<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
Answer:
μ = mg/kx
Explanation:
Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.
Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.
N = F' = kx
So, F = μN = μkx
μkx = mg
So, μ = mg/kx
Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
Pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.