A gas high school physics problem, urgent! To take the exam online and other high school physics gas

Answer: 1) According to the question, if the two ends of the glass tube ABCD are open, then the pressure in the glass tube and at both ends of the mercury column is 1 atmosphere.

Insert the A end into the mercury tank, the mercury surface pressure in the mercury tank is also 1 atmosphere, insert the A end into 5cm, then there is a section of air enclosed in the ABC segment glass tube, because the pressure on the mercury surface in the A end of the glass tube is 1 atmosphere, the right side of the mercury column in the CD segment is also 1 atmosphere, and the temperature does not change, then, the volume of the air column enclosed in the ABC will not change, because the A end is occupied by the mercury column 5cm, then the mercury column in the CD segment will move 5cm to the right. From the ideal gas equation of state pv nrt, v nrt p, then the temperature is slowly reduced, the volume v of the air column will decrease, and the mercury column in the cd segment will move to the left, think about it carefully, in the process of slowly decreasing the temperature, the pressure of the air column in the abc segment will be maintained by 1 atmosphere before the mercury column in the cd segment enters the glass tube in the bc segment. Then by the above formula:

v nrt p, the temperature t is proportional to the volume, then, the mercury column of the cd segment moves 5 cm to the left to return to the original position, then the volume is reduced: 5 (150 25 5) 1 36, then the temperature also needs to be reduced by 1 36, i.e., to 300 35 36.

The answer is: the temperature drops, and the horizontal mercury column returns to its initial position.

2) If you think about it carefully, you will know that if you continue to lower the temperature on the basis of (1) above, then the mercury column in the CD segment will continue to move to the left, and when its left end reaches the curved part of point C, the volume of the gas column in the ABC will continue to decrease, and the pressure will still be 1 atmosphere.

When the horizontal column of mercury is just all in the vertical pipe, the pressure of the air column in the ABC will be 75 - 5 70 cm Hg. (Note that if the temperature continues to be lowered, the mercury column at end A will rise above the level of the mercury tank, and the mercury surface of section A will be in the same plane as the mercury in the tank before the horizontal mercury column is just inside the vertical pipe).

Then from the ideal gas equation of state, pv nrt, we get: p1v1 t1 p2v2 t2

So: take p1 1 atmosphere 75 cm Hg.

v1＝1vt1＝300k

The volume of the gas column at the level of v2 when the mercury column is just fully entered into the vertical pipe (165 180) v

p2 70 cm Hg.

Then substitute the above formula to get:

75×1v÷300＝70×（165/180）×v÷t2

Solution: t2

That is, when the temperature continues to be lowered slowly, the mercury column in the horizontal section just enters the vertical pipe.

1) From the ideal gas equation of state pv nrt, the volume of v nrt p is reduced: 5 (150 25 5) 1 36 temperature drops to 300 35 36. The horizontal mercury column returns to its initial position.

2）p1v1/t1＝p2v2/t2

75×1v÷300＝70×（165/180）×v÷t2t2＝

The horizontal mercury column just enters the vertical pipe.

In fact, this problem is to use pv t=constant.

Assuming that the valve does not open at 127 degrees, the pressure in a at 127 degrees is calculated by the formula. See if there is a vacuum more than the pressure in b, and the pressure is 0

If the valve is not opened at 127 degrees, then the pressure should be 4 3atm, which is greater than that, so the valve is open.

The second question is to check the pressure on both sides when the valve is closed.

So the PA must be PB

The second problem uses the Claperosaurus equation, where m is the mass and m is the molar mass. r is a constant, and since this question is the same gas, the molar mass can also be used as a constant.

So pv mt=constant, using this formula.

The formula is obtained by using the state of a bottle of gas at 27 degrees: 1ATM*V (M Total*300K)=K

Using the gas flow rate of bottle A after bottle B to obtain the state of bottle A: (PB+total-MB)*400K)=K

The state of B bottle after the gas flow rate of bottle A is used to obtain three formulas: PB*V (MB*400K)=K

Solve mb = 5% in total with three formulas

The liquid pressure increases with the increase of depth, when the vial is pressed from 5 meters to 6 meters, the pressure at the mouth of the bottle increases, and more water is pressed into the bottle, because the weight of the bottle at 5 meters plus the weight of the water inside the bottle is just equal to the buoyancy of the bottle and reaches a balanced state, so when more water is squeezed into the bottle, the total weight of the bottle is greater than the buoyancy of the bottle, and the bottle as a whole is subjected to the resultant force of vertical downward, so the downward movement is accelerated.

The greater the depth ... So the greater the hydraulic pressure, the volume of the air will decrease, resulting in a decrease in buoyancy, and the buoyancy is less than gravity, so it sinks.

The premise of this question is that the entire glass bottle is under the water surface, when the depth is 5 meters, the gravity and buoyancy are equal (buoyancy is calculated by the buoyancy calculation formula, where the volume is the volume of the glass bottle, that is, the volume of the gas), when the depth is 6 meters, according to the isothermal change of the gas, the pressure becomes larger, the volume decreases, that is, the buoyancy decreases, and the gravity is greater than the buoyancy, so the sinking is accelerated.

1) When the belt is placed freely, the internal and external pressure is wide, etc. So p1=p0

2) Use the formula "PV=NRT" to the right of the unchanged, v becomes half. So p2=2*p0

3) The pressure formula "f=ps", so f=2*p0*s

Because the pressure inside the syringe must be consistent with the pressure outside the syringe (i.e., atmospheric pressure) in order for the piston to be stationary.

The pressure in the syringe, before being put into hot water, is the same as the atmospheric pressure, that is, the pressure p is constant.

According to the gas formula p1v1 t1=p2v2 t2, that is, p1=p2, and after putting hot water, the gas temperature t2 rises, the gas volume v2 increases, so the piston will move upward.

In the first word, the piston will move.

The title says that friction is not counted, which means that it is the ideal condition. Just consider the increase in the internal energy of the gas when heated. As the title suggests, the temperature of the gas increases, and because there is no friction, the gas is only subjected to the pressure given by the atmosphere and the gravity of the piston, and it remains the same.

According to the Claebron equation pv=nrt, it is concluded that when the volume of the gas increases, it is natural to jack the piston to do work.

The most important thing is to pay attention to the constant pressure. It's a motif and there will be a lot of variations.

The piston will move upward, the volume will definitely change, and the closed container will not change the volume and pressure.

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

In two processes, the volume remains unchanged and the volume changes. Then the tensile force does not change, which is an isobaric change.

The equation for the first process: p0 t0=(p0-mg s) t for the second process: v t=

t is the temperature after the end of the first process.

Then just calculate t1 and you're good to go.

The equilibrium conditions and the equation of state of the ideal gas are mainly used.

The ideal gas equation is pt=nrv

The equations of the two states can be compared.

A, B are obviously different gases, staring at the lower surface of mercury, Pa + M slip and S = Pb, ie.

pa + 20cmhg = pb

pb = 95cmhg

Valve plane analysis of mercury:

pa + m2s = pb - m let and 2s i.e.

pa + 10cmHg = pb -10cmHg = 85cmHg This means that the pressure on the upper and lower surfaces is equal.

Therefore, when the valve is opened, the upper and lower parts of mercury overflow at the same time, and the overflowing flow is equal, until it is balanced again, and the balance is satisfied.

Pa + L = Pb - L = 75 cmHg (it is convenient to use L to represent LCMHG).

The magnitude of the change in Pa and Pb is the same, and only one equation is required, since Pa = Pa*10 (20-L) = 750 (20-L).

So, 750 (20-l) +l = 75

The solution is l= cm

Total outflow 20-2*l =

The process is roughly as above.