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A certain quality of air is enclosed in a U-shaped glass tube which is still placed in a vertical plane by mercury. The upper end of the right tube is open and long enough. The silver surface in the right tube is higher than that in the left tube by h. The reason why h becomes larger is () A. The ambient temperature rises B. The atmospheric pressure rises strongly by C. Mercury is added to the right tube along the tube wall. A certain mass of air is enclosed in a glass tube which is still placed in a vertical plane by mercury. The mercury surface in the tube is higher than that outside the tube by h. Which of the following measures can make h larger () A. Press the glass tube upward slightly b. Reduce the external atmospheric pressure c. Take the nozzle as the center, and rotate it slightly in the vertical plane by a small angle D. The whole device falls freely. .U-shaped glass tube free fall The U-shaped glass tube falls freely. This is option D of the first question.
1. For the sealed gas A, if the ambient temperature rises, assuming that the length of the gas column is constant (constant volume change), the pressure increases (Charlie's law), pushing the mercury surface downward, making the left mercury surface decrease, and the right mercury surface rise ==> h becomes larger ==> A pair;
Pressure of gas = atmospheric pressure + h (pressure expressed by the height of cm mercury column): B. The atmospheric pressure increases, if h is not changed, the gas pressure increases, and the gas column becomes shorter (Boyle's law)==> The left mercury surface rises, the right mercury surface decreases,==> h decreases,==> B error; C. The right pipe injects mercury, if it is equal to the volume, the gas pressure increases, and the gas column becomes shorter, because the atmospheric pressure is strong, so according to the above formula: p=p0+h, p increase can only be caused by the increase of h ==> C pair; D. Free fall, when stable, the gas column pressure is equal to the external atmospheric pressure, but the original pressure is greater than the atmospheric pressure, so the gas pressure decreases, ==> Gas column length ==" h becomes larger ==> D pair.
2. The design should be that the glass tube with the upper end closed and the lower end opening is sealed with some gas by mercury, and the opening is inserted into the mercury groove downward: a. If the glass tube is pressed down, the limit condition shall be considered. If the top of the glass tube is pressed down to the same height as the mercury surface in the mercury tank (even if the top of the tube is lower than the mercury surface), the mercury surface in the tube will be reduced ==> h; otherwise, if the glass tube is lifted up, h will be increased.
The gas pressure p=p0-h, b. If p0 decreases, assuming the length of the gas column is not changed, then p decreases ==> the length of the gas column, and the mercury surface inside the pipe decreases. Because the mercury surface outside the pipe can be considered as unchanged, h decreases ==> b;
C. If h does not change with the rotation, the gas pressure in the tube does not change, but the gas column becomes longer ==" gas pressure decreases because p0 does not change ==" h becomes larger ==" c pairs;
Some of that gas is seal with mercury in a u-shaped glass tube, The height difference difference of mercury on both sides of the pipe is 10cm and the external atmospheric pressure is known to be 1 standard atmospheric pressure, what is the pressure of the enclosed gas equal to Pa?
1 Atm =76 cmHg
Therefore:
If the mercury on the side of the enclosed gas is high, the pressure =76-10=66 cmHg;
If the mercury on the side of the enclosed gas is low, the pressure =76+10=86 cmHg.
1 Atm =76 cmHg
So:
If the mercury on the side of the enclosed gas is high, the pressure =76-10=66 cmHg;
If the mercury on the side of the enclosed gas is low, the pressure =76+10=86 cmHg.
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