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Kinetic Theory

Ideal gas equations

A glass bulb of volume 400 {~cm}^{3} is connected to another o

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A glass bulb of volume $400 cm^{3}$ is connected to another of volume $20 cm^{3}$ by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of $2 0^{\circ} C$ and $1.00 atm$ . The larger bulb is immersed in steam at $10 0^{\circ} C$ and the smaller in melting ice at $0^{\circ} C$ . Find the final common pressure.

Name: Video answer for A glass bulb of volume 400 cm3 is connected to another of volume 20 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20∘C and 1.00 atm. The larger bulb is immersed in steam at 100∘C and the smaller in melting ice at 0∘C. Find the final common pressure.
Uploaded: 2023-01-31T16:09:44.000Z
Duration: 8 min
Description: A glass bulb of volume 400 cm3 is connected to another of volume 20 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20∘C and 1.00 atm. The larger bulb is immersed in steam at 100∘C and the smaller in melting ice at 0∘C. Find the final common pressure.

Views: 5,233 students

Updated on: Jan 31, 2023

Book:

Waves And Thermodynamics (DC Pandey)

Text solutionVerified

P_{i} = 1 atm T_{i} = 2 0^{\circ} C = 293 K V_{i} = 0.6 L ∴ \frac{PiVi}{T _{i}} = \frac{0.6}{293}

let final press be

P

T_{A} = 373 K T_{B} = 273 K ∴ \frac{PV _{A}}{T _{A}} + \frac{PV _{B}}{T _{B}} = \frac{P _{i} V _{i}}{T _{i}} \Rightarrow P (\frac{0.4}{373} + \frac{0.2}{273}) = \frac{0.6}{293} \Rightarrow P (\frac{2}{373} + \frac{1}{273}) = \frac{3}{293} \Rightarrow P = \frac{3 \times 373 \times 273}{293 \times 919} \approx 1.134 atm

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Jack Discussed

A glass bulb of volume

400 cm^{3}

is connected to another of volume

20 cm^{3}

by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of

2 0^{\circ} C

and

1.00 atm

. The larger bulb is immersed in steam at

10 0^{\circ} C

and the smaller in melting ice at

0^{\circ} C

. Find the final common pressure.

14 mins ago

Discuss this question LIVE

14 mins ago

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Question Text	A glass bulb of volume $400 cm^{3}$ is connected to another of volume $20 cm^{3}$ by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of $2 0^{\circ} C$ and $1.00 atm$ . The larger bulb is immersed in steam at $10 0^{\circ} C$ and the smaller in melting ice at $0^{\circ} C$ . Find the final common pressure.
Updated On	Jan 31, 2023
Topic	Kinetic Theory
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	238
Avg. Video Duration	8 min

Text solutionVerified

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