Question 8.6: A 25.0-mL bubble is released from a diver's air tank at a pr...
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.)
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.
T_{1} = 11 °C + 273 = 284 K
T_{2} = 18 °C + 273 = 691 K
| ANALYZE THE PROBLEM | Given | Need | Connect |
| P_{1} = 4.00 atm P_{1}=1.00 atm V_{1}=25.0 L T_{1} = 284 K T_{2} = 291 K
Factors that do not change: n |
\boxed{V_{2}} | combined gas law, \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} |
STEP 2 Rearrange the gas law equation to solve for the unknown quantity.
Using the combined gas law, we solve for V_{2} by multiplying both sides by T_{2} and dividing both sides by P_{2}.
\frac{P_{1}V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}=\frac{\cancel{P_{2}}\boxed{V_{2} }}{\cancel{T_{2}}} \times \frac{\cancel{T_{2}}}{\cancel{P_{2}}}
\boxed{V_{2}} =V_{1} \times \frac{P_{1}}{P_{2}} \times \frac{T_{2}}{T_{1}}
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.
\boxed{V_{2}} = 25.0 mL \times \underset{\begin{array}{l}\text{Pressure}\\\text{factor}\\\text{increases}\\\text{volume}\end{array}}{\frac{4.00 \cancel{atm}}{1.00 \cancel{atm}}} \times \underset{\begin{array}{l}\text{Temperature }\\\text{factor}\\\text{increases}\\\text{volume}\end{array}}{\frac{291 \cancel{K}}{284 \cancel{K}}} = 102 mLHowever, 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.