THE clapeyron equation is a mathematical expression proposed by the French physicist-chemist and civil engineer Benoit-Pierre-Émile Clapeyron and was formulated to describe the behavior of a perfect gas. Note the following equation:
P.V = n. RT
P = pressure generated by gas on the walls of the container;
V = volume occupied by the gas and can be expressed in liters or cubic meters;
n = number of mol (amount of matter in the gas);
Note: The number of moles is expressed by the relationship between the mass of the gas (m) and its molar mass (M):
n = m
M
R = general gas constant proposed by Clapeyron and depends on the pressure unit used (in atm, it is 0.082; in mmHg, it is 62.3; in KPa, it is 8.31);
T = temperature at which the gas is subjected (always used in the Kelvin unit).
Note: To transform a given temperature in degrees Celsius to Kelvin, just add its value to 273.
Thus, through the use of the clapeyron equation, we can determine several variables referring to a gas, as you can see in each of the examples proposed below:
1st Example: (Uefs-BA) A 24.6 L container contains 1.0 mol of nitrogen exerting a pressure of 1.5 atm. Under these conditions, the gas temperature on the Kelvin scale is:
a) 30 b) 40 c) 45 d) 300 e) 450
T = ?
n = 1 mol
R = 0.082 atm. L/mol. K (because the pressure is in atm)
Volume = 24.6 L
P = 1.5 atm
Entering the data provided in the clapeyron equation, we can determine the required temperature:
P.V = n. RT
1.5,24.6 = 1.0.082.T
36.9 = 0.082T
36,9 = T
0,082
T = 450 K
2nd Example: (Unimep-SP) At 25 ºC and 1 atm, dissolve 0.7 liters of carbon dioxide in one liter of distilled water. This amount of CO2 stands for:
(Data: R = 0.082 atm.l/mol.k; Atomic masses: C = 12; 0 = 16).
a) 2.40 g
b) 14.64 g
c) 5.44 g
d) 0.126 g
e) 1.26 g
T = 25 °C, which added to 273 results in 298 K
m = ?
R = 0.082 atm. L/mol. K (because the pressure is in atm)
Volume = 0.7 L
P = 1 atm
Entering the data provided in the clapeyron equation, we can determine the required mass:
P.V = n. RT
1.0,7 = m .0,082.298
44
0,7 = m.24,436
44
0.7.44 = m.24.436
30.8 = m.24.436
30,8 = m
24,436
m = 1.26 g (approximately)
3rd Example: (Fesp-PE) To 75 OC and 639 mmHg, 1.065 g of a substance occupy 623 ml in the gaseous state. The molecular mass of the substance is equal to:
a) 58 b) 0.058 c) 12.5 d) 18.36 e) 0.0125
T = 75 °C, which added to 273 results in 348 K
m = 1.065 g
R = 62.3 mmHg. L/mol. K (because the pressure is in mmHg)
Volume = 623 mL, which divided by 1000 results in 0.623 L
P = 639 mmHg
M = ?
Entering the data provided in the clapeyron equation, we can determine the required molecular mass:
P.V = n. RT
PV = m .R.T
M
639.0,623 = 1,065.62,3.348
M
398,097 = 23089,626
M
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398.097M = 23089.626
M = 23089,626
398,097
M = 58 u
4th Example: (UFRJ) It is necessary to store a certain amount of gaseous oxygen (O2). The gas mass is 19.2 g at a temperature of 277 OC and at a pressure of 1.50 atm. The only container capable of storing it will have approximately the volume of:
Data: O = 16, R = 0.082 atm. L/mol. K
a) 4.50L b) 9.00L c) 18.0L d) 20.5L e) 36.0L
T = 277 ºC, which added to 273 results in 550 K
m = 19.2 g
P = 1.5 atm
R = 0.082 atm. L/mol. K (since the pressure was supplied in atm)
Volume = ?
Note: Initially, we must calculate the molar mass of the oxygen gas, multiplying the number of atoms by the mass of the element and then adding the results:
M = 2.16
M = 32 g/mol
Entering the data provided in the Clapeyron's equation, we can determine the required volume:
P.V = n. RT
PV = m .R.T
M
1.5.V = 19,2.0,082.550
32
1.5.V = 865,92
32
1.5.V.32 = 865.92
48V = 865.92
V = 865,92
48
18.04 L (approximately)
5th Example: (Unified-RJ) 5 mol of an ideal gas, at a temperature of 27 ºC, occupy a volume of 16.4 liters. The pressure exerted by this amount of gas is:
Given: R = 0.082 atm. L/mol. K
a) 0.675 atm b) 0.75 atm c) 6.75 atm d) 7.5 atm e) 75 atm
T = 27 °C, which added to 273 results in 300 K
n = 5 mol
R = 0.082 atm. L/mol. K
Volume = 16.4 L
P = ?
Entering the data provided in the clapeyron equation, we can determine the required pressure:
P.V = n. RT
P.16.4 = 5.0,082,300
P.16.4 = 123
P = 123
16,4
P = 7.5 atm
6th Example: (Unirio-RJ) 29.0 g of a pure and organic substance, in the gaseous state, occupy a volume of 8.20 L, at a temperature of 127 °C and a pressure of 1520 mmHg. The molecular formula of the likely gas is: (R = 0.082. atm .L/mol K)
a) C2H6 b) C3H8 c) C4H10 d) C5H12 e) C8H14
T = 127 °C, which added to 273 results in 400 K
m = 29 g
R = 62.3 mmHg. L/mol. K (because the pressure is in mmHg)
Volume = 8.2 L
P = 1520 mmHg
M = ?
To determine the molecular formula in this exercise, enter the data provided in clapeyron equation to determine the molar mass:
P.V = n. RT
1520.8,2 = 29 .62,3.400
M
12464 = 722680
M
12464M = 722680
M = 722680
12464
M = 57.98 g/mol
Next, we must determine the molecular mass of each alternative provided (by multiplying the number of atoms by the mass of the element and then adding the results) to see which one matches the mass found previously:
a) M = 2.12 + 6.1
M = 24 + 6
M = 30 g/mol
b) M = 3.12 + 8.1
M = 36 + 8
M = 44 g/mol
c) M = 4.12 + 10
M = 48 + 10
M = 58 g/mol, that is, the molecular formula of the compound is C4H10.
By Me. Diogo Lopes Dias
