A sample of H2 gas (12.28 g) occupies 100.0 L at 400.0 K and 2.00 atm. How much volume will a 9.49 g sample of H2 occupy at 353 K and 2.00 atm?
a. 68.2 L
b. 54.7 L
c. 147 L
d. 109 L
e.77.3 L
Is it a. 68.2 L?
Looks ok to me.
it is a.68.2
Well, I gotta say, this question really has a lot of gas! But don't worry, I won't let it go to my head.
To solve this problem, we can use the Ideal Gas Law equation, which is PV = nRT. But wait, let me tell you a secret... nRT is never the same without U. 😉
So, first, let's figure out the number of moles of H2 gas in the initial sample. We can use the formula n = m/M, where m is the mass and M is the molar mass of H2. The molar mass of H2 is 2 g/mol, so n = 12.28 g / 2 g/mol = 6.14 mol.
Now, let's use the equation PV = nRT to find the initial volume (V1). We have P1 = 2.00 atm, T1 = 400.0 K, and n = 6.14 mol. Rearranging the equation, we get V1 = nRT1 / P1.
Plugging in the values, we have V1 = (6.14 mol)(0.0821 L·atm/(K·mol))(400.0 K) / 2.00 atm = 98.84 L.
Now, that we know the initial volume, let's find the final volume (V2) using the same equation PV = nRT, but this time with P2 = 2.00 atm and T2 = 353 K. Rearranging, we get V2 = nRT2 / P2.
Plugging in the values, we have V2 = (9.49 g / 2 g/mol)(0.0821 L·atm/(K·mol))(353 K) / 2.00 atm ≈ 68.21 L.
So, the final volume of the 9.49 g sample of H2 gas at 353 K and 2.00 atm is approximately 68.21 L. Looks like you nailed it, my friend! So, yes, your answer of 68.2 L, option a, is correct! Keep up the good work!
To solve this problem, we can use the combined gas law formula:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Given:
P1 = 2.00 atm (initial pressure)
V1 = 100.0 L (initial volume)
T1 = 400.0 K (initial temperature)
P2 = 2.00 atm (final pressure)
T2 = 353 K (final temperature)
We need to find V2 (final volume).
Rearranging the formula to solve for V2:
V2 = (P2 * V1 * T2) / (P1 * T1)
Plugging in the values:
V2 = (2.00 atm * 100.0 L * 353 K) / (2.00 atm * 400.0 K)
V2 = 176600 / 80000
V2 = 2.2075 L
Therefore, the volume of the 9.49 g sample of H2 at 353 K and 2.00 atm is approximately 2.21 L.
Since none of the provided answer choices match exactly, we need to round the answer to the nearest tenth. The closest answer choice is e. 77.3 L, but it is still not the correct answer.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
First, let's calculate the number of moles of H2 in the initial sample using its molar mass (2.016 g/mol):
n1 = mass / molar mass
n1 = 12.28 g / 2.016 g/mol
n1 ≈ 6.096 mol
Now, we can rearrange the ideal gas law equation to solve for the final volume (V2) using the information given:
V1 / T1 = V2 / T2
V2 = (V1 * T2) / T1
Substituting the values we have:
V2 = (100.0 L * 353 K) / 400.0 K
V2 ≈ 88.25 L
Finally, we can compare the calculated volume with the given multiple-choice options to find the nearest answer. From the available options, the closest volume is 77.3 L (option e), so the correct answer is:
e. 77.3 L