Problem 79 - University Physics with Modern Physics

Question

At a temperature of \(295 . \mathrm{K},\) the vapor pressure of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) is \(60.7 \mathrm{kPa}\). Suppose \(1.000 \mathrm{~g}\) of gaseous pentane is contained in a cylinder with diathermal (thermally conducting) walls and a piston to vary the volume. The initial volume is \(1.000 \mathrm{~L},\) and the piston is moved in slowly, keeping the temperature at \(295 \mathrm{~K}\). At what volume will the first drop of liquid pentane appear?

Step-by-Step Solution

Answer

Answer: The volume at which the first drop of liquid pentane appears is 0.556 L.

Step 1: Calculate the number of moles of pentane

To find the number of moles of pentane, we can use the given mass and the molecular weight of pentane. The molecular weight of pentane (C5H12) is (5 * 12.01) + (12 * 1.01) = 72.15 grams/mole. For 1.000 grams of pentane, we can find the number of moles using the formula: n = (mass) / (molecular weight) n = (1.000g) / (72.15 grams/mole) = 0.01386 moles of pentane.

Step 2: Use the ideal gas law to calculate the initial pressure

The ideal gas law is given by: PV = nRT Where P is the pressure (in kPa), V is the volume (in L), n is the number of moles, R is the ideal gas constant, and T is the temperature (in K). We know n, R, T, and the initial volume, so we can solve for the initial pressure, P: P = (nRT) / V Using R = 8.314 (L * kPa) / (K * mol): P = (0.01386 moles * 8.314 L * kPa / (K * mol) * 295 K) / 1.000 L = 34.075 kPa

Step 3: Determine the pressure at which the gas liquefies

We are given that the vapor pressure of pentane at the given temperature is 60.7 kPa. This is the point at which the gas pressure equals the vapor pressure and liquefaction begins. To maintain the temperature constant, we need to increase the pressure until it reaches 60.7 kPa.

Step 4: Calculate the volume at which the pressure reaches the vapor pressure

Using the ideal gas law again, we can solve for the volume at which the pressure reaches 60.7 kPa: PV = nRT V = (nRT) / P V = (0.01386 moles * 8.314 L * kPa / (K * mol) * 295 K) / 60.7 kPa = 0.556 L So, the volume at which the first drop of liquid pentane appears is 0.556 L.

Key Concepts

Understanding Vapor Pressure

Vapor pressure is an essential concept in thermodynamics that describes the pressure exerted by a vapor in equilibrium with its liquid or solid phase. For any given substance, this pressure is influenced by temperature, as higher temperatures typically increase vapor pressure.

• Vapor pressure is unique to each substance and varies with temperature changes.
• When the pressure of a gas in a closed system equals the vapor pressure, the substance begins to change from gas to liquid or vice versa.

In the context of our exercise, pentane has a vapor pressure of 60.7 kPa at 295 K. This means that if the internal gas pressure in the system reaches this vapor pressure, pentane will start to condense into liquid. This principle is crucial for determining when condensation occurs in closed systems where temperature is kept constant, as described in the exercise.

Applying the Ideal Gas Law

The ideal gas law is a key principle in understanding the behavior of gases under various conditions. It is represented by the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (8.314 L * kPa / (K * mol)), and
• \(T\) is the temperature in Kelvin.

To determine changes in gas behavior, such as volume or pressure at constant temperature and moles, the ideal gas law provides a straightforward way to calculate these values. In the exercise, we use this law twice: first to find the initial pressure within the diathermal cylinder and then to determine the volume when the gas pressure equals the vapor pressure, signaling the onset of liquefaction.

Exploring Molecular Weight

Molecular weight, also known as molar mass, is the sum of the atomic masses of all the atoms in a molecule. In our case, it's relevant for pentane \(\mathrm{C}_5\mathrm{H}_{12}\).

• The molecular weight of pentane can be calculated by summing 5 carbons and 12 hydrogen atoms.
• Carbon has an atomic mass of 12.01 and hydrogen 1.01.
• Thus, the molecular weight is \((5 \times 12.01) + (12 \times 1.01) = 72.15 \text{ grams/mole}\).

Knowing the molecular weight allows us to calculate the number of moles from a given mass, which is essential for using the ideal gas law. For instance, in the exercise, 1.000 g of pentane corresponds to roughly 0.01386 moles, as determined by division of the mass by the molecular weight.

Understanding Liquefaction

Liquefaction is the process by which a gas transforms into a liquid. This occurs when the pressure on the gas is increased beyond its vapor pressure, or when the temperature drops, provided that this pressure is met.

• As the pressure of a gas increases, the forces between particles become strong enough to change the gas into a liquid.
• Vapor pressure must be achieved for liquefaction to begin.

In the exercise, liquefaction occurs when pentane's pressure within the cylinder matches the vapor pressure, which is 60.7 kPa at 295 K. At this point, gas particles start to collect as liquid within the system. Understanding liquefaction is crucial for various practical applications such as refrigeration, where controlling phase change is fundamental.