Which Gas Shows Greater Pressure Increase When Compressed Adiabatically?

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SUMMARY

In the discussion regarding the pressure increase of gases compressed adiabatically, it is established that helium (He) exhibits a greater pressure increase than oxygen (O2) when both are compressed to one-third of their original volume. This phenomenon is attributed to the lower molecular mass of He, which allows for faster molecular motion and a higher rate of pressure increase. The ideal gas law (PV = nRT) was initially considered, but the Van der Waals equation is recommended for more accurate results in non-ideal conditions. The adiabatic exponent gamma, which differs between mono-atomic and diatomic gases, plays a crucial role in determining the final pressure.

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  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of the Van der Waals equation for real gases
  • Familiarity with adiabatic processes in thermodynamics
  • Concept of the adiabatic exponent (gamma) and its significance
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  • Study the Van der Waals equation and its applications in real gas behavior
  • Explore the concept of adiabatic processes and their equations
  • Learn about the differences in specific heats (Cp and Cv) for various gases
  • Investigate the implications of molecular mass on gas behavior under compression
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Students and professionals in physics, particularly those studying thermodynamics, as well as engineers and scientists working with gas compression and behavior in various applications.

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Gas, pressure, physics, what??

Question: Consider two cylinders of gas identical in all respects except that one contains O2 and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater pressure increase?

The "answer" is apparently that He shows the greater pressure increase since it is lighter. Gases with lower molecular masses will travel at a faster rate. Therefore, their pressure will increase at a faster rate.

Using the ideal gas law PV = nRT, I assumed that V, n, R, and T were all identical since the question mentions "identical in ALL RESPECTS". So just through this, I figured that the pressure has to be identical. Did I do something wrong?

Also, if the "answer" is actually correct, which equation would prove it?
 
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For an ideal gas, this is true. However, you're not dealing with ideal gases, so you need to use the http://en.wikipedia.org/wiki/Van_der_Waals_equationVan .
 
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This ia an adiabatic process. You can find the final pressure from the equation for this process:

pV^gamma=constant.

The adiabatic exponent gamma is different for the two gases. The reason is not the atomic weight but the fact that He is mono-atomic whereas O2 is diatomic.
For ideal gas, gamma = Cp/Cv where Cp and Cv are the specific heats at constant pressure and constant volume, respectively.
 
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