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Chapter 5: Problem 46

A sample of oxygen gas is collected over water at \(25^{\circ} \mathrm{C}\) (vp \(\mathrm{H}_{2} \mathrm{O}(l)=23.8 \mathrm{~mm} \mathrm{Hg}\) ). The wet gas occupies a volume of \(7.28 \mathrm{~L}\) at a total pressure of \(1.25\) bar. If all the water is removed, what volume will the dry oxygen occupy at a pressure of \(1.07 \mathrm{~atm}\) and a temperature of \(37^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Answer: The volume of the dry oxygen gas will be 7.64 liters at a pressure of 1.07 atm and a temperature of 37°C.

Step by step solution

01

Convert all units to the proper format

We need to make sure that all pressures, volumes, and temperatures are in the correct units. For this problem, let's use atm for pressure, L for volume, and K for temperature. Convert 1.25 bar to atm: 1 bar = 0.9869 atm, so 1.25 bar = 1.25 * 0.9869 = 1.2336 atm. Next, we'll need to convert the temperature to kelvin: T1 = 25°C + 273.15 = 298.15 K, and T2 = 37°C + 273.15 = 310.15 K.
02

Find the partial pressure of oxygen gas

Since the total pressure of the mixture is given by the sum of the pressures of both oxygen and water vapor, we can find the partial pressure of the oxygen gas as follows: Total pressure = partial pressure of oxygen gas + partial pressure of water vapor 1.2336 atm = partial pressure of oxygen gas + 23.8 mmHg First, we need to convert the water vapor pressure to atm: 23.8 mmHg * (1 atm/760 mmHg) = 0.0313 atm. Now, we can find the partial pressure of oxygen gas: partial pressure of oxygen gas = 1.2336 atm - 0.0313 atm = 1.2023 atm.
03

Use the Ideal Gas Law to find the number of moles of oxygen gas

Now that we have the partial pressure of oxygen gas, we can use the Ideal Gas Law to find the number of moles of oxygen gas present in the wet gas. PV = nRT We know that P = 1.2023 atm, V = 7.28 L, R = 0.0821 (L * atm)/(K * mol), and T1 = 298.15 K. 1.2023 atm * 7.28 L = n * 0.0821 (L * atm)/(K * mol) * 298.15 K Solve for n: n = 0.2997 moles of O2 gas.
04

Use the combined gas law to find the volume of the dry oxygen gas

Now that we know the number of moles of oxygen gas, we can use the combined gas law to determine the volume of the dry oxygen gas at the given temperature and pressure: P1V1/T1 = P2V2/T2 We know that P1 = 1.2023 atm, V1 = 7.28 L, T1 = 298.15 K, P2 = 1.07 atm, and T2 = 310.15 K. (1.2023 atm * 7.28 L) / 298.15 K = (1.07 atm * V2) / 310.15 K Solve for V2: V2 = 7.64 L. So, the volume of the dry oxygen gas will be 7.64 L at a pressure of 1.07 atm and a temperature of 37°C.

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