Final Temperature of a monatomic gas

In summary, the molar energy of a monatomic gas that obeys van der Waals' equation is given by E=\frac{3}{2}RT - \frac{a}{V}. When one mole of this gas at temperature T_1 and volume V_1 expands adiabatically into a vacuum, the final temperature can be found using the equation T_f= \frac{a}{C_v}\Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}+T_i, where C_v is the specific heat at constant volume. This is derived from the fact that the change in energy, \Delta E, is equal to zero in an adiabatic process.
  • #1
patrickmoloney
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4

Homework Statement


The molar energy of a monatomic gas that obeys van der Waals' equation is given by
[tex]E=\frac{3}{2}RT - \frac{a}{V}[/tex]

where [itex]V[/itex] is the molar volume at temperature [itex]T[/itex] and [itex]a[/itex] is a constant.
Initially, one mole of such a gas is at temperature [itex]T_1[/itex] and occupies a volume [itex]V_1[/itex]. The gas is allowed to expand adiabatically into a vacuum, so that it occupies a total volume of [itex]V_2[/itex]. What is the final temperature of the gas?

Homework Equations

The Attempt at a Solution


This is what I have, I'm not sure if it's correct.

[tex]\begin{align*}
\Delta E & = \frac{3}{2}R\Delta T - \frac{a}{\Delta V} \\
& = \frac{3}{2}R(T_f - T_i) - \frac{a}{(V_f - V_i)} \\
& = \frac{3}{2}RT_f - \frac{3}{2}RT_i - \frac{a}{(V_f - V_i)}
\end{align*}
[/tex]

re-arranging for [itex]T_f[/itex],

[tex]T_f= \frac{2}{3R}(\Delta E + \frac{a}{V_f - V_i})+T_i[/tex]
 
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  • #2
If ##A = 1/B##, then ##\Delta A \neq 1/\Delta B##.

Also, from the problem statement, you know something about ##\Delta E##.
 
  • #3
DrClaude said:
If ##A = 1/B##, then ##\Delta A \neq 1/\Delta B##.

Also, from the problem statement, you know something about ##\Delta E##.
The system is adiabatic so the there is no heat transfer. Meaning [itex]\Delta E = 0[/itex]
 
  • #4
patrickmoloney said:
The system is adiabatic so the there is no heat transfer. Meaning [itex]\Delta E = 0[/itex]
That is not sufficient to conclude that ##\Delta E=0##
 
  • #5
Chestermiller said:
That is not sufficient to conclude that ##\Delta E=0##
The system is adiabatically insulated, no heat flows into the system; [tex]Q=0[/tex] Furthermore, the system does no work in the process; [tex]W=0[/tex] Thus it follows by the first law of thermodynamics that the total energy of the system is conserved; [tex]\Delta E =0[/tex]
 
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  • #6
What about this:

[tex] \begin{align*}
\Delta E & = \frac{3}{2}RT - \frac{a}{V} = 0 \\
& = C_{v}T - a \Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}
\end{align*}
[/tex]
and therefore
[tex]\Delta T = \frac{a}{C_v}\Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}[/tex]

and

[tex] T_f= \frac{a}{C_v}\Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}+T_i[/tex]
 
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1. What is the final temperature of a monatomic gas?

The final temperature of a monatomic gas is determined by the initial temperature, pressure, and volume of the gas, as well as the number of moles of gas and any changes in these variables during the process.

2. How is the final temperature of a monatomic gas calculated?

The final temperature of a monatomic gas can be calculated using the ideal gas law, which states that the product of pressure and volume is directly proportional to the product of temperature and number of moles of gas. This can be expressed as P1V1/T1 = P2V2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

3. What factors can affect the final temperature of a monatomic gas?

The final temperature of a monatomic gas can be affected by changes in the initial pressure, volume, or temperature of the gas, as well as the addition or removal of heat through a process such as compression or expansion.

4. Why is the final temperature of a monatomic gas important to understand?

The final temperature of a monatomic gas is important to understand in order to accurately predict the behavior of gases in various processes, such as in engines or industrial processes. It can also be used to determine the efficiency of a process and make adjustments for optimal performance.

5. Can the final temperature of a monatomic gas ever be negative?

No, the final temperature of a monatomic gas cannot be negative as temperature is a measure of the average kinetic energy of particles, and it is not physically possible for particles to have negative kinetic energy. However, the final temperature can approach zero as the gas expands and cools, approaching absolute zero.

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