Question 8.4: Helium gas is used to inflate the abdomen during laparoscopi...
Helium gas is used to inflate the abdomen during laparoscopic surgery. A sample of helium gas has a volume of 5.40 L and a temperature of 15 °C. What is the final volume, in liters, of the gas after the temperature has been increased to 42 °C when the pressure and amount of gas do not change?
STEP 1 State the given and needed quantities .We place the gas data in a table by writing the initial temperature and volume as T_{1} and V_{1} and the final temperature and volume as T_{2} and V_{2} . We see that the temperature increases from 15 °C to 42 °C.
Using Charles’s law,
\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} No change in pressure and amount of gas (Charles’s law)
we predict that the volume increases.
T_{1} = 15 °C + 273 = 288 K
T_{2} = 42 °C + 273 = 315 K
| ANALYZE THE PROBLEM | Given | Need | Connect |
| T_{1} = 288 K T_{2} = 315 K V_{1} = 5.40 L
Factors that do not change: P and n |
\boxed{V_{2}} | Charles’s law, \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}
Predict: T increases, V increases |
STEP 2 Rearrange the gas law equation to solve for the unknown quantity. In this problem, we want to know the final volume (V_{2}) when the temperature increases.
Using Charles’s law, we solve for V_{2} by multiplying both sides by T_{2}.
\frac{V_{1}}{T_{1}} \times T_{2}=\frac{\boxed{V_{2} }}{\cancel{P_{2}}} \times \cancel{P_{2}}
\boxed{V_{2}} =V_{1} \times \frac{T_{2}}{T_{1}}
STEP 3 Substitute values into the gas law equation and calculate .From the table, we see that the temperature has increased. Because temperantre is directly related to volume, the volume must increase. When we substitute in the values, we see that the ratio of the temperatures (temperature factor) is greater than 1, which increases the volume as predicted.
\boxed{V_{2}} = 5.40 L \times \underset{\begin{array}{l}\text{Temperature factor}\\\text{increases volume}\end{array}}{\frac{315 \cancel{K}}{288 \cancel{K}}} = 5.91 L