Mitemo yegesi yaBoyle, Charles law naGay-Lussac haishande kune ese mamiriro egasi, saka ongororo yedu inova yakaoma. Saka, takaratidza muenzaniso wegasi wakakodzera. Gasi rakakodzera haripo muhupenyu hwezuva nezuva; gasi rakakodzera ndiro fomu rakakwana rekubatsira ongororo. Kuvapo kwepfungwa iyi yegasi rakakodzera kunotibatsirawo mukuongorora hukama huripo pakati pemitemo mitatu yegasi.
Hukama huripo pakati pekupisa, vhoriyamu, uye kumanikidzwa kwegesi
Nekutarisa mitemo mitatu yegesi iri pamusoro apa, tinogona kuwana hukama hwakazara pakati pekupisa, vhoriyamu, uye kumanikidzwa kwegesi.
P1 = initial pressure (Pa or N/m2), P2 = final pressure (Pa or N/m2), Mukati1 = initial volume (m3), Mukati2 = final volume (m3), T1 = initial temperature (K), T2 = end temperature (K)
(Pa = pascal, N = Newton, m2 = square meter, m3 = cubic meter, K = Kelvin)
The relationship between gas mass (m) and volume (V)
The more air that is put into the rubber balloon, the more bloated the balloon is. In other words, the larger the gas mass, the greater the volume of the balloon. We can say that the mass of gas (m) is directly proportional to the volume of gas (V).
Number of moles (n)
1 mol = mass of a substance equal to the molecular mass of the substance. Mass and molecular oxygen mass are different.
Example 1, the oxygen molecular mass (O2) = 16 u + 16 u = 32 u (each oxygen molecule contains 2 Oxygen atoms, where each Oxygen atom has mass 16 u). Thus, 1 mole O2 has mass 32 grams. Or the molecular mass of O2 = 32 grams / mol = 32 kg/kmol.
Example 2, the molecular mass of carbon monochide gas (CO) = 12 u + 16 u = 28 u (each molecule of carbon monooksida contains 1 carbon atom (C) and 1 oxygen atom (O) Mass 1 carbon atom = 12 u and mass 1 atom Oxygen = 16 u. 12 u + 16 u = 28 u). Thus, 1 mole of CO has a mass of 28 grams. Or the molecular mass of CO = 28 grams/mol = 28 kg/kmol.
Example 3, the molecular mass of the carbon dioxide gas (CO2) = [12 u + (2 x 16 u)] = [12 u + 32 u] = 44 u (each carbon dioxide molecule contains 1 carbon atom (C) and 2 oxygen atoms (O). The mass of 1 Carbon atom = 12 u and the mass of 1 oxygen atom = 16 u). Thus, 1 mole of CO2 has mass 44 grams. Or the molecular mass of CO2 = 44 grams/mol = 44 kg/kmol.
The number of moles (n) of a substance = the mass ratio of the substance to its molecular mass. Mathematically:
Mubvunzo 1:
Calculate the number of moles at 64 g of O2
O2 mass = 64 grams
The molecular mass of O2 = 32 grams/mol
Mubvunzo wechi2:
Calculate the number of moles at 280 grams of CO
CO mass = 280 grams
The molecular mass of CO = 28 grams/mol
Mubvunzo wechi3:
Calculate the number of moles at 176 grams of CO2
CO2 mass = 176 grams
The molecular mass of CO2 = 44 grams/mol
Universal gas constants (R)
Based on research by scientists, it was found that if the number of moles (n) is used to express the size of a substance, then the constant of comparison for each gas has the same value. The corresponding constant is the universal gas constant (R).
R = 8.315 J / mol. K
= 8315 kJ / kmol. K
= 0.0821 (L.atm) / (mol. K)
= 1.99 cal / mol. K
(J = Joule, K = Kelvin, L = liter, atm = atmosphere, cal = calorie)
IDEAL GAS LAW (in the number of moles)
The above comparison can be converted into equations by entering the number of mol (n) and universal gas constants (R).
PV = n RT
This equation is called the ideal gas law or ideal gas state equation.
P = gas pressure (N/m2)
V = volume of gas (m3)
n = number of moles (mole)
R = universal gas constants (R = 8.315 J/mol. K)
T = absolute temperature of the gas (K)
In solving the problem, you will find the term STP. STP is the Standard Temperature and Pressure.
Tembiricha yakajairika (T) = 0 oC = 273 K
Kumanikidzwa kwakajairika (P) = 1 atm = 1.013 x 105 N / m2 = 1.013 x gumi2 kPa = 101 kPa
In solving the gas law issues, the temperature should be expressed in the Kelvin (K) scale. If the gas pressure is still a measuring pressure, change it first into absolute pressure. Absolute pressure = atmospheric pressure + measuring pressure. If the known is atmospheric pressure (no measuring pressure), solve the problem.
Mubvunzo wechi1:
At atmospheric pressure (101 kPa), the temperature of carbon dioxide gas = 20 oC and its volume = 2 liters. If the pressure is converted to 201 kPa and the temperature is increased to 40 °C, calculate the final volume of the carbon dioxide gas.
mhinduro
P1 = 101 kPa
P2 = 201 kPa
T1 = 20 oC + 273 K = 293 K
T2 = 40 oC + 273 K = 313 K
V1 = 2 marita
Mubvunzo wechi2:
Determine the volume of 2 moles of gas on STP (this gas is the ideal gas)
mhinduro
Mubvunzo wechi3:
Volume of oxygen gas at STP = 20 m3. Calculate the oxygen gas mass.
mhinduro
Volume 1 mol gas at STP = 22.4 liter = 22.4 dm3 = 22.4 x gumi-3 m3 (22.4x10-3 m3/mol)
Volume of oxygen gas at STP = 20 m3
The mass of oxygen molecules = 32 grams / mol (mass 1 mole oxygen = 32 grams). Thus, the oxygen gas mass is:
mass (m) = number of mol (n) x molecular mass
mass = (893 mol) x (32 grams / mol) = 28576 grams = 28.576 kg
Mubvunzo wechi4:
A tank contains 4 liters of oxygen (O2). The oxygen temperature is = 20 oC and measured pressure is = 20 x 105 N / m2. Determine the mass of oxygen (Molecular mass of Oksijeni = 32 kg/kmol = 32 gram/mol).
mhinduro
P = P atm + P measured = (1 x 105 N / m2) + (20 x 105 N / m2) = 21 x 105 N / m2
T = 20oC + 273 = 293 K
V = 4 liters = 4 dm3 = 4 x gumi-3 m3
R = 8.315 J/mol. K = 8.315 Nm/mol. K
Molecular mass of Oksijeni = 32 kg/kmol = 32 gram/mol
O2 mass ?
IDEAL GAS LAW (In the number of molecules)
If we state the size of the substance not in the form of mass (m), but in the number of moles (n), then the universal gas constant (R) applies to all gases. It was first discovered by Amadeo Avogadro (1776-1856), an Italian scientist.
Avogadro says that when the volume, pressure, and temperature of each gas are equal, each gas has the same number of molecules.
The Avogadro hypothesis corresponds to the fact that the constant R is the same for all gases. Here are some proofs:
First, if we solve the problem using the ideal gas law equation (PV = nRT), we will find that when the number of moles (n) is equal,
the pressure and temperature are also similar, then the volume of all gases will be the same if we use universal gas constants R = 8.315 J / mol. K). In STP, any gas having the same number of moles (n) will have the same volume. Volume 1 mol of gas at STP = 22.4 liters. Volume 2 mol of gas = 44.8 liters. Volume 3 mol of gas = 67.2 liters. And so on. This applies to all gas.
Second, the number of molecules in 1 mol is the same for all gases. The number of molecules in 1 mol = number of molecules per mole = Avogadro number (NA). So, the Avogadro number is the same for all gas.
Avogadro’s number obtained through measurement:
NA = 6.02 x 1023 molecule / mol
To get the total number of molecules (N), multiply the number of molecules per mole (NA) by the number of moles (n).
Total number of molecules (N) = number of molecules per mole (NA) x number of moles (n)
P = Pressure, V = Volume, N = Total number of molecules, k = Boltzmann constant (k = 1.38 x 10-23 J / K), T = Temperature
Vhoriyamu Units
1 liter (L) = 1000 milliliters (mL) = 1000 cubic centimeters (cm3)
1 liter (L) = 1 cubic decimeter (dm3) = 1 x 10-3 m3
Pressure dzakabatana
1 N/m2 = 1 Pa
1 atm = 1.013 x 105 N / m2 = 1.013 x gumi5 Pa = 1.013 x 102 kPa = 101.3 kPa (usually used 101 kPa)
Pa = pascal
atm = atmosphere
