A large tank is filled with methane gas at a concentration of 0.740 kg/m^3. The

Question

A large tank is filled with methane gas at a concentration of 0.740 kg/m^3. The valve of a 1.40-m pipe connecting the tank to the atmosphere is inadvertently left open for 19.0 hours. During this time, 6.30 Times 10^-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 2.10 Times10^-5 m^2/s. What is the cross-sectional area of the pipe? Assume that the concentration of methane in the atmosphere is zero. Number Units the tolerance is +/-2%

Explanation / Answer

Diffusion connstant(D) = 2.10*10-5 m2/s

Diffusion flux = -D*change_in_concentration/distance = 2.10*10-5*0.74/1.4 kg/(m2*s) = 1.11*10-5kg/(m2*s)

So amount of methane leaked from A area in 19hrs will be Diffusion_flux*A*time

=> 6.3*10-4 = 1.11*10-5*A*19*60*60 => A = 6.3*10-4 /(1.11*10-5*19*60*60) = 0.829*10-3 m2 Answer

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