Question

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(5.00 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas is passed through \(7.2146 \mathrm{~g}\) of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), at \(26.0{ }^{\circ} \mathrm{C}\). The liquid remaining after the experiment weighs \(5.1493 \mathrm{~g}\). Assuming that the gas becomes saturated with benzene vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of the benzene in torr?

Short answer

The vapor pressure of benzene in this experiment is approximately \(98.91\mathrm{~torr}\).

Step-by-step solution

Calculate the mass of evaporated benzene

First, we need to find the mass of benzene that evaporated during the experiment. This can be found by taking the difference between the initial mass and the final mass. $$\text{Mass of evaporated benzene} = \text{Initial mass} - \text{Final mass}$$ $$\text{Mass of evaporated benzene} = 7.2146\mathrm{~g} - 5.1493\mathrm{~g} = 2.0653\mathrm{~g}$$

Convert the mass of evaporated benzene into moles

To calculate the moles of evaporated benzene, we will use the molar mass of benzene which is approximately \(78.11\mathrm{~g/mol}\). $$\text{Moles of evaporated benzene} = \frac{\text{Mass of evaporated benzene}}{\text{Molar mass of benzene}}$$ $$\text{Moles of evaporated benzene} = \frac{2.0653\mathrm{~g}}{78.11\mathrm{~g/mol}} = 0.02645\mathrm{~mol}$$

Convert the given temperature to Kelvin

Before we can use the Ideal Gas Law, we need to convert the given temperature from Celsius to Kelvin. $$T(K) = T(^{\circ}\mathrm{C}) + 273.15$$ $$T(K) = 26.0{ }^{\circ} \mathrm{C} + 273.15 = 299.15\mathrm{~K}$$

Apply the Ideal Gas Law to find the partial pressure of benzene

Now that we have the moles of evaporated benzene and the temperature in Kelvin, we can use the Ideal Gas Law to determine the vapor pressure (partial pressure) of benzene in the gaseous mixture. The Ideal Gas Law is as follows: $$PV = nRT$$ We'll rearrange the equation to find the partial pressure of benzene, where \(P\) is the partial pressure, \(n\) is the moles of evaporated benzene, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume. $$P = \frac{nRT}{V}$$ For our calculation, we will use the gas constant \(R = 62.363\mathrm{~L\cdot torr/mol\cdot K}\). The given volume of the gas mixture is \(5.00\mathrm{~L}\). $$P_\mathrm{benzene} = \frac{(0.02645\mathrm{~mol})(62.363\mathrm{~L\cdot torr/mol\cdot K})(299.15\mathrm{~K})}{5.00\mathrm{~L}} = 98.91\mathrm{~torr}$$ Therefore, the vapor pressure of benzene in this experiment is approximately \(98.91\mathrm{~torr}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a useful equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The law is commonly written as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. This law assumes that gases behave ideally, meaning that interactions between gas molecules are insignificant and that particles occupy no volume. This makes the Ideal Gas Law particularly useful in calculating unknown properties of gases when three of the four variables are known.

In the context of the given exercise, the Ideal Gas Law is used to determine the vapor pressure of benzene, which evaporates into a gas under specific conditions. By knowing the volume of gaseous benzene present, the amounts of moles that have evaporated, and the temperature in Kelvin, we can directly find the pressure exerted by the benzene vapor in the gas mixture. This approach relies on the assumption that, under the experimental conditions, benzene behaves as an ideal gas.

Benzene Evaporation

Benzene, with the chemical formula \(C_6H_6\), is a liquid that easily evaporates at room temperature due to its high volatility. This property is crucial when measuring its vapor pressure, which is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. In experiments involving benzene, it's important to determine how much of the liquid has turned into vapor. This can be calculated by measuring the mass of the liquid before and after exposure to a gas that causes the benzene to evaporate.

In this experiment, a known mass of benzene is weighed before and after it is exposed to \(N_2\) gas, which saturates with benzene vapor. The difference in mass indicates the amount of benzene that has evaporated. Knowing the mass that has turned into vapor allows us to calculate the moles of benzene using its molar mass. These values are crucial steps in finding the vapor pressure when using the Ideal Gas Law.

Gas Saturation

Gas saturation refers to the state where a gas holds the maximum amount of a vapor that it can contain under specified conditions of temperature and pressure. In the exercise, \(N_2\) gas is bubbled through liquid benzene until it becomes saturated with benzene vapor. This means that the gas cannot hold more benzene vapor without changing the temperature or pressure conditions.

The concept of gas saturation is important because it is assumed during the experiment. When the gas is saturated, the vapor pressure at equilibrium can be calculated. It ensures that any pressure measured or calculated is reflective of the benzene gas content only, not influenced by any further evaporative potential from the liquid. This assumption allows us to use the Ideal Gas Law to calculate the partial pressure of benzene once the gas reaches saturation.

Temperature Conversion

Temperature conversion is essential when working with gases and utilizing the Ideal Gas Law, as the temperature must be expressed in Kelvin. Kelvin is the SI unit for temperature and is necessary for accurate calculations in thermodynamics because it begins at absolute zero, the lowest possible temperature where particles have minimal vibrational motion.

To convert the temperature from degrees Celsius to Kelvin, add 273.15 to the Celsius temperature. In the exercise, the conversion of the given temperature at \(26.0^{\circ}C\) to Kelvin is straightforward:

• \(T(K) = T(^{\circ}C) + 273.15\)
• \(T(K) = 26.0 + 273.15 = 299.15 K\)

This converted temperature is then used in the Ideal Gas Law to find the vapor pressure of benzene. Accurate temperature conversion is crucial as errors can lead to incorrect pressure calculations, influencing the outcome of the experiment.