Resources previous | 4 of 5 Avogadro\'s law states that If the volume of a certa

Question

Resources previous | 4 of 5 Avogadro's law states that If the volume of a certain gas is changed from Vi to the corresponding change in number of moles will be from ni to n2. For such a case, Avogadro's law can also be expressed as n2 where ni and n2 are the initial and final numbers of moles of the gas and Vi and ½ are the initial and final volumes of the gas, respectively. In an ideal gas, particles are considered to interact only when they collide, with no loss in energy or speed occurring during the collision. Studying ideal gases allows for simplifications, such as Avogadro's law. Part B Imagine that the gas shown in the simulation is an ideal gas such as helium. Notice that the final number of moles of gas is 1.00 mol for each experimental run. You can find the final volume of the gas using the y axis of the graph shown. Consider an experimental run at 273 K where the initial number of moles (ni) is actually 1.00 mol. and the final number of moles (n2) is 2.00 mol Use the simulation to find the volume (Vi) of 1.00 mol of helium at 273 K, and calculate the final volume (V2) Express the volume to three significant figures, and include the appropriate units. Hints | ½: | Taille 1 Units Submit My Answers Give Up Part C At 4 00 L an expandable vessel contains 0.864 mol of oxygen gas How many liters of o must be added at constant temperature and pressure if you need a total of 1 72 mol of oxy in the vessel? gen gas Express the volume to three significant figures, and include the appropriate units.

Explanation / Answer

B) As per Avagadro's Law, at constant tempreature and pressure, moles of an ideal gas is directly proportional to its volume

Thus,

n1/V1 = n2/V2

or, 1/V1 = 2/V2

or, V1/V2 = 1/2 = 0.5

***For exact value of V1 & V2, the pressure at which the experiment is runned is to be provided.

C) Since temperature and pressure during the entire expirement remains constant, thus Avagadro's law is applicable.

Thus, n1/V1 = n2/V2

or, 0.864/4 = 1.72/V2

or, V2 = 7.963 litres

Thus, the volume to be added = 7.963-4 = 3.963 litres

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.