Free expansion process and temperature change

[Amaelle]

Good day all!
I have the following exercise
One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)

My problem is with the joule expansion : in the joule expansion W12=0 and because we have an adiabatic process Q=0
so according to the first law of thermodynamics dU12=0 which means T2=T1 which I totally agree with

On the other hand we have an adiabatic process which means T1V1^(k-1)=T2V2(^k-1) so T2=T1*(V1/V2)^(k-1) so accordingly T2 ≠T1?!

THIS IS THE POINT I'M STRUGGLING WITH AND ANY HELP WOULD BE HIGHLY APPRECIATED THANKS!

 

[TSny]

On the other hand we have an adiabatic process which means T1V1^(k-1)=T2V2(^k-1)

This equation does not apply to every adiabatic process of an ideal gas. Do you know what other condition must be met? Does a free expansion meet this additional requirement?

 

[Amaelle]

thanks for your prompt reply!
I really don't know, this is why I'm asking
I would be very grateful if you could elaborate more

 

[ehild]

Good day all!
I have the following exercise
One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)

My problem is with the joule expansion : in the joule expansion W12=0 and because we have an adiabatic process Q=0
so according to the first law of thermodynamics dU12=0 which means T2=T1 which I totally agree with

On the other hand we have an adiabatic process which means T1V1^(k-1)=T2V2(^k-1) so T2=T1*(V1/V2)^(k-1) so accordingly T2 ≠T1?!

The equation
T1V1^(k-1)=T2V2(^k-1)
is valid for a quasistatic, , "equilibrium" process, and free expansion is not that. The intensive parameters, temperature, pressure are defined for the static case. During free expansion, the pressure outside is zero, the piston moves fast, neither pressure nor temperature is defined inside the gas .

 

[Amaelle]

Awesome! thanks a million! to both of you ehild and Tsny

 

[Chestermiller]

thanks for your prompt reply!
I really don't know, this is why I'm asking
I would be very grateful if you could elaborate more

It applies to an adiabatic reversible process. Joule expansion is not a reversible expansion.

 

[Amaelle]

It applies to an adiabatic reversible process. Joule expansion is not a reversible expansion.

Thanks Chestermiller! you nail it as usual!

 

[Apashanka]

One kilo-mole of air expands adiabatically from an initial state p1=1 bar, T1=340 K to a final state V2= 2V1. Compute the temperature of the final state, for the following cases: Joule expansion (gas free expansion with W12=0)

For free expansion the change in the internal energy is 0 and as the process is adiabatic dQ is 0 and from that we conclude that dW (work done) is also 0.
now dU(T,v)=C_vdT+(∂U/∂v)_Tdv ,
Now the final temp. is C_v(T_f-T_i)=-∫(∂U/∂v)_Tdv

 

[Chestermiller]

For free expansion the change in the internal energy is 0 and as the process is adiabatic dQ is 0 and from that we conclude that dW (work done) is also 0.
now dU(T,v)=C_vdT+(∂U/∂v)_Tdv ,
Now the final temp. is C_v(T_f-T_i)=-∫(∂U/∂v)_Tdv

This is basically correct, except that, at 340 K and 1 bar, air behaves essentially as an ideal gas, in which case the temperature change would be very close to zero.