In a gaseous mixture of CH4 and C2H6, there are twice as many molecules of CH4 as C2H6. The partial pressure of CH4 is 40 mmHg. What is the partial pressure (in mmHg) of C2H6?
PV = nRT or P = nRT/V.
R , T, and V are constant; therefore, P = k*n so if n is 2 times then P is ......?
Well, let's calculate it! If there are twice as many molecules of CH4 as C2H6, that means the ratio of CH4 to C2H6 is 2:1.
Given that the partial pressure of CH4 is 40 mmHg, we can use this ratio to find the partial pressure of C2H6.
Since the ratio is 2:1, we can say that the total number of molecules in the mixture is 3.
40 mmHg of pressure is contributed by 2 parts of CH4, so each part would contribute 40/2 = 20 mmHg.
Therefore, the partial pressure of C2H6 would be 20 mmHg.
So, the partial pressure of C2H6 is 20 mmHg. It's a gas-tastic ratio!
To find the partial pressure of C2H6, we can use the information given about the ratio of CH4 to C2H6 molecules.
Let's assume there are 'x' molecules of C2H6 in the mixture. According to the given information, there are twice as many molecules of CH4 as C2H6. Therefore, the number of CH4 molecules would be '2x'.
The total pressure of the mixture is the sum of the partial pressures of the individual gases. So, we need to find the partial pressure of C2H6.
The partial pressure of a gas is directly proportional to the number of moles (or molecules) of that gas present. So, if we have the ratio of the number of moles or molecules, we can use it to determine the partial pressure.
Since we know that the ratio of CH4 to C2H6 molecules is 2:1, we can write:
(2x molecules of CH4) / (x molecules of C2H6) = 2/1
Cross-multiplying, we get:
2x = x
Now, let's find the partial pressure of C2H6. We know that the partial pressure of CH4 is 40 mmHg.
According to Dalton's Law of Partial Pressures, the partial pressure of a gas is directly proportional to its mole fraction. Since both CH4 and C2H6 are part of the same gaseous mixture, their mole fractions add up to 1.
The mole fraction of C2H6 can be calculated as:
Mole fraction of C2H6 = (Number of moles of C2H6) / (Total number of moles of all gases)
Since we are given the ratio of CH4 to C2H6 molecules (2:1), we can write:
Total number of moles of all gases = (2x + x) = 3x
Mole fraction of C2H6 = x / (3x) = 1/3
According to Dalton's Law of Partial Pressures, the partial pressure of C2H6 is equal to the mole fraction of C2H6 multiplied by the total pressure of the mixture.
Partial pressure of C2H6 = (Mole fraction of C2H6) * (Total pressure of the mixture)
Partial pressure of C2H6 = (1/3) * (Total pressure of the mixture)
Since the total pressure of the mixture is not given, we cannot determine the exact partial pressure of C2H6.
To find the partial pressure of C2H6 in a gaseous mixture of CH4 and C2H6, we can use the concept of mole ratios.
Let's assume that there are "x" molecules of C2H6 in the mixture. Given that there are twice as many molecules of CH4 as C2H6, the number of molecules of CH4 would be 2x.
Now, we know that the partial pressure (P) of a gas in a mixture is directly proportional to the number of moles (n) of that gas. Therefore, we can write:
P(CH4) / n(CH4) = P(C2H6) / n(C2H6)
Substituting the given values:
40 mmHg / (2x) = P(C2H6) / x
Cross-multiplying:
40 mmHg * x = (2x) * P(C2H6)
Simplifying the equation:
40 mmHg * x = 2x * P(C2H6)
Dividing both sides by 2x:
40 mmHg = P(C2H6)
Therefore, the partial pressure of C2H6 in the gaseous mixture is 40 mmHg.