Cooling via Expansion: He vs Ar Gas

[manuj]

Hi everyone, I have encountered a situation described as follows in my research and would appreciate it if someone could help me out -

A ball of metal vapors ( Pressure = 10^10 Pa and T = 10^4 K) is suddenly exposed to He gas at 1 atm. I wanted to know how would the Temperature change due to expansion. Also if instead of He gas, I have Ar gas how would the temperature change.

Would really appreciate it if someone could suggest a solution.

 

[berkeman]

Hi everyone, I have encountered a situation described as follows in my research and would appreciate it if someone could help me out -

A ball of metal vapors ( Pressure = 10^10 Pa and T = 10^4 K) is suddenly exposed to He gas at 1 atm. I wanted to know how would the Temperature change due to expansion. Also if instead of He gas, I have Ar gas how would the temperature change.

Would really appreciate it if someone could suggest a solution.

Question moved to Homework Help.

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[manuj]

Hi

Well, I will try to give a glimpse of what is happening. In an experiment that I am carrying out - I shoot a laser through a particle and compress it to a high pressure and T (10^10 Pa and 10^4 K), by generating a shockwave. The shockwave moves ahead leaving the particle suddenly exposed to He at 1 atm. Now I figured that the metal must expand to 1 atm very quickly and consequently cool during the process. However I am not able to find the governing equations for real gases expansion. And specially if they are expanded to a particular gas (He or Ar in my case) at 1 atm. Most of the solutions that I found out were for the ones where a an ideal gas is exposed to vacuum. I know the solution of an ideal gas to vacuum but don't know what is the scenario when the vapos are similar to a real gas and are exposed to a gas at 1 atm.

 

[Redbelly98]

I'm no thermodynamics expert, but could one just ignore the 1 atm of He or Ar for most of the expansion? The initial pressure is 10^5 atmospheres.

 

[Count Iblis]

You need to specify if the helium gas at 1 atm is held at constant volume or at constant pressure. You can solve the problem simply by using consevation of energy. You write down te total internal energy of the initial state, and then if the volume stays constant the total internal eenrgy of the final state will be the same. If the system is kept at constant pressure, the system will perform work and the total internal energy in the final state will be less due to work performed against a pressure of 1 atm.

 

[manuj]

Thanks, He is at constant pressure (1 atm ), as in it is a large reservoir of He. Can i get a governing equation.

 

[Count Iblis]

Denote the number of metal atoms by N and the number of helium atoms by M. The initial internal energy is:

E_i = 3/2 N k T_1 + 3/2 M k T_2

where T_1 is the temperature of the metal vapor and T_2 is the temperature of the helium gas.

The final internal energy is given by:

E_f = 3/2 (N+M) k T

where T is the final temperature.

If the whole system is not at constant volume but kept at constant pressure, the energy is not conserved because the system will do work on the reservoir which keeps the system at P = 1 atm pressure. The work done will be P Delta V, where Delta V is the increase in the volume:

The initial volume is:

V_i = N k T_1/P_1 + M k T_2/P

The final volume is:

V_f = (N+M)k T/P

So, we have:

Delta V = (N+M)k T/P - N k T_1/P_1 - M k T_2/P

Final internal energy is thus

E_f = E_i - P Delta V =

E_i - (N+M)k T + N k T_1 P/P_1 + M k T_2

Insert the expressions for E_i and E_f and solve for T.

 

[manuj]

thank u guys