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 = 291 K
Table 1
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}}=\frac{P_{2}V_{2}}{T_{2}} \\ \frac{P_{1}V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} =\frac{\cancel{P_{2}}V_{2}}{\cancel {T_{2}}} \times \frac{\cancel {T_{2}}}{\cancel{P_{2}}} \\ 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.
V_2=25.0 \mathrm{~mL} \times \underset{{\begin{array}{c}\text { Pressure } \\ \text { factor } \\ \text { increases } \\ \text { volume }\end{array}}}{\frac{4.00 \space \cancel{\mathrm{atm}}}{1.00 \space \cancel{\mathrm{atm}}}} \times\underset{{\begin{array}{c}\text { Temperature } \\ \text { factor } \\ \text { increases } \\ \text { volume }\end{array}}}{\frac{291 \space \cancel{\mathrm{K}}}{248 \space \cancel{\mathrm{K}}}}=102 \mathrm{~mL}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.
Table 1 :
| ANALYZE THE PROBLEM |
Given | Need | Connect |
| P_{1} = 4.00 atm P_{2} = 1.00 atm V_{1} = 25.0 mL T_{1} = 284 K T_{2} = 291 K Factors that do not change: n |
V_{2} | combined gas law, \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} |