Question 11.7: Calculate the number of moles of gas in a basketball inflate......
Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Step-by-Step
GIVEN:
P = 24.2 psi
V = 3.2 L
t = 25 °C
FIND: n
RELATIONSHIPS USED
PV = nRT (Ideal gas law, presented in this section)
PV = nRT
n\,=\,{\frac{P V}{R T}} \\ P\,=\,24.2\,\cancel{\mathrm{psi}}\,\times\,{\frac{1\,\mathrm{a}\mathrm{tm}}{14.7\,\cancel{\mathrm{psi}}}}\;=\;1.6\underline{4}62\,\mathrm{atm}T = t + 273
= 25 + 273 = 298 K
n={\frac{1.6\underline{4}62 \cancel{atm} \times 3.2 \mathrm{\cancel{L}}}{0.0821{\frac{\mathrm{\cancel{L}}\cdot\mathrm{\cancel{atm}}}{\mathrm{mol}\cdot\mathrm{\cancel{K}}}}\times298 \mathrm{\cancel{K}}}}= 0.22 mol
The answer has the correct units, moles. The value of the answer is a bit more difficult to judge. Again, knowing that at standard temperature and pressure (T = 0 °C or 273.15 K and P = 1 atm), 1 mol of gas occupies 22.4 L can help (see Check step in Example 11.6). A 3.2 L sample of gas at STP would contain about 0.15 mol; therefore at a greater pressure, the sample should contain a bit more than 0.15 mol, which is consistent with the answer.
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