avogadro's law problems with solutions (dave hook)

Relationships between properties of gases can be measured by changing one property while holding the others constant. Avogadro's Law relates the properties of gases to the amount of gas present. Avogadro's law states that the volume of a gas at a given temperature and pressure is directly proportional to the amount of gas, in moles. Thus, V is proportional to n, the number of moles of the gas, AT CONSTANT TEMPERATURE and PRESSURE.

Avogadro said that quantity and volume are directly related:

V 'alpha' n 'rArr' V/n is a constant

Or for the same sample of gas at constant temperature and pressure,

V1/ n1 = V2 / n2

Let us take an example of avogadro's law problems with solutions:

If we have 8 molecules of H2 and 4 molecules of O2, we get 12 molecules and a combined volume. If we then react the mixture, we end up in 8 molecules of gaseous water molecules, which occupy the same volume as the 8 molecules of any other gas.

8 H2 (g) + 4 O2 (g) ? 8 H2O (g)

Avogadro Law Concept Based Questions: avogadro's law problems with solutions

1. Which sample represents the smallest number of moles?
a) 1L H2 at STP
b) 1L Argon at STP
c) 1L of H2 at 27 ? C and 760mm of Hg

Solution:

Let us consider each situation.
At STP, 22.4 L volume will have 1 mole of the gas.
So, in a) 1L of H2, there will be (1L) (1 mole) / 22.4L = 0.0446 moles
˜ 0.045 moles

Similarly in b) 22.4 L of Ar would correspond to 1 mole.

So, 1L would be = (1L) (1 mole) /22.4L = 0.045 moles

In c) we should use the ideal gas equation: PV = nRT
P = 760 mm of Hg V = 1 L T = 27 ? C = (27 +273) K
= 1 atm = 300 K

Therefore, n = PV/ RT
R is the gas constant that has a value of 0.0821 atm L mol -1 K-1
n = [(1 atm) (1 L)] / [(0.0821 atm L mol -1 K-1) (300 K)]

= 0.0406 moles
So the correct option is c) 1L H2 at 27 ? C and 760 mm of Hg.

2. Which of the samples would have the same number of particles?

a) 1L He at STP and 1L O2 at STP
b) 2L He at STP and 1L of He at STP
c) 1L He at 27 ? C and 760 mm of Hg; and 2L He at STP

Solution:

a) In 22.4 L of gas, there will be 1 mole of the gas

So, in 1L of He there will be (1L) (1 mole) /22.4L = 0.0446 moles
˜ 0.045 moles

In 1L of O2, the number of moles will be the same, 0.045 moles.

Now, we know that in every mole, there will be Avogadro number of particles = 6.023 x 10 ^23 particles.

So in 0.045 moles, there will be
(0.045 moles) x (6.023 x 10 ^23 particles)/ 1 mole
= 2.7 x 10 22 number of particles

So, this is the required answer: a) 1L He at STP and 1L O2 at STP

Application based questions on Avogadro's Law: avogadro's law problems with solutions

1. Suppose we have 12.2 L sample containing 0.5 moles oxygen gas at a pressure of 1 atm and temperature of 25 ? C. If all this oxygen were converted into ozone, at the same temperature and pressure, what would be the volume of ozone?

Solution:
The equation for this reaction can be given as follows-

3O2 ? 2 O3
(Oxygen) (Ozone)

In this balanced equation 3 moles of oxygen produce 2 moles of ozone.
In the question given, 0.5 moles of oxygen are reacting. So the number of moles of ozone that it would result

= (0.5 ? 2) / 3
= 0.33 moles of ozone

So, V1 = 12.2 L n1 = 0.5
V2 =? n2 = 0.33

So from Avogadro's Law equation, V1/n1 = V2 / n2,
V2 = (V1 n2) / n1
V2 = (12.2 L) (0.33 moles) / (0.5 moles)
= 8.133 L

2. 11.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas would there be in a 20 L sample?
Solution:

V1 = 11.2 L n1 = 0.5 moles
V2 = 20 L n2 =?

From the Avogadro's law,
n2 = (V2 n1)/ (V1)
n2 = (20 L) (0.5 moles) / (11.2 L)
= 0.89 moles

3. Consider the following chemical equation:

2NO2 (g) ? N2O4 (g)

If 25 mL of NO2 gas is completely converted to N2O4 gas, under the same conditions, what volume will the N2O4 occupy?

Solution:
Here, from the equation given, 2 moles of NO2 (g) forms 1 mole of N2O4 (g).

So, V1 = 25 mL n1 = 2 mole
V2 =? n2 = 1mole

V2 = (V1 n2) / n1

V2 = (25 mL) (1 mole) / (2 moles)
= 12.5 mL

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