Delta V and here the gas particles are moving against vacuum that is there is no
external pressure acting against the gas molecules for which it'll have to do some work against it right because in
this case P external is zero there is no work done by the system and W is equal to zero from here you can see that the
two important thermodynamic parameters he transferred and work done by the system are Zer during the free expansion
of an ideal gas into vacuum so what does this say about the change in internal energy Delta U or the change in internal
energy is the sum of the heat added to the system plus the work done on the system now we know that W is positive if
work is done on the system and negative when work is done by the system but in any case we don't have to concern
ourselves with the signs of either Q or W because in a free expansion both of these terms are zero isn't it now since
Q + W is equal to 0 change in internal energy is also equal to zero that means internal energy remains constant Now
Delta U can also be written as n CV delta T since n and CV are constants and cannot be zero if Delta U is equal to
zero then delta T also has to be zero and that means the temperature of the system remains constant in other words
the final temperature is same as the initial temperature okay so what else can can you comment about the final
conditions well after expansion the gas molecules that were confined within the small box have now occupied the entire
volume of the container as a result the final volume is greater than the initial volume and consequently the final
pressure will be much lower than the initial pressure this is because when the number of gas particles and
temperature remain constant pressure becomes inversely proportional to volume which is why pressure decreases and as
the gas particles expand and occupy the entire container or as the volume increases so the final conditions of the
system would be temperature is same it's constant so final temperature is same as the initial temperature T1 the final
pressure PF would be less than the initial pressure because the final volume has now become much larger than
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