Question 5.11: In order to determine the rate of photosynthesis (the conver...
In order to determine the rate of photosynthesis (the conversion by plants of carbon dioxide and water into glucose and oxygen), the oxygen gas emitted by an aquatic plant is collected over water at a temperature of 293 K and a total pressure of 755.2 mmHg. Over a specific time period, a total of 1.02 L of gas is collected. What mass of oxygen gas (in grams) formed?
| SORT The problem gives the volume of gas collected over water as well as the temperature and the pressure. You are asked to find the mass in grams of oxygen formed. | GIVEN V = 1.02 L, P_{total} = 755.2 mmHg, T = 293 K FIND g O_{2} |
| STRATEGIZE You can determine the mass of oxygen from the amount of oxygen in moles, which you can calculate from the ideal gas law if you know the partial pressure of oxygen. Since the oxygen is mixed with water vapor, you find the partial pressure of oxygen in the mixture by subtracting the partial pressure of water at 293 K(20 °C) from the total pressure. Use the ideal gas law to find the number of moles of oxygen from its partial pressure, volume, and temperature. Finally, use the molar mass of oxygen to convert the number of moles to grams. |
CONCEPTUAL PLAN PO_{2} = P_{total} – P_{H_{2}O}(20 °C) P_{O_{2}} , V, T → n_{O_{2}} . P_{O_{2}}V = n_{O_{2}}RT n_{O_{2}} → g O_{2} . \frac{32.00 g O_{2}}{mol O_{2}} RELATIONSHIPS USED |
| SOLVE Follow the conceptual plan to solve the problem. Begin by calculating the partial pressure of oxygen in the oxygen/water mixture. You can find the partial pressure of water at 20 °C in Table 5.3.
Next, solve the ideal gas law for number of moles. Next, substitute into the ideal gas law to find the number of moles of oxygen. |
|
| TABLE 5.3 Vapor Pressure of Water versus Temperature | |||
| Temperature (°C) | Pressure (mmHg) |
Temperature (°C) | Pressure (mmHg) |
| 0 | 4.58 | 55 | 118.2 |
| 5 | 6.54 | 60 | 149.6 |
| 10 | 9.21 | 65 | 187.5 |
| 15 | 12.79 | 70 | 233.7 |
| 20 | 17.55 | 75 | 289.1 |
| 25 | 23.78 | 80 | 355.1 |
| 30 | 31.86 | 85 | 433.6 |
| 35 | 42.23 | 90 | 525.8 |
| 40 | 55.4 | 95 | 633.9 |
| 45 | 71.97 | 100 | 760.0 |
| 50 | 92.6 | ||
P_{O_{2}} = P_{total} – P_{H_{2}O}(20 °C)
= 755.2 mmHg – 17.55 mmHg
= 737.\underline{6}5 mmHg
n_{O_{2}} =\frac{ P_{O_{2}}V}{RT}
737.\underline{6}5 \cancel{mmHg} \times\frac{1 atm}{760\cancel{ mmHg }}= 0.970\underline{5}9 atm
n_{O_{2}} = \frac{P_{O_{2}}V}{RT} =\frac{ 0.970\underline{5}9 \cancel{atm} (1.02 \cancel{L})}{0.08206 \frac{\cancel{L} . \cancel{atm}}{mol .\cancel{ K}} (293 \cancel{K})}
= 4.1\underline{1}75 \times 10^{-2} mol
4.1\underline{1}75 \times 10^{-2} \cancel{mol O_{2}} \times\frac{32.00 g O_{2}}{1 \cancel{mol O_{2}}}= 1.32 g O_{2}
CHECK The answer is in the correct units. You can quickly check the magnitude of the answer by using molar volume. Under STP one liter is about 1/22 of 1 mole. Therefore the answer should be about 1/22 the molar mass of oxygen (1/22×32 = 1.45). The magnitude of the answer seems reasonable.