- #1

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.