If 100cm3 of oxygen diffuses through a porous plug at 5cm3/sec how long will it take 200cm3 of methane to diffuse through the same plug under similar conditions
rate1 = oxygen at 5 cc/sec
rate 2 = methane at x cc/sec
molar mass = mm oxygen = 32
molar mass = mm methane = 16
(rate1/rate2) = sqrt [(mm CH4/mm O2)]
(5/x)= sqrt (16/32).
25/x^2 = 16/32 = 0.5
25 = 0.5x^2
x = sqrt(25/0.5) = sqrt (50) = 7.07
so rate CH4 = 7.07 cc/sec
It takes 1 sec to diffuse 7.07 cc so it will take ? sec to diffuse 200 cc
Well, let's calculate with a humorous twist!
First things first, oxygen and methane walk into a bar. The bartender, a porous plug, says, "Hey, oxygen! You'll need to diffuse through me at 5 cm3/sec." Oxygen walks up to the plug and starts diffusing.
Now, methane, being a bit slower, says, "Hey, Mr. Porous Plug, can you tell me how long it'll take for me to diffuse?" The plug scratches its head and replies, "Sure, let me do some calculations while juggling a few atoms."
If 100 cm3 of oxygen takes 5 seconds to diffuse, then 1 cm3 takes 5/100 = 0.05 seconds. Now, for the methane to diffuse 200 cm3, we'll need to do a bit of mathematical clownery.
200 cm3 of methane, each taking 0.05 seconds, would require a total of 200 * 0.05 = 10 seconds. Ta-da!
So, methane, you'll need to patiently wait for about 10 seconds for your turn to diffuse through the same plug under similar conditions as oxygen did. Enjoy the show while you wait, and don't forget to bring some popcorn for an extra touch of gas-troentertainment!
To find the time it takes for 200cm3 of methane to diffuse through the same plug under similar conditions, we can use the concept of molar volume and the rate of diffusion.
First, let's calculate the molar volume of oxygen and methane at standard conditions.
Molar volume of a gas at standard conditions (STP) is 22.4 L/mol. This is equal to 22.4 x 1000 cm3/mol.
Now, let's find the number of moles of oxygen that diffused through the plug.
100 cm3 of oxygen is equal to 100/22.4 mol.
So, the number of moles of oxygen is 100/22.4 mol.
Next, let's convert the volume of methane to moles.
200 cm3 of methane is equal to 200/22.4 mol.
Now, let's find the ratio of the number of moles for methane to oxygen.
(200/22.4 mol)/(100/22.4 mol) = 200/100 = 2.
The ratio tells us that the volume of methane is twice the volume of oxygen that diffused. Therefore, we can conclude that it will take twice the time for methane to diffuse compared to oxygen.
Since oxygen took 1 second to diffuse (5 cm3/sec x 100/22.4 sec = 5 seconds), methane will take 2 seconds to diffuse through the same plug under similar conditions.
So, it will take 2 seconds for 200 cm3 of methane to diffuse through the same plug.
To determine the time it will take for 200cm3 of methane to diffuse through the same porous plug under similar conditions, we can use a proportion based on the rates of diffusion.
First, we need to find the time it takes for 100cm3 of oxygen to diffuse. We can divide the volume of oxygen (100cm3) by the rate of diffusion (5cm3/sec):
Time for 100cm3 of oxygen to diffuse = 100cm3 / 5cm3/sec = 20 seconds
Now, we can use this information to find the time it takes for 200cm3 of methane to diffuse. We set up a proportion based on the volumes and times:
Time for 100cm3 of oxygen / Volume of oxygen = Time for 200cm3 of methane / Volume of methane
20 seconds / 100cm3 = Time for 200cm3 of methane / 200cm3
Now we can solve for the unknown time:
20 seconds / 100cm3 = Time for 200cm3 of methane / 200cm3
Cross multiplying:
(20 seconds * 200cm3) / 100cm3 = Time for 200cm3 of methane
4000 seconds = Time for 200cm3 of methane
Therefore, it will take 4000 seconds for 200cm3 of methane to diffuse through the same porous plug at a rate of 5cm3/sec.