How to Calculate ΔUm, q, and w for the Isothermal Expansion of Nitrogen Gas?

In summary, to calculate the change in internal energy (ΔU) for the isothermal expansion of nitrogen gas, we use the equation dU(V,T)=πTdV+CvdT and substitute in the values for πT and Cv. Since this is an isothermal process, we can set dT=0 and simplify the equation to dU(V,T)=πTdV. Integrating both sides, we get ΔU=a∫dV/Vm2. Plugging in the values for a, initial and final volumes, we get a value of 0.676*n^2 J/mol2 for ΔU. To calculate q, we need to use the equation w=-nRT∫dV/V and
  • #1
lee403
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1. For a van der Waals gas πT=a/Vm2. Calculate ΔUm for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm3 to 24.8 dm3at 298K. What are values for q and w?

Homework Equations


I looked up a for N2 gas to be 1.352 atm*dm6/mol2
dU(V,T)=(∂U/∂V)TdV+(∂U/∂T)VdT
(∂U/∂V)TT
(∂U/∂T)V=Cv

3. The Attempt at a Solution

I start with (∂U/∂V)TT and substitute in πT and Cv
which gives dU(V,T)=πTdV+CvdT

Since this is isothermal dt=0 so dU(V,T)=πTdV

I integrate both sides ∫dU=∫πTdV. πT= a/Vm2.
a is a constant so it comes out of the integral leaving
ΔU=a∫dV/Vm2.= a*n2 ∫ dV/V2

Evaluating the integral
ΔU= -((a*n2)/2)*V-3.
After plugging in values for a, Vi, and Vf I get:
ΔU=-((1.352 *n2)/2)*[(1/24.83)-(1/13)]

I get 0.676 *n 2 J/mol2. I just don't know what to do with the moles.

I know how to get work which is just w=-nrt∫dV/V and the solution is -7.95 kJ of work
but without ΔU I cannot calculate q.
 
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  • #2
I actually solved my own problem. I apparently don't know how to integrate.
 
  • #3
Could you post the solution, please? I'm having trouble with this one myself.
 
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  • #4
Everything is right up until I pulled out the n^2 from the integral, just keep it as Vm. I also integrated wrong so if you get the proper solution of the integral, which is -1/Vm then you can just plug everything in. With unit conversion you get U then you can calculate q.
 
  • #5
Thank you!
 

Related to How to Calculate ΔUm, q, and w for the Isothermal Expansion of Nitrogen Gas?

1. What is Van der Waals force?

Van der Waals force is a type of intermolecular force that exists between molecules. It is caused by temporary dipoles that form in molecules due to the constant motion of electrons, creating weak attractions between molecules.

2. How does Van der Waals force relate to nitrogen gas?

Nitrogen gas is a nonpolar molecule, meaning that it has an equal distribution of charge and no permanent dipole moment. However, Van der Waals forces still exist between nitrogen molecules due to the temporary dipoles that form.

3. What is the role of Van der Waals force in the properties of nitrogen gas?

Van der Waals force contributes to the cohesive forces between nitrogen molecules, making it more difficult to separate them and increasing its boiling point and melting point. It also affects the compressibility and density of nitrogen gas.

4. How does temperature affect Van der Waals force in nitrogen gas?

As temperature increases, the motion of molecules also increases, leading to stronger temporary dipoles and stronger Van der Waals forces. This results in a decrease in the compressibility and an increase in the density of nitrogen gas.

5. Can Van der Waals force be stronger than other intermolecular forces?

In general, Van der Waals forces are weaker than other intermolecular forces such as hydrogen bonding or dipole-dipole interactions. However, in certain cases, such as between large molecules or at very close distances, Van der Waals forces can become stronger than other intermolecular forces.

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