The diffusion constant for the alcohol ethanol in water is 1.24x10^-9 m2/s. A cylinder has a cross-sectional area of 4.10 cm^2 and a length of 2.06 cm. A difference in ethanol concentration of 1.37 kg/m^3 is maintained between the ends of the cylinder.
In one hour, what mass of ethanol diffuses through the cylinder?
well, I will try to do dimensional analysis because I have no idea.
constant * cross sectional area
inversely proportional to length
1 hr = 3600 s
4.1 cm^2* 1m^2/10^4 cm^2 = 4.1*10^-4 m^2
Q = k * A/L * delta c * time
try
Q = 1.24*10^-9 m^2/s * 4.1*10^-4 m^2 /.0106 m * 1.37 kg/m^3 * 3600 s
gives kilograms in an hour
Damon was close.
m= (DAΔC)t/L = 1.24e-9*4.10e-4*1.37*3600/2.06e-2
= 1.217e-7
THANK YOU!!!!
To calculate the mass of ethanol that diffuses through the cylinder in one hour, we first need to calculate the volume of the cylinder.
The volume of a cylinder can be calculated by multiplying the cross-sectional area of the cylinder by its length.
Given:
Cross-sectional area = 4.10 cm^2 (convert to m^2)
Length = 2.06 cm (convert to m)
To convert the units:
1 cm^2 = 1 x 10^-4 m^2 (1 square cm is equal to 1 times 10 to the power of negative 4 square meters)
1 cm = 1 x 10^-2 m (1 cm is equal to 1 times 10 to the power of negative 2 meters)
Now we can convert the given values:
Cross-sectional area = 4.10 cm^2 x (1 x 10^-4 m^2/cm^2) = 4.10 x 10^-4 m^2
Length = 2.06 cm x (1 x 10^-2 m/cm) = 2.06 x 10^-2 m
Next, we can calculate the volume of the cylinder:
Volume = Cross-sectional area x Length
Volume = (4.10 x 10^-4 m^2) x (2.06 x 10^-2 m) = 8.446 x 10^-6 m^3
To calculate the mass of ethanol that diffuses through the cylinder, we can use Fick's first law of diffusion:
Mass = Diffusion constant x Area x Concentration gradient x Time
Given:
Diffusion constant = 1.24 x 10^-9 m^2/s
Area = Cross-sectional area = 4.10 x 10^-4 m^2
Concentration gradient = 1.37 kg/m^3
Time = 1 hour (convert to seconds)
To convert the units:
1 hour = 3600 seconds (1 hour is equal to 3600 seconds)
Now we can calculate the mass of ethanol:
Mass = (1.24 x 10^-9 m^2/s) x (4.10 x 10^-4 m^2) x (1.37 kg/m^3) x (3600 s)
Mass = 1.7494 x 10^-8 kg
Therefore, in one hour, approximately 1.7494 x 10^-8 kg of ethanol diffuses through the cylinder.