How to learn the chapter on physical gases in high school?

Just remember pv=nrt.

The proportion of gas in the college entrance examination is not much, like this year, there are not a few questions, and it is not difficult.

In the college entrance examination, there are more questions about mechanics and work and energy, as well as electromagnetic induction.

Generally, three aspects are used.

The first law of thermodynamics.

The Clabron equation.

Changing the internal energy is work and heat transfer.

The nature of the gas.

1.gaseous'Status Parameters:

Temperature: macroscopically, how hot or cold an object is; Microscopically, it is a sign of the intensity of the irregular movement of the vertical parts of an object.

Relationship between osmotic thermodynamic temperature and Celsius: t=t+273 {t: thermodynamic temperature (k), t: temperature in Celsius ( )

Volume v: the space that gas molecules can occupy, unit conversion: 1m3 = 103l = 106ml

Pressure p: per unit area, a large number of gas molecules frequently hit the wall of the device to produce a continuous, uniform pressure, standard atmospheric pressure:

1atm=2.Characteristics of gas molecular motion: large intermolecular voids; Except for the moment of collision, the interaction force is weak; The rate of movement of molecules is large.

3.The equation of state for an ideal gas: p1v1 t1=p2v2 t2 {pv t=constant, t is the thermodynamic temperature (k)}

Note: 1) The internal energy of an ideal gas is not related to the volume of the ideal gas, but to the temperature and the amount of matter;

2) Equation 3 is established as an ideal gas of a certain mass, and the unit of temperature should be paid attention to when using the formula, t is the temperature in Celsius ( ) and t is the thermodynamic temperature (k).

Temperature: macroscopically, how hot or cold an object is; Microscopically, a sign of the degree of intensity of the irregular movement of molecules inside an object, the thermodynamic temperature and Celsius temperature: t=t+273 {t: thermodynamic temperature (k), t: Celsius temperature ( )

Volume v: the space that gas molecules can occupy, unit conversion: 1m3 = 103l = 106ml

Pressure p: per unit area, a large number of gas molecules frequently hit the wall of the device to produce a continuous, uniform pressure, standard atmospheric pressure: 1atm=

2.Characteristics of the movement of gas molecules:

large intermolecular voids; Except for the moment of collision, the interaction force is weak; The rate of movement of molecules is large.

3.Equation of state for an ideal gas:

p1v1 t1=p2v2 t2 {pv t=constant, t is the thermodynamic temperature (k)}

Note: 1) The internal energy of an ideal gas is not related to the volume of the ideal gas, but to the temperature and the amount of matter;

2) Equation 3 is established as an ideal gas of a certain mass, and the unit of temperature should be paid attention to when using the formula, t is the temperature in Celsius ( ) and t is the thermodynamic temperature (k).

Use the formula: PV=NRT

The atmospheric pressure p is the same, the amount of air n is constant, and r is a fixed value.

Change the formula down, p nr=t v=fixed value.

Let the absolute zero degree be t, and the cross-sectional area of the glass tube is s

So: (T-30).

t=-270

His principle of temperature measurement is v1t2=v2t1

Here t is the Kelvin temperature.

Let absolute 0 degrees be t

then (90-t) 30=(30-t) 36

t=-270

This is due to the existence of static friction between piston B and the pipe wall, when piston A moves to the left, the closed gas pressure is stronger than the atmospheric pressure, and the piston B is subjected to the static friction force of the pipe wall against it, with the increase of the pressure of the closed gas, the static friction force is also increasing, and the force of piston B is always balanced, but the horizontal direction is not two forces balance, but three forces (atmospheric pressure, gas pressure in the tube, static friction) balance. When the combined force of the gas pressure and atmospheric pressure in the tube reaches and exceeds the maximum static friction, piston B is ejected.

There should be friction between B and the pipe wall or there is an elastic buckle, and there must be a large force to break away, otherwise it can't be called a gun.

There is friction between the piston B and the cylinder wall, and the static friction between the piston B and the cylinder wall also increases with the increase of the pressure in the tube. We all know that the maximum static friction is greater than the sliding friction, so when the pressure in the cylinder acting on the piston B reaches and exceeds the maximum static friction between the piston B and the cylinder wall, the piston B begins to move, and at this time the piston B becomes sliding friction between the cylinder wall, and the piston B is instantaneously subjected to a relatively large acceleration and quickly ejects.

For closed gases, the gas pressure is p1 p0 and the length of the gas column is l1 h when the tube is at rest

When the tube rotates the whole apparatus around the axis of the open vertical arm at angular velocity, the pressure of the enclosed gas is p2 p, and the length of the gas column is l2 h2

where p right is the pressure at the right end of the horizontal tube and p mercury g(h2) is the pressure generated by the mercury column in the right tube.

For closed gases, there are p1*l1 p2*l2

i.e. p0*h [p-right g(h2)h2).

After finishing, get p0 (2*p right g h) 4 ...Equation 1

For horizontal pipes the length of the fraction of mercury is l (h2): obtained by the formula of centripetal force.

F-direction (p-right, p0)*s, m*2*r

m is the mass of mercury remaining in the horizontal tube, m*l(h2)]s, s is the cross-sectional area of the tube.

r is the distance from the center of gravity of the mercury in the horizontal tube to the left end, i.e., r(h2)l(h2)]2(l2)(h4).

(p0)*s *l (h 2 )]s* 2* [l 2) (h 4 )].

After finishing the above formula, we get p0 *2*(4*l 2 h 2) 8 ...Equation 2

The angular velocity obtained from Equations 1 and 2 is 2*root number {(2*p0 gh) [ 4*l 2 h 2)].

Take the bottom of the right end of the U tube as the demarcation point, then the left:

There is an object mass m = g(l-h2)s, let s be the area of the nozzle. When rotating with angular velocity, there is a centripetal force f=m r, where r=h 2+(l-h 2) 2=l 2+h 4

Right: The pressure of the liquid p1 = gh 2, the pressure of the air, because the temperature does not change, the volume decreases by half so p = 2p0

So there is f=(p1+p2)s

Substituting has g(l-h2)s l 2+h 4)=( gh 2+2p0)s

I won't write about the results of the solution.

Personally, I think that if the air does not enter the air when it is lifted up, then let the volume of part A become 30+x and the volume of part B becomes 30+y, then X+Y=MPa1*VA1=PA2*VA2

pb1*vb1=pb2*vb2

and PA2=PB2-45

That is: 45*40=(pb2-45)*(40+x)90*30=pb2*(30+y).

x+y=m can be solved to get pb2=f(m).

y=g(m)

Then the pressure at the level is p=pb2-(15-y)=h(m) (it's too troublesome, it doesn't count).

After h(m) is compared with atmospheric pressure p0, h(m) > p0, then the liquid flows out, h(m), so I think it should be related to the distance further up. Of course, it is not perfect, that is, the premise is to control the horizontal tube atmosphere not to move first, and then let the horizontal tube move after the piston is lifted to see if it enters, and Y 15 is not considered.