题目内容
如图所示，一上端开口，下端封闭的长度为60cm的细长玻璃管，底部封有一定质量的可视为理想气体的长L1=12cm的空气柱，上部有长L2=38cm的水银柱，已知大气压强为Po=76cmHg。如果使玻璃管在竖直平面内缓慢地转动到开口向下，求在转动至水平位置和开口向下位置时管中空气柱的长度？（在转动过程中没有发生漏气，外界温度恒定，计算结果保留两位有效数字，已知
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解：设玻璃管开口向上时，空气柱的压强为
P1=P0+L2 =114cmHg···················································································· （1分）
玻璃管水平位置时，空气柱压强为Po设此时空气柱长度为,玻璃管截面积为S,
P1 L1S=PoL S······························································································· （2分）
解得L=18cm
玻璃管开口向下时，原来的水银会有一部分流出，设此时空气柱长度为L′，
P2=76-(60- L′)cmHg=16+ L′ ········································································ （2分）
则由玻意耳定律得P1 L1S = P2 L′S································································ （2分）
解得L′=30cm······························································································· （2分）
（B）
（D）