Physics, asked by KushalKumar2610, 9 months ago

An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.

Answers

Answered by Fatimakincsem
2

The length of the air column is 48 cm.

Explanation:

Here are two situations:

  • Case 1 : Atmospheric pressure + Pressure
  • Case 2 : Atmospheric pressure +  Component of the pressure due to mercury column

Solution:

P1 = Po + PHg

Let the cross sectional area of the tube be "A".

= 0.76 + 0.2 = 0.96 m Hg

T1 = T2 = T

V1  = 0.43 A

If tube is slanted , Po remais same but PHg changes.

P2 = Po + PHg cos (60°)

    = 0.76 + 0.2 x 0.5 = 0.86 mHg

P1V1 = P2V2

V2 = P1V1/P2

    = 0.96 x 0.43 A/0.86

Let the length of the air column is "l"

Al = 0.96 x 0.43 A/0.86

L = 0.48 m

L = 48 cm

Also learn about

One litre of an ideal gas is compressed isothermally at 0.72 m of hg column so that its volume becomes 0.9 l find its stress?

https://brainly.in/question/2861828

Answered by bhuvna789456
0

The length of the air column is 60°

Explanation:

Given Data

The length of the mercury column = 20 cm

The length of trapped air column = 43 cm

To find the length of the air column when tilted at an angle of 60°

Original pressure = Atmospheric pressure + Mercury pressure

P_{1}=P_{0}+P_{H g}

Let the Cross sectional area of the tube be A.

P_{1}=0.76+0.2=0.96 \mathrm{m} \mathrm{Hg}

T_{1}=T_{2}=T

\mathrm{V}_{1}=0.43 \mathrm{A}

If the tube is slanted then Pore retains the same atmospheric pressure. The PHg alone does shift.

P_{2}=P_{0}+P_{H g} \cos 60^\circ

= 0.76 + 0.2 × 0.5 = 0.86

\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}

V_{2}=\frac{P_{1} V_{1}}{P_{2}}

=\frac{0.96 \times 0.43 A}{0.86}

Let the Air column length be l.

A I=\frac{P_{1} V_{1}}{P_{2}}

=\frac{0.96 \times 0.43 A}{0.86}

l=0.48 m

l=48 cm  

Therefore the length of the air column is 48 cm when the tube is tilted slowly in a vertical plane through an angle of 60°.

Similar questions