Answer to Question #133839 in Mechanics | Relativity for Jessica

Question

A large tank is filled with methane gas at a concentration of 0.760 kg/m3. The valve of a 1.40-m pipe connecting the tank to the atmosphere is inadvertantly left open for 13.0 hours. During this time, 8.20x10^-4 kg of methane diffuses out of the tank, leaving the concentration of methane in the tank essentially unchanged. The diffusion constant for methane in air is 2.10x10^-5 m2/s. What is the cross-sectional area of the pipe? Assume that the concentration of methane in the atmosphere is zero.

Answer 1

To get the cross-sectional area ,we isolate A and get

$A= \frac{ml}{D(∆C)t}$

$=\frac{8.2\times10^{-4}\times140}{2.10\times10^{-5}\times(0.760-0)\times13\times3600}=1.5\times10^{-3}m^{2}$

For this item ,we use flicks law of diffusion.The equation gives us the mass of solute that diffuses in time through the solvent gas and is given by $m=\frac{DA(∆C)t}{L}$

For the problem ,m is the mass of methane ,L is the length of the tube ,A is cross sectional area,D is the diffusion constant and ∆C is the change in concentration ,or $C_t ank-C_air$ and t is time in seconds.

To get the cross-sectional area , we isolate A and get.

$A=\frac{ml}{D(∆C)t}$

$=\frac{8.2\times10^{-4}\times140}{2.10\times10^{-5}(0.760-0)\times13\times3600}=1.5\times10^{-3}m^{2}$