<u>Gay Lussac’s law</u> state that the pressure and absolute temperature of a fixed quantity of a gas are directly proportional under constant volume conditions.
<h2>Further Explanation
</h2><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>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.
<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, Gay-Lussac’s law, pressure, volume, absolute temperature, ideal gas
<h3>Learn more about:
</h3>
- Gay-Lussac’s law: brainly.com/question/2644981
- Charles’s law: brainly.com/question/5016068
- Boyles’s law: brainly.com/question/5016068
- Dalton’s law: brainly.com/question/6491675
Level: High school
Subject: Chemistry
Topic: Gas laws
Sub-topic: Gay-Lussac’s law
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Explanation:
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
Answer:
c
Explanation:
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