Advertisements
Advertisements
Question
How does Avogadro's law explain Gay - lussac's law of combining volumes?
Advertisements
Solution
Gay-Lussac experimentally determined that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas.
According to Avogadro's law, if :
1 volume of hydrogen contains n molecules of the gas then 1 volume of chlorine also contains n molecules of the gas. Therefore 2 volume of hydrogen chloride contain 2n molecules of the gas.
\[\ce{H2 + Cl2 → 2HCI}\]
1 vol 1 vol 2 vol (by Gay-Lussac)
n n 2n (By Avogadro)
but hydrogen and chlorine are diatomic.
So, 2 atoms + 2 atoms → 2 molecules
1 atom + 1 atom → 1 molecule
i.e. 1 molecule of hydrogen chloride is formed when 1 atom of hydrogen combines with 1 atom of chlorine. Thus Avogadro's law explains Gay-Lussac's law of combining volumes.
APPEARS IN
RELATED QUESTIONS
Propane burns in air according to the following equation:
C3H8 + 5O2 → 3CO2 + 4H2O
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is `1/5`th of air)
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Caclulate the composition of resulting mixture.
\[\ce{2NO + O2 → 2NO2}\]
A mixture of hydrogen and chlorine occupying 36 cm3 was exploded. On shaking it with water, 4 cm3 of hydrogen was left behind. Find the composition of the mixture.
LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
The reaction: \[\ce{4N2O + CH4 -> CO2 + 2H2O + 4N2}\] takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N2O) required to give 150 cm3 of steam.
The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.
- Which sample of gas contains the maximum number of molecules?
- If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
- If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
- Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate the volume of gaseous product formed.
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate composition of the resulting mixture.
