Reversible Process: Final Temp Calc of Argon Mass 12.0g

[Gil-H]

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]

pV_i=nRT_i

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

T_f = pV_f/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?

 

[ideasrule]

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?