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Avogadro' s law example problems with answers pdf
As a result of the EU's General Data Protection Regulation (GDPR).We are not permitting internet traffic to Byju's website from countries within European Union at this time.No tracking or performance measurement cookies were served with this page. Question: Three balloons are filled with different amounts of an ideal gas. One balloon
is filled with 3 moles of the ideal gas, filling the balloon to 30 L.a) One balloon contains 2 moles of gas. What is the volume of the balloon?b) One balloon encloses a volume of 45 L. How many moles of gas are in the balloon?Answer: Avogadro’s law says the volume (V) is directly proportional to the number of molecules of gas (n) at the
same temperature.n ∝ VThis means the ratio of n to V is equal to a constant value.Since this constant never changes, the ratio will always be true for different amounts of gas and volumes.where ni = initial number of molecules Vi = initial volume nf = final number of molecules Vf = final volume.Part a) One balloon has 3 moles of gas in 30
L. The other has 2 moles in an unknown volume. Plug these values into the above ratio:Solve for Vf(3 mol)Vf = (30 L)(2 mol) (3 mol)Vf = 60 L⋅mol Vf = 20 LYou would expect less gas to take up a smaller volume. In this case, 2 moles of gas only took up 20 L.Part b) This time, the other balloon has a known volume of 45 L and an unknown
number of moles. Start with the same ratio as before:Use the same known values as in part a, but use 45 L for Vf.Solve for nf(3 mol)(45 L) = (30L)nf 135 mol⋅L = (30L)nf nf = 4.5 molesThe larger volume means there is more gas in the balloon. In this case, there are 4.5 moles of the ideal gas in the larger balloon.An alternative method
would be to use the ratio of the known values. In part a, the known values were the number of moles. There was the second balloon had 2/3 the number of moles so it should have 2/3 of the volume and our final answer is 2/3 the known volume. The same is true of part b. The final volume is 1.5 times larger so it should have 1.5 times as
many molecules. 1.5 x 3 = 4.5 which matches our answer. This is a great way to check your work. Avogadro’s law is a specific version of the ideal gas law. It says equal volumes at equal temperatures of an ideal gas all have the same number of molecules. This Avogadro’s law example problem will show how to use Avogadro’s law to find
the number of moles in a given volume or the volume of a given number of moles.Avogadro’s Law ExampleThree balloons filled with different amounts of an ideal gas.Question: Three balloons are filled with different amounts of an ideal gas. One balloon is filled with 3 moles of the ideal gas, filling the balloon to 30 L.a) One balloon
contains 2 moles of gas. What is the volume of the balloon?b) One balloon encloses a volume of 45 L. How many moles of gas are in the balloon?Solution:Avogadro’s law says the volume (V) is directly proportional to the number of molecules of gas (n) at the same temperature.n ∝ VThis means the ratio of n to V is equal to a constant
value.Since this constant never changes, the ratio will always be true for different amounts of gas and volumes.whereni = initial number of moleculesVi = initial volumenf = final number of moleculesVf = final volume.Part a) One balloon has 3 moles of gas in 30 L. The other has 2 moles in an unknown volume. Plug these values into the
above ratio:Solve for Vf(3 mol)Vf = (30 L)(2 mol)(3 mol)Vf = 60 L⋅molVf = 20 LYou would expect less gas to take up a smaller volume. In this case, 2 moles of gas only took up 20 L.Part b) This time, the other balloon has a known volume of 45 L and an unknown number of moles. Start with the same ratio as before:Use the same known
values as in part a, but use 45 L for Vf.Solve for nf(3 mol)(45 L) = (30L)nf135 mol⋅L = (30L)nfnf = 4.5 molesThe larger volume means there is more gas in the balloon. In this case, there are 4.5 moles of the ideal gas in the larger balloon.An alternative method would be to use the ratio of the known values. In part a, the known values were
the number of moles. There was second balloon had 2⁄3 the number of moles so it should have 2⁄3 of the volume and our final answer is 2⁄3 the known volume. The same is true of part b. The final volume is 1.5 times larger so it should have 1.5 times as many molecules. 1.5 x 3 = 4.5 which matches our answer. This is a great way to
check your work.Related Posts Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame. The ChemTeam is not sure when, but probably sometime in the early 1800s. Gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since
volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract. If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases. Why? Suppose the amount is increased. This means there are more
gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase (for an instant), becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure
will return to its original value. The mathematical form of Avogadro's Law is: This means that the volume-amount fraction will always generate a constant if the pressure and temperature remain constant. Let V1 and n1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n2, then
the volume will change to V2. We know this: And we know this: Since k = k, we can conclude: This equation will be very helpful in solving Avogadro's Law problems. You will also see it rendered thusly: V1 / n1 = V2 / n2 Sometimes, you will see Avogadro's Law in cross-multiplied form: V1n2 = V2n1 Avogadro's Law is a direct mathematical
relationship. If one gas variable (V or n) changes in value (either up or down), the other variable will also change in the same direction. The constant K will remain the same value. Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged
temperature and pressure)? Solution: I'll use V1n2 = V2n1 (5.00 L) (1.80 mol) = (x) (0.965 mol) x = 9.33 L (to three sig figs) Example #2: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same.
How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.) Solution: 1) Convert grams of He to moles: 2.00 g / 4.00 g/mol = 0.500 mol 2) Use Avogadro's Law: V1 / n1 = V2 / n2 2.00 L / 0.500 mol = 2.70 L / x x = 0.675 mol 3) Compute grams of He added:
0.675 mol − 0.500 mol = 0.175 mol (0.175 mol) (4.00 g/mol) = 0.7 grams of He added Example #3: A balloon contains a certain mass of neon gas. The temperature is kept constant, and the same mass of argon gas is added to the balloon. What happens? (a) The balloon doubles in volume. (b) The volume of the balloon expands by more
than two times. (c) The volume of the balloon expands by less than two times. (d) The balloon stays the same size but the pressure increases. (e) None of the above. Solution: We can perform a calculation using Avogadro's Law: V1 / n1 = V2 / n2 Let's assign V1 to be 1 L and V2 will be our unknown. Let us assign 1 mole for the amount
of neon gas and assign it to be n1. The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n2) after the Ar is added. (I picked 1.5 because neon weighs
about 20 g/mol and argon weighs about 40 g/mol.) 1 / 1 = x / 1.5 x = 1.5 answer choice (c). Example #4: A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas. More gas is then added to the container until it reaches a final volume of 18.10 L. Assuming the pressure and temperature of the gas remain constant,
calculate the number of moles of gas added to the container. Solution: V1 / n1 = V2 / n2 5.120 L 18.10 L –––––––– = –––––– 8.500 mol x x = 30.05 mol 30.05 − 8.500 = 21.55 mol (to four sig figs) Notice the specification in the problem to determine moles of gas added. The Avogadro Law calculation gives you the total moles required
for that volume, NOT the moles of gas added. That's why the subtraction is there. Example #5: If 0.00810 mol neon gas at a particular temperature and pressure occupies a volume of 214 mL, what volume would 0.00684 mol neon gas occupy under the same conditions? Solution: 1) Notice that the same conditions are the temperature
and pressure. Holding those two constant means the volume and the number of moles will vary. The gas law that describes the volume-mole relationship is Avogadro's Law: V1 V2 ––––– = –––––– n1 n2 2) Substituting values gives: 214 mL V2 ––––––––– = –––––––––– 0.00810 mol 0.00684 mol 3) Cross-multiply and divide for the
answer: V2 = 181 mL (to three sig figs) When I did the actual calculation for this answer, I used 684 and 810 when entering values into the calculator. 4) You may find this answer interesting: Dividing PV1 = n1RT by PV2 = n2RT, we get V1/V2 = n1/n2 V2 = V1n2/n1 V2 = [(214 mL) (0.00684 mol)] / 0.00810 mol V2 = 181 mL In case you
don't know, PV = nRT is called the Ideal Gas Law. You'll see it a bit later in your Gas Laws unit, if you haven't already. Example #6: A flexible container at an initial volume of 6.13 L contains 7.51 mol of gas. More gas is then added to the container until it reaches a final volume of 13.5 L. Assuming the pressure and temperature of the gas
remain constant, calculate the number of moles of gas added to the container. Solution: 1) Let's start by rearranging the Ideal Gas Law (which you'll see a bit later or you can go review it right now): PV = nRT V/n = RT / P R is, of course, a constant. 2) T and P are constant, as stipulated in the problem. Therefore, we can write this: k = RT /
P where k is some constant. 3) Therefore, this is true: V/n = k 4) Given V and n at two different sets of conditions, we have: V1 / n1 = k V2 / n2 = k 5) Since k = k, we have this relation: V1 / n1 = V2 / n2 6) Insert data and solve: 6.13 / 7.51 = 13.5 / n (6.13) (n) = (13.5) (7.51) n = [(13.5) (7.51)] / 6.13 n = 16.54 mol (this is not the final
answer) 7) Final step: 16.54 − 7.51 = 9.03 mol (this is the number of moles of gas that were added) Example #7: A container with a volume of 25.47 L holds 1.050 mol of oxygen gas (O2) whose molar mass is 31.9988 g/mol. What is the volume if 7.210 g of oxygen gas is removed from the container, assuming the pressure and
temperature remain constant? Solution #1: 1) Initial mass of O2: (1.050 mol) (31.9988 g/mol) = 33.59874 g 2) Final mass of O2: 33.59874 − 7.210 = 26.38874 g 3) Final moles of O2: 26.38874 g / 31.9988 g/mol = 0.824679 mol 4) Use Avogadro's Law: V1/n1 = V2/n2 25.47 L / 1.050 mol = V2 / 0.824679 mol V2 = 20.00 L Solution #2: 1)
Let's convert the mass of O2 removed to moles: 7.210 g / 31.9988 g/mol = 0.225321 mol 2) Subtract moles of O2 that got removed: 1.050 mol − 0.225321 mol = 0.824679 mol 3) Use Avogadro's Law as above. Solution #3: 1) This solution depends on seeing that the mass ratio is the same as the mole ratio. Allow me to explain by using
Avogadro's Law: 2) Replace moles with mass divided by molar mass: V1 V2 –––––––––– = –––––––––– mass1 / MM mass2 / MM 3) Since the molar mass is of the same substance (oxygen in this case), they cancel out leaving us with this: V1 V2 –––– = –––– mass1 mass2 4) Solve using the appropriate values 25.47 L V2
–––––––– = –––––––– 33.59874 g 26.38874 g V2 = 20.00 L Example #8: What volume (in L) will 5.5 g of oxygen gas occupy if 2.2 g of the oxygen gas occupies 3.0 L? (Under constant pressure and temperature.) Solution: 1) State the ideal gas law: P1V1 P2V2 ––––– = ––––– n1T1 n2T2 Note that it is the full version which includes
the moles of gas. Usually a shortened version with the moles not present is used. Since grams are involved (which leads to moles), we choose to use the full version. 2) The problem states that P and T are constant: 3) Cross-multiply and rearrange to isolate V2: V2n1 = V1n2 V2 = (V1) (n2 / n1) 4) moles = mass / molecular weight: n =
mass / mw V2 = (V1) [(mass2 / mw) / (mass1 / mw)] 5) mw is a constant (since they are both the molecular weight of oxygen), which means it can be canceled out: V2 = (V1) (mass2 / mass1) 6) Solve: V2 = (3.0 L) (5.5 g / 2.2 g) V2 = 7.5 L Example #9: At a certain temperature and pressure, one mole of a diatomic H2 gas occupies a
volume of 20 L. What would be the volume of one mole of H atoms under those same conditions? Solution: One mole of H2 molecules has 6.022 x 1023 H2 molecules. One mole of H atoms has 6.022 x 1023 H atoms. The number of independent "particles" in each sample is the same. Therefore, the volumes occupied by the two samples
are the same. The volume of the H atoms sample is 20 L. By the way, I agree that one mole of H2 has twice as many atoms as one mole of H atoms. However, the atoms in H2 are bound up into one mole of molecules, which means that one molecule of H2 (with two atoms) counts as one independent "particle" when considering gas
behavior. Bonus Example: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.50 L? (The
temperature was held constant.) Solution: 1) The two variables are the volume and the amount of gas (temp and press are constant). The gas law that relates these two variables is Avogadro's Law: V1 V2 ––––– = –––––– n1 n2 2) We convert the grams to moles: 2.00 g / 4.00 g/mol = 0.500 mol 3) Now, we use Avogadro's Law: 2.00 L
2.50 L –––––––– = –––––– 0.500 mol x x = [(0.500 mol) (2.50 L)] / 2.00 L x = 0.625 mol 4) This is the total moles to create the 2.50 L. We need to convert back to grams: (4.00 g/mol) (0.125 mol) = 0.500 g Notice that I subtracted 0.500 mol from 0.625 mol and used 0.125 mol in the calculation. This is because I want the amount added,
not the final ending amount. avogadro's law example problems with answers pdf
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