Question 8.7: A weather balloon with a volume of 44 L is filled with 2.0 m...
A weather balloon with a volume of 44 L is filled with 2.0 moles of helium. What is the final volume, in liters, if helium are added to give a total of 5.0 moles of helium, if the pressure and temperature do not change?
STEP 1 State the given and needed quantities .We list those properties that change, which are volume and amount (moles). Because there is an increase in the number of moles of gas, we can predict that the volume increases.
| ANALYZE THE PROBLEM | Given | Need | Connect |
| V_{1} = 44 mL n_{1} = 2.0 moles n_{2} = 5.0 moles
Factors that do not change: P and T |
\boxed{V_{2}} | Avogadro’s law, \frac{V_{1}}{n_{1}} =\frac{V_{2}}{n_{2}}
Predict: n increases, V increases |
STEP 2 Rearrange the gas law equation to solve for the unknown quantity.
Using Avogadro’s law, we can solve for V_{2} by multiplying both sides of the equation by n_{2}.
\frac{V_{1}}{n_{1}} \times n_{2}=\frac{\boxed{V_{2} }}{\cancel{n_{2}}} \times \cancel{n_{2}}
\boxed{V_{2}} =V_{1} \times \frac{n_{2}}{n_{1}}
STEP 3 Substitute values into the gas law equation and calculate . When we substitute in the values, we see that the ratio of the moles(mole factor) is greater than 1, which increases the volume as predicted.
\boxed{V_{2}} = 44 L \times \underset{\begin{array}{l}\text{mole factor}\\\text{increases volume}\end{array}}{\frac{5.0 \cancel{moles}}{2.0 \cancel{moles}}} = 110 L