Entropy of Vaporization Calculation for 5 Moles of Argon at 87.5 K and 150 K

In summary, we calculate the change in entropy for 5 moles of liquid argon undergoing vaporization at its normal boiling point (87.5 K) and subsequent heating to 150 K under constant volume conditions. Using the standard enthalpy change of vaporization for argon, we use Trouton's Rule to calculate the entropy of vaporization at 87.5 K and then use the ideal gas law to calculate the entropy change from 87.5 K to 150 K. The total entropy change is found to be 405.04 J K-1.
  • #1
dancingdodo27
6
1

Homework Statement



5 moles of liquid argon undergoes vaporization at its normal boiling point (87.5 K) and the resulting argon gas is subsequently heated to 150 K under constant volume conditions. Calculate the change of entropy for this process. The standard enthalpy change of vaporization, ∆ ⊖ is +6.5 kJ mol-1 for argon. Treat gaseous argon as an ideal gas in your calculations.

Homework Equations


ΔS= ΔvapH/ T
ΔS= q/T
ΔS= nCvln(Tf/Ti)

The Attempt at a Solution


For entropy of vaporisation at 87.5 K
ΔvapH= 6.5 kJ mol-1 for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol-1
Using Trouton's Rule ΔS1= 32500/87.5 = 371.43 J K-1

Entropy from 87.5 K to 150 K
ΔS2= 5⋅ 3/2R⋅ ln(150/87.5)
ΔS2= 33.61 J K-1

Total entropy change= ΔSsys= ΔS1 + ΔS2
ΔSsys= 371.43 + 33.61 = 405.04 J K-1
 
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  • #2
dancingdodo27 said:

Homework Statement



5 moles of liquid argon undergoes vaporization at its normal boiling point (87.5 K) and the resulting argon gas is subsequently heated to 150 K under constant volume conditions. Calculate the change of entropy for this process. The standard enthalpy change of vaporization, ∆ ⊖ is +6.5 kJ mol-1 for argon. Treat gaseous argon as an ideal gas in your calculations.

Homework Equations


ΔS= ΔvapH/ T
ΔS= q/T
ΔS= nCvln(Tf/Ti)

The Attempt at a Solution


For entropy of vaporisation at 87.5 K
ΔvapH= 6.5 kJ mol-1 for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol-1
Using Trouton's Rule ΔS1= 32500/87.5 = 371.43 J K-1

Entropy from 87.5 K to 150 K
ΔS2= 5⋅ 3/2R⋅ ln(150/87.5)
ΔS2= 33.61 J K-1

Total entropy change= ΔSsys= ΔS1 + ΔS2
ΔSsys= 371.43 + 33.61 = 405.04 J K-1
Your methodology looks correct. I haven't checked your arithmetic.
 

Related to Entropy of Vaporization Calculation for 5 Moles of Argon at 87.5 K and 150 K

1. How is entropy of vaporization calculated?

The entropy of vaporization is calculated using the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization and the gas constant. It is represented by the formula ΔS = ΔHvap / T, where ΔS is the change in entropy, ΔHvap is the enthalpy of vaporization, and T is the temperature in Kelvin.

2. What is the significance of 5 moles of Argon in this calculation?

In this calculation, 5 moles of Argon is the amount of gas being analyzed. It is a unit of measurement used in chemistry to represent a specific quantity of a substance. In this case, it is used to determine the change in entropy for a specific amount of Argon gas.

3. Why are the temperatures of 87.5 K and 150 K used?

The temperatures of 87.5 K and 150 K are used in this calculation because they represent the boiling and melting points of Argon, respectively. These specific temperatures are used to determine the change in entropy during the phase transition from liquid to gas.

4. What is the purpose of calculating the entropy of vaporization for Argon?

The purpose of calculating the entropy of vaporization for Argon is to understand the thermodynamic behavior of the gas during the phase transition from liquid to gas. It can also provide valuable information about the energy required for the phase change and the stability of the gas at different temperatures.

5. Are there any limitations to this calculation?

Yes, there are limitations to this calculation. It assumes ideal gas behavior and does not take into account any non-idealities or deviations from the ideal gas law. It also assumes that the Argon gas is pure and does not consider any impurities or interactions with other substances. Additionally, the calculation is not applicable to phase transitions that occur at temperatures below the melting point of Argon.

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