Question #62475

a mercury barometer is defective. it contains some air in the space above the mercury. when an aacurate barometer reads 770mm, the defective one reads 760mm. when the accurate barometer reads 750mm, the defective one reads 742mm.
(a) what is the length of the air column, when the aacurate barometer reads 770mm?
(b) what is the reading of the accurate barometer when the defective one reads 752mm?
assume that the temperature remains constant.
1

Expert's answer

2016-10-20T11:29:10-0400

Answer on Question #62475-Physics-Molecular Physics-Thermodynamics

A mercury barometer is defective. it contains some air in the space above the mercury. when an accurate barometer reads 770mm, the defective one reads 760mm. when the accurate barometer reads 750mm, the defective one reads 742mm.

(a) What is the length of the air column, when the accurate barometer reads 770mm?

(b) What is the reading of the accurate barometer when the defective one reads 752mm?

Assume that the temperature remains constant.

Solution

Let the area of cross-section of the barometer tube be A and x cm the length of air column above 760 mm mark.

(a) The pressure of air in space above mercury, P1P_{1} is reading of accurate barometer - reading of defective barometer:


P1=770760=10 mmP_{1} = 770 - 760 = 10 \text{ mm}


The volume of air is V1=xAV_{1} = xA.

Similarly, P2=750742=8 mmP_{2} = 750 - 742 = 8 \text{ mm}

The volume of air is


V2=(x+76.074.2)A=(x+1.8)A.V_{2} = (x + 76.0 - 74.2)A = (x + 1.8)A.


According to the Boyle's law


P1V1=P2V2(10)(xA)=(8)((x+1.8)A)x=7.2 cm.\begin{array}{l} P_{1}V_{1} = P_{2}V_{2} \\ (10)(xA) = (8)\big((x + 1.8)A\big) \\ x = 7.2 \text{ cm}. \end{array}


(b) Suppose the reading of accurate barometer be P0P_0, then the pressure of air is P=P0752P = P_0 - 752.

The volume of air is


V=(x+76.075.2)A=(x+0.8)A.V = (x + 76.0 - 75.2)A = (x + 0.8)A.


According to the Boyle's law


P1V1=P2V2(10)(xA)=(P0752)((x+0.8)A)x=7.2 cm.(10)(7.2A)=(P0752)((7.2+0.8)A)P0=761 mm.\begin{array}{l} P_{1}V_{1} = P_{2}V_{2} \\ (10)(xA) = (P_{0} - 752)\big((x + 0.8)A\big) \\ x = 7.2 \text{ cm}. \\ (10)(7.2A) = (P_{0} - 752)\big((7.2 + 0.8)A\big) \\ P_{0} = 761 \text{ mm}. \end{array}


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