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kondor19780726 [428]
3 years ago
11

An underwater air bubble has an excess inside pressure of 13 pa. what is the excess pressure inside an air bubble with twice the

radius?
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0
Pressure = 2 * pi * sigma * radius where sigma is the pressure constant of the medium or the material.

Thus, Pressure/Radius = constant

Based on this:
(pressure*radius) in case 1 = (pressure*radius) in case 2
(13 * R) = (P * 2R)
P = 13/2 = 6.5 pa
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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

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a) The final pressure of an isothermal expansion is given by:

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T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

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