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The height of the original gas column in the cup is L 1 =6cm, p1= P0;

Later, the cup was turned upside down, and the air column height L 2 =6cm+0.035cm=6.035cm.

Let the cross-sectional area of the cup be S.

The gas in the cup changes isothermally, which is obtained by Boyle's law:

P 1 L 1 S=P 2 L 2 S

Solution: P2 = L1L2p1= 66.035×1.013×105pa ≈1.007×105pa.

The pressure generated by the water column in the cup is p water = rhogh =1.0×1.03× 9.8× 6×1.02pa = 588pa = 0.00588×1.05pa.

The pressure of the gas in the cup plus the pressure of the water column is P= P 2+P water =1.007×1.00588×1.005pa ≈1.01.005pa = p 0.

That is, the pressure of the gas in the cup plus the pressure of the water column is just equal to atmospheric pressure, which is in equilibrium, so the water will not flow out.