Prove non adiabatic reversible expansion and ΔS

In summary, for a monatomic ideal gas undergoing a transformation from 0.00°C and 2.00 atm to -40.0°C and 0.400 atm, it can be shown that this is not an adiabatic reversible expansion by calculating ΔS for the change. This requires considering a reversible path that includes a constant pressure step and a constant temperature step. The equations for ΔU, pV, and ΔS were used to derive the necessary equations for constant pressure and temperature.
  • #1
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Homework Statement


One mole of a monatomic ideal gas is transformed from 0.00°C and 2.00 atm to
-40.0°C and 0.400 atm. Show that this is NOT an adiabatic reversible expansion and
calculate ΔS for the change. (Hint: you will need to consider a reversible path that
includes a constant pressure step and a constant temperature step.)

Homework Equations


ΔU=q+w
pV=nRT
ΔS=nC(p,m)ln(Vf/Vi)

The Attempt at a Solution


For adiabatic reversible expansion, q=0, so i have to show that q= a number.
Found Vi= 11.2L and Vf=47.8L via ideal gas law

ΔS=(1mol)(5/2(8.314J/Kmol)ln(47.2/11.2)
=30.15J/K
 
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  • #2
Hi Nick_273! :smile:

I'm afraid it does not suffice to calculate ΔS for the change in volume.
You also need to account for the change in either pressure or temperature.

There's a few more relevant equations that you could derive, but you can also find them here:
http://en.wikipedia.org/wiki/Table_of_thermodynamic_equations

In particular you can find the equations for ΔS here when p is constant, but also when T is constant.
You need these 2 to follow the hint.
 

1. What is non-adiabatic reversible expansion?

Non-adiabatic reversible expansion refers to a thermodynamic process in which a gas expands without gaining or losing any heat. This means that the expansion occurs at a constant temperature, and the system is in thermal equilibrium with its surroundings throughout the process.

2. How is non-adiabatic reversible expansion different from adiabatic expansion?

In adiabatic expansion, the gas expands without gaining or losing any heat, but the expansion also occurs without any transfer of energy as work. This means that the temperature of the gas will change during the expansion. In non-adiabatic reversible expansion, the gas expands at a constant temperature, but work is still being done on or by the system.

3. What is the significance of proving non-adiabatic reversible expansion?

Proving non-adiabatic reversible expansion is important because it helps us understand and predict the behavior of gases in various thermodynamic processes. It also allows us to calculate and analyze the changes in entropy (ΔS) during the expansion, which is crucial in understanding the efficiency of various systems.

4. How do you determine the change in entropy (ΔS) during non-adiabatic reversible expansion?

The change in entropy (ΔS) during non-adiabatic reversible expansion can be determined by using the formula ΔS = nRln(V2/V1), where n is the number of moles of gas, R is the gas constant, and V2/V1 is the ratio of final and initial volumes.

5. Can non-adiabatic reversible expansion occur in real-world systems?

Yes, non-adiabatic reversible expansion can occur in real-world systems under certain conditions, such as in a piston-cylinder setup with a perfectly insulated container. However, in most cases, there will be some heat transfer and the process will not be completely reversible.

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