Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to its temperature if<u> the temperature and the number of particles are constant.</u>
<h3>Further Explanation</h3><h3>Boyles’s law </h3>
- This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
- Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.
- Mathematically; Volume α 1/Pressure
Vα1/P
- Therefore, constant k, is = PV
<h3>Other gas Laws</h3><h3>Gay-Lussac’s law </h3>
- It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
- Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.
<h3>Charles’s law</h3>
- It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
- Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.
<h3>Dalton’s law </h3>
- It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
- Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.
Keywords: Gas law, Boyles's law, pressure, volume, absolute temperature, ideal gas
<h3>Learn more about:</h3>
Level: High school
Subject: Chemistry
Topic: Gas laws
Sub-topic: Boyle's Law
The answer is A I hope it was right
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
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