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Chapter 8: Problem 133

A \(2.00-\mathrm{L}\) sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of \(785\) torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(785\) torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The vapor pressure of water at 25°C is 21.8 torr.

Step by step solution

01

Write down the given values.

We are given the following information: - Volume of the gas mixture (O₂ and H₂O vapor): V_total = 2.00 L - Total pressure: P_total = 785 torr - Temperature: T = 25°C = 298K - Volume of dry O₂ gas: V_O₂ = 1.94 L
02

Determine the moles of dry O₂ gas using the Ideal Gas Law.

First, we need to calculate the number of moles of dry O₂ gas using the Ideal Gas Law: \(PV = nRT\) Here, we are looking for n (number of moles), and we are given the pressure (P), volume (V), and temperature (T). The gas constant R = 62.363 L torr / K mol for pressure given in torr. Rearranging the equation to find n: \(n = \frac{PV}{RT}\) Using the values provided for dry O₂ gas, we have: P = 785 torr, V = 1.94 L, and T = 298K Plugging these values into the equation: \(n = \frac{(785\,\text{torr})(1.94\,\text{L})}{(62.363\,\text{L torr / K mol})(298\,\text{K})}\)
03

Calculate the number of moles of dry O₂ gas.

Now, we can calculate the number of moles of dry O₂ gas: \(n = \frac{(785\,\text{torr})(1.94\,\text{L})}{(62.363\,\text{L torr / K mol})(298\,\text{K})} = 0.0815\,\text{mol}\)
04

Calculate the partial pressure of O₂ in the total pressure.

Next, we can use the ideal gas law again to find the partial pressure of O₂ gas in the total pressure: \(P_{\text{O}_2} = \frac{nRT}{V}\) Using the values we calculated previously for dry O₂ gas: \(P_{\text{O}_2} = \frac{(0.0815\,\text{mol})(62.363\,\text{L torr / K mol})(298\,\text{K})}{2.00\,\text{L}}\)
05

Calculate the partial pressure of O₂ gas.

Now, we can calculate the partial pressure of O₂ gas: \(P_{\text{O}_2} = \frac{(0.0815\,\text{mol})(62.363\,\text{L torr / K mol})(298\,\text{K})}{2.00\,\text{L}} = 763.2\,\text{torr}\)
06

Calculate the vapor pressure of water using Dalton's Law of Partial Pressures.

According to Dalton's Law of Partial Pressures, the total pressure is equal to the sum of the partial pressures of the gases: \(P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2\text{O}}\) Now, we can solve for the vapor pressure of water, \(P_{\text{H}_2\text{O}}\): \(P_{\text{H}_2\text{O}} = P_{\text{total}} - P_{\text{O}_2}\) Using the values we know: \(P_{\text{H}_2\text{O}} = 785\,\text{torr} - 763.2\,\text{torr}\)
07

Calculate the vapor pressure of water at 25°C.

Finally, we can calculate the vapor pressure of water at 25°C: \(P_{\text{H}_2\text{O}} = 785\,\text{torr} - 763.2\,\text{torr} = 21.8\,\text{torr}\) The vapor pressure of water at 25°C is 21.8 torr.

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