A 25.0-mL bubble is released from a diver's air tank at a pressure of 4.00 atm and a temperature of 11 °C. What is the volume, in milliliters, of the bubble when it reaches the ocean surface where the pressure is 1.00 atm and the temperature is 18 °C? (Assume the amount of gas in the bubble does

Question

A 25.0-mL bubble is released from a diver's air tank at a pressure of 4.00 atm and a temperature of 11 °C. What is the volume, in milliliters, of the bubble when it reaches the ocean surface where the pressure is 1.00 atm and the temperature is 18 °C? (Assume the amount of gas in the bubble does not change.)

Accepted Answer

STEP 1 State the given and needed quantities .We list the properties that change, which are the pressure, volume, and temperature. The temperatures in degrees Celsius must be changed to kelvins.

[latex]T_{1}[/latex] = 11 °C + 273 = 284 K
[latex]T_{2}[/latex] = 18 °C + 273 = 691 K

ANALYZE THE PROBLEM

Given

Need

Connect

[latex]P_{1} = 4.00 atm P_{1}=1.00 atm V_{1}=25.0 L T_{1} = 284 K T_{2} = 291 K[/latex]

Factors that do not change: n

[latex] \boxed{V_{2}}[/latex]

combined gas law, [latex] \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/latex]

STEP 2 Rearrange the gas law equation to solve for the unknown quantity.
Using the combined gas law, we solve for [latex]V_{2}[/latex] by multiplying both sides by [latex]T_{2}[/latex] and dividing both sides by [latex]P_{2}[/latex].

[latex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}\boxed{V_{2}}}{T_{2}} [/latex]

[latex] \frac{P_{1}V_{1}}{T_{1}} imes \frac{T_{2}}{P_{2}}=\frac{\cancel{P_{2}}\boxed{V_{2} }}{\cancel{T_{2}}} imes \frac{\cancel{T_{2}}}{\cancel{P_{2}}}[/latex]

[latex] \boxed{V_{2}} =V_{1} imes \frac{P_{1}}{P_{2}} imes \frac{T_{2}}{T_{1}}[/latex]

STEP 3 Substitute values into the gas law equation and calculate . From the data table, we determine that both the pressure decrease and the temperature increase will increase the volume.

[latex] \boxed{V_{2}} = 25.0 mL imes \underset{\begin{array}{l} ext{Pressure}\ ext{factor}\ ext{increases}\ ext{volume}\end{array}}{\frac{4.00 \cancel{atm}}{1.00 \cancel{atm}}} imes \underset{\begin{array}{l} ext{Temperature }\ ext{factor}\ ext{increases}\ ext{volume}\end{array}}{\frac{291 \cancel{K}}{284 \cancel{K}}} = 102 mL[/latex]

However, in situations where the unknown value is decreased by one change but increased by the second change, it is difficult to predict the overall change for the unknown.

Question 8.6: A 25.0-mL bubble is released from a diver's air tank at a pr...

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