200cm3 of oxygen diffused through a porous plug in 50 secs,how long will 80cm3 of methane(CH4) take to diffuse through the same porous plug under the same condition.
google "Graham's law of diffusion"
To solve this question, we can use Graham's law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's first find the molar masses of oxygen (O2) and methane (CH4):
- Molar mass of oxygen (O2):
O = 16.00 g/mol (atomic mass of oxygen)
Molar mass of O2 = 16.00 g/mol * 2 = 32.00 g/mol
- Molar mass of methane (CH4):
C = 12.01 g/mol (atomic mass of carbon)
H = 1.01 g/mol (atomic mass of hydrogen)
Molar mass of CH4 = 12.01 g/mol + 1.01 g/mol * 4 = 16.05 g/mol
Now, we can calculate the ratio of the square roots of the molar masses:
Square root of molar mass of O2 : Square root of molar mass of CH4
= √(32.00 g/mol) : √(16.05 g/mol)
Now, simplify the equation:
√(32.00 g/mol) : √(16.05 g/mol)
= 5.657 : 4.005
≈ 1.414 : 1 (rounded to 3 decimal places)
From the ratio, we can determine that methane (CH4) diffuses approximately 1.414 times slower than oxygen (O2).
Since it took 50 seconds for 200 cm³ of oxygen to diffuse, we can calculate the time it would take for 80 cm³ of methane to diffuse using the following ratio:
Time taken for oxygen to diffuse : Time taken for methane to diffuse
= 1 : 1.414
Let's solve for the time taken for methane:
50 seconds : Time taken for methane to diffuse
= 1 : 1.414
Cross-multiplying:
50 seconds * 1.414 = Time taken for methane to diffuse
Time taken for methane to diffuse = 70.7 seconds (rounded to 1 decimal place)
Therefore, it would take approximately 70.7 seconds for 80 cm³ of methane (CH4) to diffuse through the same porous plug under the same conditions.
To solve this problem, we will use Graham's Law of diffusion. According to Graham's Law, the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.
The formula for Graham's Law is:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
Let's break down the information given in the problem:
In the first scenario:
Volume1 = 200 cm3
Gas1: Oxygen (O2)
Rate1 = ?
Molar mass1 = 32 g/mol (molar mass of O2)
In the second scenario:
Volume2 = 80 cm3
Gas2: Methane (CH4)
Rate2 = ?
Molar mass2 = 16 g/mol (molar mass of CH4)
Since the porous plug and conditions are the same in both scenarios, the rates of diffusion can be compared.
Now we will set up the equation using Graham's Law:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
Substituting the values:
Rate1 / Rate2 = √(16 g/mol / 32 g/mol)
Rate1 / Rate2 = √(1/2)
Simplifying, we get:
Rate1 / Rate2 = 1 / √2
Now, plug in the values using cross-multiplication:
Rate1 = Rate2 * 1 / √2
Rate2 = Rate1 * √2
Since we know the rate of diffusion for gas1 (oxygen) is 200 cm3/50 sec = 4 cm3/sec, we can calculate the rate for gas2 (methane) as:
Rate2 = 4 cm3/sec * √2
Rate2 = 4 cm3/sec * 1.414
Rate2 = 5.656 cm3/sec (rounded to three decimal places)
To find out how long it will take for 80 cm3 of methane to diffuse through the same porous plug, we divide the volume by the rate:
Time = Volume2 / Rate2
Time = 80 cm3 / 5.656 cm3/sec
Time ≈ 14.14 seconds (rounded to two decimal places)
Therefore, it will take approximately 14.14 seconds for 80 cm3 of methane (CH4) to diffuse through the same porous plug under the same conditions.