Question 5.10: A 12.5-L scuba diving tank is filled with a helium–oxygen (h...
A 12.5-L scuba diving tank is filled with a helium–oxygen (heliox) mixture containing 24.2 g of He and 4.32 g of O_{2} at 298 K. Calculate the mole fraction and partial pressure of each component in the mixture and the total pressure of the mixture.
| SORT The problem gives the masses of two gases in a mixture and the volume and temperature of the mixture. You are asked to find the mole fraction and partial pressure of each component, as well as the total pressure. | GIVEN m_{He} = 24.2 g, m_{O_{2}} = 4.32 g, V = 12.5 L, T = 298 K FIND \chi_{He}, \chi_{O_{2}}, P_{He}, P_{O_{2}}, P_{total} |
| STRATEGIZE The conceptual plan has several parts. To calculate the mole fraction of each component, you must first determine the number of moles of each component. Therefore, in the first part of the conceptual plan, you convert the masses to moles using the molar masses. In the second part, calculate the mole fraction of each component using the mole fraction definition. Calculate the total pressure from the sum of the moles of both components. (Alternatively, you could calculate the partial pressures of the components individually, using the number of moles of each component. Then you could sum them to obtain the total pressure.) Last, use the mole fractions of each component and the total pressure to calculate the partial pressure of each component. |
CONCEPTUAL PLAN m_{He} → n_{He} m_{O_{2}} → n_{O_{2}} . \frac{1 mol He}{4.00 g He} \frac{1 mol O_{2}}{32.00 g O_{2}} \chi_{He} =\frac{ n_{He}}{n_{He} + n_{O_{2}}} ; \chi_{O_{2}} = \frac{n_{O_{2}}}{n_{He} + n_{O_{2}}} P_{total} = \frac{(n_{He} + n_{O_{2}}) RT}{V} P_{He} = \chi_{He} P_{total} ; P_{O_{2}} = \chi_{O_{2}} P_{total} RELATIONSHIPS USED \chi_{a} = n_{a} /n_{total} (mole fraction definition) P_{total}V = n_{total} RT (ideal gas law) P_{a} = \chi_{a} P_{total} |
| SOLVE Follow the plan to solve the problem. Begin by converting each of the masses to amounts in moles.
Calculate each of the mole fractions. |
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24.2 \cancel{g He} \times\frac{1 mol He}{4.00 \cancel{g He}} = 6.05 mol He
4.32 \cancel{g O_{2}} \times\frac{1 mol O_{2}}{32.00\cancel{ g O_{2}}}= 0.135 mol O_{2}
\chi _{He} = \frac{n_{He}}{n_{He} + n_{O_{2}}}= \frac{6.05 \cancel{mol}}{6.05 \cancel{mol} + 0.135 \cancel{mol}} = 0.97\underline{8}17
\chi _{O_{2}} = \frac{n_{O_{2}}}{n_{He} + n_{O_{2}}}= \frac{0.135\cancel{ mol}}{6.05 \cancel{mol} + 0.135 \cancel{mol}} = 0.021\underline{8}27
P_{total} = \frac{(n_{He} + n_{O_{2}})RT}{V}2
= \frac{(6.05\cancel{mol}+0.135\cancel{mol})\left(0.08206\frac{\cancel{L}.atm}{\cancel{mol}.\cancel{K}} \right)(298\cancel{K}) }{12.5\cancel{L}}
= 12.\underline{0}99 atm
P_{He} =\chi _{ He} P_{total }= 0.97\underline{8}17 \times 12.\underline{0}99 atm
= 11.8 atm
P_{O_{2}} = \chi _{O_{2}} P_{total} = 0.021\underline{8}27 \times 12.\underline{0}99 atm
= 0.264 atm
CHECK The units of the answers are correct, and the magnitudes are reasonable.