Question

Suppose that a 0.10-ml bubble of 0.00274 moles of methane is formed from decaying vegetation on an ocean floor where the temperature is 4.00 degrees C and the pressure is 623 atm. Calculate the volume (in ml) of the bubble at the ocean surface at 1.0 atm and 20.0 degrees C.

Answer 1

i see what we need to do

Answer 2

we can apply the ideal gas law, have you heard of it?

Answer 3

Yeah, but you have to do conversions don't you?

Answer 4

\[pV = nRT\]

we're looking for the new volume under a different set of conditions. that is V2 when the bubble reaches the ocean surface at a different pressure and temperature.

now, our formula becomes this.

\[\frac{ P_{1}V_{1} }{ T_{1} } = \frac{ P_{2}V_{2?} }{ T_{2} }\]

Now, we're looking for the final volume, so we re-arrange our equation and solve for the final volume.

\[\frac{ P_{1}*V_{1}*T_{2} }{ T_{1}*P_{2} } = V_{2}\]

Now we have an equation that we have solved for V2 we just need to plug everything in but we need to make sure that


Temperature is in kelvins
Pressure is in atm
Volume is in liters

Answer 5

Let's first put the information we have together before we plug the values into the equation.

Answer 6

P1 = 623 atm
T2 = 277 Kelvins
V1 = 1.0*10^{-4} L

P2 = 1.0 atm
T2 = 293 Kelvins


only after we have solved our equation for the variable we need. In this case the final volume of our bubble do we plug in all the numbers.

\[\frac{ P_{1}*V_{1}*T_{2} }{ T_{1}*P_{2} } = V_{2}\]

\[\frac{ 623~atm*1.0*10^{-4}~L*293~K }{ 273 K*1.0~atm } = Liters~V_{2}\]

Answer 7

This equation is rested on the fact that the number of moles stays constant. just to give you a better picture of the units.


\[\frac{ P_{1}*V_{1}*T_{2} }{ T_{1}*P_{2} } = V_{2}\]

if you notice all the units we don't want cancel out and we're left with liters in the numerator.

\[\frac{ 623~\cancel\atm*1.0*10^{-4}~L*293~\cancel\K }{ 273\cancel\K*1.0~\cancel\atm } = Liters~V_{2}\]