Integral: Solve (nRT)/(V-nb)dV from Vi to Vf

In summary, for the given integral of (nRT)/(V-nb)dV from Vi to Vf, the suggested substitution is u = V-nb. The resulting differential is du/dv = 1, making it safe to integrate by ln. The final solution is ln|Vf-nb| - ln|Vi-nb|. It is important to note that the limits of integration should also be written in terms of u.
  • #1
Riles246
12
0
1. The problem statement, all variables and given

integral of : (nRT)/(V-nb)dV from Vi to Vf[tex]\int[/tex] [tex]^{Vf}_{Vi}[/tex] (nRT)/(V-nb)dV

I am really stuck on this one!
Do you need to move the denominator up first?
 
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  • #2
Assuming that everything except V is a constant for this problem, try the substitution

u = V -nb
 
  • #3
You need to tell us which are constants and which are functions of V in this question. Otherwise, the problem becomes very simple when you factor nRT out.
 
  • #4
Since the integral says "dV" doesn't that mean it only needs to be integrated with respect to the variable V while holding all other variables constant?
 
  • #5
Assuming everything else is constant.

factor the nRT out. Let u = V-nb.
du/dv = 1 if nb is a constant.
This means that when u differentiate the denominator u get the numerator and its safe to straightaway ln the denominator.


I think it should be ln|Vf-nb| - ln|Vi-nb|

i might be wrong though.

Btw, is it just a practice question or are u doing this for chem? just curious. ;).
 
  • #6
Also don't forget if you substitute u = V - nb, you must also write the limits of integration in terms of u.
 
  • #7
I am doing these problems to touch up on my Calc for a class I am taking (physical chem)
 

Related to Integral: Solve (nRT)/(V-nb)dV from Vi to Vf

1. What is the formula for "Integral: Solve (nRT)/(V-nb)dV from Vi to Vf"?

The formula is (nRT) ln(V-nb) + C, where n is the number of moles, R is the ideal gas constant, T is the temperature, b is the excluded volume, V is the volume, Vi is the initial volume, and Vf is the final volume.

2. What is the significance of the integral in this equation?

The integral represents the sum of infinitesimal changes in volume dV over the entire range from Vi to Vf, and is used to calculate the total work done by a gas as it expands or contracts within this range.

3. How is this integral derived?

This integral is derived from the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the work done is calculated by integrating the pressure-volume curve of an ideal gas.

4. What are the units for the variables in this equation?

The units for n are moles, R is typically given in units of Joules/mol*K, T is in units of Kelvin, b is in units of m3/mol, and V, Vi, and Vf are in units of m3. The unit for the integral itself will be in Joules.

5. How can this integral be used in real-world applications?

This integral, along with the ideal gas equation (pV=nRT), can be used to calculate the work done in various processes involving gases, such as in an internal combustion engine or in a refrigeration system. It is also important in understanding the thermodynamics of gases and their behavior under different conditions.

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