Molar volume (+ equation of state)
1. Calculate the volume occupied by 2.55 mol of methane at a temperature of 20°C and a pressure of 100.5 kPa.
Solution:
n = 2.55 mol; T = 20°C, which corresponds to 293.15 K; p = 100500 Pa; R = 8.314 J·mol-1·K-1
We solve the problem using the state equation rearranged into the form:
Under the given conditions, methane occupies a volume of 61.8 L.
2. A gas holder with a volume of 50.00 m³ contains hydrogen (H2) at a temperature of 25°C and a pressure of 101300 Pa. Calculate the amount of substance of hydrogen in the gas holder.
Solution:
V = 50 m3; p = 101300 Pa; R = 8.314 J·mol-1·K-1; T = 25°C which corresponds to 298.15 K.
We rearrange the state equation into the form:
The gas holder contains 2.043 kmol of hydrogen.
3. The reaction produced 19.72 dm3 of acetylene under normal conditions. Calculate its amount of substance.
Solution:
We use the adjusted relation for the molar volume of a substance under normal conditions:
Substitute into the relation:
V (C2H2) = 19.72 dm3; Vmn (C2H2) = 22.41 dm3
The reaction produced 0.88 mol of acetylene.
(The problem can also be solved logically. One mole of an ideal gas occupies 22.41 L. Our amount occupies only 19.72 L. To calculate the amount of substance, just take the ratio 19.72 / 22.41, which equals 0.88 mol.
This can also be shown in a proportion:
1 mol ..........................22.41 dm3
x mol...........................19.72 dm3
The reaction produced 0.88 mol of acetylene.
4. In a 20-liter pressure vessel, hydrogen is under a pressure of 101.97 kPa. Calculate how the pressure in the vessel changes
- a) when cooled from 20°C to -20°C
- b) when heated from 20°C to 50°C
Solution:
In both cases, it is a change of pressure depending on temperature at constant volume, so we use Charles's law.
and from that
The pressure of hydrogen decreases to 88.05 kPa when cooled to -20°C and increases to 112.4 kPa when heated to 50°C.
5. By how much does the pressure of oxygen decrease if its volume is tripled at constant temperature?
Solution:
We need to find the change in pressure depending on volume at constant temperature, so we use Boyle’s law.
The resulting pressure will be one-third of the original pressure, so the pressure decreases by 2/3.
6. The volume of a gas at 27°C is 50 ml. To what temperature must the gas be heated at constant pressure for the volume to increase to 200 ml?
Solution:
This is a change in volume depending on temperature at constant pressure, so we use Gay-Lussac’s law.
V1 = 50 ml; T1 = 300.15 K;
V2 = 200 ml; T2= ?
The gas must be heated to a temperature of 127.05 °C.
7. The volume of hydrogen in a vessel above water at 27°C and a pressure of 107.59 kPa is 124 ml. Calculate the volume that dry hydrogen would occupy under normal conditions. The partial pressure of water vapor at 27°C is 3.59 kPa.
Solution:
The volume of hydrogen under normal conditions is calculated using the relation
However, we need to know the pressure of hydrogen in the initial state, p(H2). To calculate it, we use Dalton’s law. According to this law, the total pressure in the vessel is equal to the sum of the partial pressures of hydrogen and water vapor.
from which
The calculated partial pressure of hydrogen is then substituted along with the known values of initial volume and gas temperature into the first equation rearranged into the form
Dry hydrogen would occupy a volume of 115.82 ml under normal conditions.