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?

 

[Blueshift5]

Your calculation seems to be correct, but it is important to note that in a reversible process, the system is always in equilibrium with its surroundings. This means that the pressure and temperature of the system will change gradually as it expands, unlike in an irreversible process where the pressure and temperature may change abruptly. Therefore, your final temperature calculation may be slightly off due to the assumption of a constant pressure. To account for this, you may need to use an equation such as the ideal gas law for a reversible process, which takes into account the gradual changes in pressure and temperature. Additionally, it may be helpful to consider the specific heat capacity of argon and how it may affect the final temperature. Overall, your approach is on the right track, but it is important to consider the concept of reversibility in your calculations.