Question 5.6: Calculate the number of moles of gas in a 3.24 L basketball ...
Calculate the number of moles of gas in a 3.24 L basketball inflated to a total pressure of 24.3 psi at 25 °C. (Note: The total pressure is not the same as the pressure read on the type of pressure gauge used for checking a car or bicycle tire. That pressure, called the gauge pressure, is the difference between the total pressure and atmospheric pressure. In this case,if
atmospheric pressure is 14.7 psi, the gauge pressure would be 9.6 psi. However, for calculations involving the ideal gas law, you must use the total pressure of 24.3 psi.)
| SORT The problem gives you the pressure, the volume, and the temperature and asks you to find the number of moles of gas. | GIVEN P = 24.3 psi, V = 3.24 L, T(°C) = 25 °C FIND n |
| STRATEGIZE The conceptual plan shows how the ideal gas law provides the relationship between the given quantities and the quantity you need to find. | CONCEPTUAL PLAN P, V, T → n . PV = nRT RELATIONSHIP USED PV = nRT (ideal gas law) |
| SOLVE To solve the problem, first solve the ideal gas law for n. Before substituting into the equation, convert P and T into the correct units. Finally, substitute into the equation and calculate n. |
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PV = nRT
n = \frac{PV}{RT}
P = 24.3\cancel{ psi} \times\frac{1 atm}{ 14.7 \cancel{psi}} = 1.6\underline{5}31 atm
(Since rounding the intermediate answer would result in a slightly different final answer, we mark the least significant digit in the intermediate answer, but don’t round until the end.)
T(K) = 25 + 273 = 298 K
n =\frac{ 1.6\underline{5}31 \cancel{atm} \times 3.24 \cancel{L}}{0.08206\frac{ \cancel{L} . \cancel{atm}}{mol.\cancel{K}} \times 298 \cancel{K}}
= 0.219 mol
CHECK The units of the answer are correct. The magnitude of the answer (0.219 mol) makes sense because, as you will see in the next section, one mole of an ideal gas under standard temperature and pressure (273 K and 1 atm) occupies 22.4 L.
At a pressure that is 65% higher than this, the volume of 1 mol of gas would be proportionally lower. Since this gas sample occupies 3.24 L, the answer of 0.219 mol is reasonable.