Finding the number of moles of an ideal gas in a capillary

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Homework Help Overview

The problem involves determining the number of moles of an ideal gas in a capillary with a temperature gradient described by the equation ##T=T_0e^{-kx}##. The context is rooted in thermodynamics and the ideal gas law, with a focus on how temperature and volume relate under constant pressure conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the number of moles using the ideal gas law and considers the implications of a constant pressure scenario. They express concerns about a potential mistake in their reasoning regarding the relationship between volume and temperature.
  • Some participants question the interpretation of the differential equations used and suggest considering the number of moles contained in small segments of the capillary.
  • Others suggest that the independent variable should be the distance along the capillary rather than treating temperature and volume changes independently.

Discussion Status

The discussion is ongoing, with participants providing insights into the interpretation of the differential equations and the relationship between the variables involved. There is a focus on clarifying the reasoning behind the calculations and the assumptions made in the original post.

Contextual Notes

Participants note that the change in temperature is negative, which raises questions about the implications for volume change. The original poster is also grappling with the justification for certain derivatives in the context of the problem.

Potatochip911
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Homework Statement


The temperature across the capillary with constant cross-sectional area and length L is given by ##T=T_0e^{-kx}##. Assuming an ideal gas and constant pressure show the number of moles to be: $$n=\frac{PV(e^{kL} - 1)}{RkLT_0}$$

Homework Equations


##PV=nRT##

The Attempt at a Solution



The equation of state can be expressed as ##g(P,V,T) = 0## but since pressure is given to be constant we have ##g(V,T) = 0## therefore we can express the volume as ##V=V(T)## from which we can get the differential for V as $$dV = (\frac{\partial V}{\partial T})dT = \frac{nR}{P}dT\\ V =\frac{nR}{P} \int_{T_i}^{T_f}dT = \frac{nR}{P}(T_f-T_i)$$

Using ##T=T_0e^{-kx}## it is evident that ##T_f = T_0e^{-kL}## and ##T_i = T_0## therefore $$V = \frac{nRT_0}{P}(e^{-kL}-1)\Longrightarrow n=\frac{PV}{nRT_0(e^{-kL}-1)}$$

Which clearly isn't the correct answer. I'm curious as to what the mistake is following this reasoning.

Another method I've attempted is for an ideal gas ##n=n(P,V,T)## but pressure is constant therefore ##n=n(V,T)## and we obtain the differential $$dn = \left(\frac{\partial n}{\partial V}\right)_{P,T} dV + \left(\frac{\partial n}{\partial T}\right)_{P,V} dT$$

Now I know from the the answer that ##\left(\frac{\partial n}{\partial T}\right)_{P,V}## must equal 0 since the answer is obtained from integrating ##dn = \left(\frac{\partial n}{\partial V}\right)_{P,T} dV ## but I can't seem to justify why ##\left(\frac{\partial n}{\partial T}\right)_{P,V}## should be equal to zero without setting both of them equal to zero.
 
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Potatochip911 said:
I'm curious as to what the mistake is following this reasoning.
You need to understand what your differential equation says. You have taken a constant n and expressed how the volume changes if you change the temperature. Since the change in temperature was negative, you got a negative volume change.
Consider segments length dx and the number of moles they contain.
 
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haruspex said:
You need to understand what your differential equation says. You have taken a constant n and expressed how the volume changes if you change the temperature. Since the change in temperature was negative, you got a negative volume change.
Consider segments length dx and the number of moles they contain.
Edit: Ok so that's the logic behind taking ##dn/dV## and then multiplying the ##dV## over and integrating, but shouldn't this also come from my second method in the main post?
 
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Potatochip911 said:
Edit: Ok so that's the logic behind taking ##dn/dV## and then multiplying the ##dV## over and integrating, but shouldn't this also come from my second method in the main post?
Your second method considers how n varies as T and V change independently. I cannot think what that means in the context of the question.
As I posted, your independent variable should be distance along the capillary.
 
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