High School Physics Gas Pressure requires an analysis process and a calculation process

It is assumed that the length does not change after the change, that is, the volume of the gas does not change, and the column of mercury above the liquid level also does not change.

Let the length of the original mercury column be h1 and h2 respectively; After the change, the volume remains unchanged, and the total length is h1+h2, and the total pressure in the tube relative to the page is h0=h1+p2. p2=h1+p1 h0=h1+h2+p1

After mixing, the gas p1v1+p2v2=p3(v1+v2) and then the total pressure p'=p3+h1+h2

p'-p0=（p1v1+p2v2）/（v1+v2）-p1（p1v1+（h1+p1）v2）/（v1+v2）-p1h1v2/(v1+v2)

Therefore, the pressure after mixing is stronger than the atmospheric pressure, and the length of the gas will become larger (the gas pressure will be smaller, and the mercury column pressure will be smaller).

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The known condition of the question: "The atmospheric pressure is h0" is obviously contradictory to the h0 of the height of the mercury column (upper section) in the tube in the figure, if it is really as stated in the title, then the height of the mercury column in the lower section of the tube is equal to 0 (that is, its mercury surface is equal to the mercury surface outside the tube).

Solution: Change the atmospheric pressure h0 to p0. (Analysis of questions that do not affect the topic).

Analysis: Using the hypothetical method, it is assumed that no mercury from outside the tube enters the tube, and no mercury comes out of the tube.

In the diagram, the state is state 1, the height of the mercury column at the bottom of the tube is h1, the length of the upper air column is l1, and the length of the lower air column is l2.

Then the pressure of the lower air column is p2 p0 h1 and the pressure of the upper air column is p1 p2 h0 p0 (h1 h0).

When the two sections of air column are combined, it is state 2, then the height of the mercury column in the tube is (h1 h0), and the length of the air column in the tube is set to l

Then the pressure of the air column is p p0 (h1 h0).

Since the total number of moles of the two segments of the air column is constant, there is p1*l1 p2*l2 p*l (from the Clapelon equation).

i.e. [ p0 (h1 h0)]*l1 (p0 h1)*l2 [ p0 (h1 h0)]*l

Because p0 h1 p0 h1 h0 i.e. (p0 h1) p0 (h1 h0).

So [ p0 (h1 h0)]*l1 [p0 (h1 h0)]*l2 [ p0 (h1 h0)]*l

L1 L2 L or L L1 L2).

In this way, the length of the air column l after the two air columns are combined into one is greater than the sum of the original two air column lengths (l1 l2).

Since the above results are obtained under the condition that the total length of mercury in the tube remains unchanged (i.e., it is considered to be deduced under the condition of l l1 l2), the previous assumption is not valid. The result will be l (l1, l2).

Conclusion: The length l of the air column after the two stages of air column are combined into one is less than the sum of the length of the original two stages of air column (L1 L2).

i.e. l (l1, l2).

Not the same, the sum of the two paragraphs is greater.

For this kind of problem, it is best to assume that the length of the air column remains unchanged after being combined, that is, it is equal to the sum of the original two segments, and then analyze whether it is reasonable and how the combined air column will move.

If the pressure of the L1 section is set to P1, then the pressure of the L2 section P2=P1-H0 will be compressed and the length will decrease when the pressure of the L2 section increases.

Refine it. Let the pressure of L1 gas be H1, the pressure of L2 gas be H2, and the length of the mercury column in the lower section be H3

h1+h0=h2

h2+h3=h0

i.e. h1 + h3 = 0

Because the gas pressure is always positive, the actual lower mercury column should be concave downward;

Now the two sections are combined, h1v1=n1rt, h2v2=n2rt, and the addition has h2(v1+v2)>(n1+n2)rt=h1v1+h2v2>h1(v1+v2).

If merged in the upper instant, the volume is the sum of the two segments, and the pressure is set to h4 because the temperature does not change, i.e., (n1+n2)rt=h4(v1+v2).

then h2>h4>h1, then h4+h0+h3>h1+h0+h3=h0, that is, the gas will discharge a part of the mercury, that is, the volume increases, that is, the length increases;

If merged in the lower instant, the volume is the sum of the two segments, and the pressure is set to h5n1+n2)rt=h5(v1+v2).

then h2>h5>h1, then h5+h0+h3>h0 that the gas will discharge a part of the mercury, that is, the volume increases, that is, the length increases;

In summary, the length will increase.

As shown in the figure, the pressure of L2 is greater than L1, then after mixing, the total pressure generated by the gas is less than the pressure of L2, and the total pressure generated by the gas liquid in the test tube is unchanged (equal to one atmosphere), then it must be that the total height of the mercury column rises, so that the length of the gas column decreases, and thus the sum of the gas columns decreases.

The formula for calculating the pressure of a physically confined gas in high school is PV=NRT: The magnitude of the gas pressure is proportional to the amount of gas (n), the temperature of the gas (t), and inversely proportional to the volume of gas (v) r is the general gas trapped slag constant, which is about.

1. Gas pressure

Refers to the hydrostatic pressure exerted by a gas at a certain point. The cause of this is the continuous, irregular impact of a large number of gas molecules on the walls of the container.

2. Atmospheric pressure

Atmospheric pressure refers to the pressure produced by the air at a certain location on the earth. The air on the surface of the earth is subject to gravity, which produces atmospheric pressure The density of the air layer above the earth is not equal, the air near the surface layer is denser, and the air in the upper layers is thin and less dense.

3. Causes of atmospheric pressure

The earth is surrounded by a thick layer of air, which is mainly composed of nitrogen, oxygen, carbon dioxide, water vapor and helium, neon, argon and other gases, usually the whole of this layer of air is called the atmosphere It is densely distributed around the earth, with a total thickness of 1000 kilometers, and all objects immersed in the atmosphere must be subjected to the pressure of the atmosphere acting on it, just as the objects immersed in water are subjected to the pressure of water.

Fourth, the relationship between atmospheric pressure and altitude

Experiments have shown that within 2,000 meters of sea level, for every 12 meters of elevation, the atmospheric pressure decreases by one millimeter of Hg, or about 133 Pa. Using this feature, the atmospheric pressure value of a certain place can be measured within 2000 meters of the sea dial, and the sea dial height of a certain place can be calculated.

There seems to be no necessary connection between pressure and volume, but there is a formula in the university pv=nrt, where p is the pressure, v is the volume, n is the amount of the gaseous substance, r is a constant, and t is the temperature of the gas.

Boyle's Law: P1V1=P2V2 (Isothermal Isomolar Gas)Charlie's Law: P1 T1=P2 T2 (Equal Equimolar Gas)Guy Lussac's Law:

v1 t1=p2 t2 (isobaric equimolar gas) Claberon equation: pv=nrt (for any ideal gas) Inference 1:

p1 v1= p2 v2 (isothermal isomolar gas) p1 t1= p2 t2 (isomolar gas) v1 t1= p2 t2 (isobaric equimolar gas) Inference 2:

For any ideal gas: ( for the density of the atmosphere).

p1/ρ1t1=p2/ρ2t2

It is used to calculate the atmospheric pressure in the mountains.

pv=nrt

c=v/tp1v1/t1=p2v2/t2

P: Pressure. v: volume of gas.

n: The amount of gaseous substance.

r: constant, t, t: temperature t=t+

C: Specific heat. pv=nrt is true in all ideal gas states.

Since PV t=nr, for the same ideal gas system, in the case of different temperatures t, pressure p, and volume v, since the quantity n of the substance does not change, r is a constant, it satisfies the following equation of p1v1 t1=p2v2 t2.

In addition, this equation also applies between different ideal gases where the amount of matter is equal.

I think I have the most complete answer.

The liquid level in the tube is stationary, so the pressure is balanced. The downward pressure is inside P, and the upward pressure is outside P + PH, so P inside = P outside + PH = P0 + Gh.

Based on the horizontal line on the way, the gas pressure p is inside the tube, and the atmospheric pressure + the pressure gh above the horizontal line outside the tube

So p=p0+ gh

pa=65cmhg

pb=70cmhg

Analysis of the force The upper surface of the liquid in the tube is subjected to the pressure of the air column in the tube, the atmospheric pressure p0s and h ps, the direction of p0s and h ps is downward, and the pressure of the gas in the test tube is p0+ gh

The pressure calculation is different for different forms of matter, as follows:

Solid pressure: p=f s

Liquid pressure: p= gh

Gas pressure: PV=Nrt (P is the gas pressure, V is the volume, N is the amount of matter, R is the constant, T is the Kelvin temperature).

The ratio of the magnitude of the pressure on the object to the area of the force is called the pressure, and the pressure is used to compare the effect of pressure, the greater the pressure, the more obvious the effect of pressure. The formula for calculating the pressure is: p=f s, the unit of pressure is pascal, and the symbol is pa.

The methods of increasing the pressure are: increasing the pressure while the force area is the same or decreasing the force area when the pressure is not the same. The methods of reducing the pressure are: reducing the pressure under the condition that the force area remains the same or increasing the force area when the pressure is not the same.

The pressure calculation is different for different forms of matter, as follows:

Solid pressure: p=f s

Liquid pressure: p= gh

Gas pressure: PV=Nrt (P is the gas pressure, V is the volume, N is the amount of matter, R is the constant, T is the Kelvin temperature).

The ratio of the magnitude of the pressure on the object to the area of the force is called the pressure, and the pressure is used to compare the effect of pressure, the greater the pressure, the more obvious the effect of pressure. The formula for calculating the pressure is: p=f s, the unit of pressure is pascal, and the symbol is pa.

The methods of increasing the pressure are: increasing the pressure while the force area is the same or decreasing the force area when the pressure is not the same. The methods of reducing the pressure are: reducing the pressure under the condition that the force area remains the same or increasing the force area when the pressure is not the same.

First calculate the force f, the solid is f = gh, and then calculate the force area, s, so the pressure can be obtained, p gh s

<> conclusion satisfies the following conditions: there is no external force on the piston in the process from 1 to 2; From 2 to 1, the piston is subjected to a constant force f to the left, p2 is equal to the atmospheric pressure, and when Yu Sen tells the piston to move to 1, it is stationary; Both processes are carried out slowly.

First look at 1 to 2, the volume of the gas increases, and the positive work is done to the outside world, and when the gas pressure is the same as the atmospheric pressure, the volume stops increasing, that is, the 2 state. Let the cross-sectional area be s and the gas pressure is p when the volume is v, and under this condition, the piston does work when the piston moves dl vertically through a small section: dw=fdl=psdl=pdv, which is the derivation process of the purple formula in the figure above.

Integrating in the interval of 1 to 2, we find the area of the trapezoid-like pattern in which the positive work of the gas in the process from 1 to 2 is obtained.

Looking at 2 to 1 again, the volume of the gas decreases, the outside world does positive work on the gas, and the gas does negative work on the outside world. Let the spring or cross-sectional area be s, and from the state 1 at rest, we can get: f=(p1-p2) to do :

w=fl=(p1-p2)sl=(p1-p2)(v2-v1), the area of the rectangle enclosed by p1, p2, v1, and v2 of the image. The amount of negative work done by the gas to the outside world is equal to the amount of positive work done by the outside world to the gas.

If the conclusion is different from your teacher's, it may be that the conditions are different. You understand the process of deriving the conclusion from my condition, and then use your teacher's condition to deduce it yourself.

The formula for calculating gas pressure is PV=NRT, and the gas pressure generally refers to the hydrostatic pressure exerted by the gas to a certain point, and the reason for the gas pressure is the continuous and irregular impact of a large number of gas molecules on the container wall.

Air pressure generally refers to the hydrostatic pressure exerted by the gas to a certain point, and the gas pressure is caused by the continuous and irregular impact of a large number of gas molecules on the container wall. According to the ideal gas law pv=nrt: the magnitude of the gas pressure is proportional to the amount of the gas (n), the temperature of the gas (t), and inversely proportional to the volume of the gas (v) r is the universal gas constant.

The reason why atmospheric pressure is generated can be explained from different angles

1. The air is affected by gravity, and the air is fluid, so there is pressure in all directions. Due to the attraction of the earth to the air, the air is pressed on the ground, and it must be supported by the ground or other objects on the ground, and these objects that support the atmosphere and the ground are affected by the atmospheric pressure. The atmospheric pressure per unit area is the atmospheric pressure.

2. It can be explained by the first paragraph of molecular motion. This is because gases are made up of a large number of molecules that move irregularly, and these molecules are bound to constantly collide with objects immersed in the air. With each collision, the gas molecules have to give an impact to the surface of the object, and the result of the continuous collision of a large number of air molecules is reflected in the pressure of the atmosphere on the surface of the object.

3. Atmospheric pressure is formed, if the number of molecules contained in the unit volume is more, the more air molecules collide with the surface of the object in the same time, and the pressure generated is also greater.