Isothermal and reversible expansion work problem (Physical Chem)

In summary, the problem involves finding the work and heat for the reversible and isothermal condensation of 2 mol of CH3OH (g) to liquid at 64*C. The standard enthalpy of condensation at this temperature is -35.3 kJ/mol. The volume of the gas and liquid can be calculated using the ideal gas equation and an assumption about density, and the work can be calculated using the formula w = -nRT * ln(Vf/Vi). However, the ideal gas equation may not be completely accurate and the Van der Waals equation may need to be considered for a more accurate calculation.
  • #1
LakeMountD
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Homework Statement


A sample of 2.00 mol. CH3OH (g) is condensed isothermally and reversibly to liquid at 64*C. The standard enthalpy of condensation of methanol @ 64*C is -35.3 kJ/mol. Find w and q for the reaction.


Homework Equations


w = -nRT * ln(Vf/Vi) (although I don't have volume so I don't know how to find it)


The Attempt at a Solution



Honestly I can't find any other reversible/isothermal condition work formulas to use. I am sure I am just missing how to calculate volume here. I have n, R, and T but am not given V. Not sure if this can be calculated?
 
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  • #2
What is the volume of 2 mol of methanol at 64C? Make an assumption about it's density and calculate it (I assume you know how to convert moles to mass). Same thing goes for the volume of 2 moles of gas (methanol before condensation). The ideal gas equation might not be completely accurate but it is a good place to start.
 
  • #3
Well, I am stuck on the same problem. I don't think that the ideal gas expression is the correct one... also, what do you get if you do 64 kJ/mol * 2 mol ? (I think this is the enthalpy for the reaction). And then, should you go back to the Van der waals equation for the real gas and implement that in your calculation of work ?
 

Related to Isothermal and reversible expansion work problem (Physical Chem)

1. What is isothermal and reversible expansion work?

Isothermal and reversible expansion work is a type of thermodynamic process in which a gas expands at a constant temperature and in a reversible manner. This means that the expansion of the gas is done slowly and without any energy losses, allowing the system to return to its original state if the process is reversed.

2. How is the work done during isothermal and reversible expansion calculated?

The work done during isothermal and reversible expansion can be calculated using the equation W = -nRTln(V2/V1), where n is the number of moles of the gas, R is the gas constant, T is the temperature, and V1 and V2 are the initial and final volumes of the gas.

3. What are some real-life examples of isothermal and reversible expansion?

One example of isothermal and reversible expansion is the expansion of a gas in a cylinder with a movable piston. As the gas expands, the piston moves slowly and without any energy losses, allowing the gas to return to its original state if the process is reversed. Another example is the expansion of a balloon filled with helium at a constant temperature.

4. How does isothermal and reversible expansion work differ from other types of expansion work?

Isothermal and reversible expansion work differs from other types of expansion work in that it is done at a constant temperature and in a reversible manner. This means that the process is slow and there are no energy losses, making it an ideal process for calculating work done in thermodynamic systems.

5. What is the significance of isothermal and reversible expansion work in physical chemistry?

Isothermal and reversible expansion work is significant in physical chemistry because it allows for the calculation of work done in thermodynamic systems at a constant temperature. This helps in understanding the behavior of gases and their properties, and is also used in various industrial processes such as refrigeration and gas compression.

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