Reversible Process: Final Temp Calc of Argon Mass 12.0g

AI Thread Summary
To calculate the final temperature of 12.0 g of argon expanded reversibly and adiabatically from 1.0 L to 3.0 L, the ideal gas law is initially applied, yielding a pressure of 6.7 atm. However, the discussion highlights that the approach must consider the adiabatic process, which requires a different equation for accurate results. The user is prompted to identify the appropriate adiabatic equation to apply in this scenario. The oversight in the initial calculation is linked to not incorporating the temperature in the pressure calculation. Correctly addressing the adiabatic nature of the expansion is essential for determining the final temperature accurately.
Gil-H
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Homework Statement


Calculate the final temperature of a sample of argon of mass 12.0 g that is expanded
reversibly and adiabatically from 1.0 L at 273.15 K to 3.0 L.


Homework Equations





The Attempt at a Solution



n = 12[g]/40[g][mol]-1 = 0.3 [mol]

pVi=nRTi

p = 0.3[mol]0.082[L][atm][K]-1[mol]-1/1[L] = 6.7 [atm]

Tf = pVf/nR = 6.7[atm]3[L]/0.3[mol]0.082[L][atm][K]-1[mol]-1 = 817.07 [K]

What is wrong? What have I overlooked?
I think it has something to do with the term 'reversibly', but how?
 
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Gil-H said:
p = 0.3[mol]0.082[L][atm][K]-1[mol]-1/1[L] = 6.7 [atm]

How can that be correct? You haven't multiplied in the temperature.

To solve this problem, you need to use an equation that's specifically meant for adiabatic expansion/contraction. Do you know the equation?
 
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