# Ideal Gas Law and a gas bubble

## Homework Statement

A gas bubble with a volume of 0.10 cm^3 is formed at the bottom of a 10.0 cm deep container of mercury. If the temperature is 27C at the bottom of the container and 37C at the top of the container, what is the volume of the bubble just beneath the surface of the mercury? assume that the surface is at atmospheric pressure.

P1V1/T1=P2V2/T2
P=Po+pgh

## The Attempt at a Solution

P2=Po+pgh
P2= 1.01x10^5Pa + (13.6x10^3kg/m3)(9.81m/s)(1.0x10^-3m)
P2= 101130Pa

P1V1/T1=P2V2/T2 > V2=P1V1T2/T1P2

V2=(1.01x10^5Pa)(1.0X10^-3 m^3)(300K)/(310K)(101130Pa)

V2=9.66x10^-4 m^3

The answer in the back differs with mine and I have tried to redo the problem many times.

## Answers and Replies

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Hi greyknight, welcome to PF

Firstly, What do your labels 1 and 2 refer to? Which is bottom and which is top?

Secondly,

P2= 1.01x10^5Pa + (13.6x10^3kg/m3)(9.81m/s)(1.0x10^-3m)
What is 1.0x10^-3 m?

Finally,
V2=(1.01x10^5Pa)(1.0X10^-3 m^3)(300K)/(310K)(101130Pa)
What is 1.0x10^-3 m^3?

Label 1 is the top and label 2 is the bottom.

1.0x10^-3 m is the depth of the container. Originally 10.0 cm but I converted it to meters.

1.0x10^-3 m^3 is the volume of the bubble on top converted from 0.10 cm^3.

Label 1 is the top and label 2 is the bottom.
Well, in that case you are trying to find the volume V2 at the bottom, but the question asks you for the volume at the surface. Bubbles rise up!

1.0x10^-3 m is the depth of the container. Originally 10.0 cm but I converted it to meters.
Check this conversion. 1x10^-3 m = 1/1000 m = 1mm.

1.0x10^-3 m^3 is the volume of the bubble on top converted from 0.10 cm^3.
Check this conversion also. How do you convert cm^3 to m^3?

1.0x10^-3 m is the depth of the container. Originally 10.0 cm but I converted it to meters.
10 cm = 0.1 m

1.0x10^-3 m^3 is the volume of the bubble on top converted from 0.10 cm^3.
0.1 cubic centimeter = 1.0e-7 cubic meter

I divided 0.10 by 100x100x100 or 0.10/100^(3)

Wow, that was just a lot of sloppy math by me. Thanks, naresh. :)

do you get the answer?