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At a temperature of \(295 .\) 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?

Short Answer

Expert verified
Answer: 0.562 L

Step by step solution

01

Calculate the number of moles of pentane

To find the number of moles of pentane, we'll use the relation: Number of moles = (mass of the substance) / (molar mass of the substance) The molar mass of pentane (C5H12) is 5 * 12.01 (mass of carbon) + 12 * 1.01 (mass of hydrogen) = 72.15 g/mol. Number of moles = (1.000 g) / (72.15 g/mol) = 0.01386 mol
02

Use the ideal gas law to find the initial pressure

The ideal gas law is given by: PV = nRT where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given the initial volume (V1 = 1.000 L) and the temperature (T = 295 K). The ideal gas constant, R, in units of L*kPa/mol*K is 8.314/1000 = 0.008314 L*kPa/mol*K. We can now find the initial pressure (P1) as follows: P1 * V1 = n * R * T P1 = (n * R * T) / V1 P1 = (0.01386 mol * 0.008314 L*kPa/mol*K * 295 K) / 1.000 L P1 = 34.14 kPa
03

Find the volume when the pressure is equal to the vapor pressure

At the volume when the first liquid drop appears, the pressure inside the cylinder will be equal to the vapor pressure of pentane. As the temperature is constant, we can use Boyle's Law, which states that P1 * V1 = P2 * V2. We are given the vapor pressure (P2 = 60.7 kPa) and we calculated the initial pressure (P1 = 34.14 kPa) and initial volume (V1 = 1.000 L). Let's find the final volume (V2): V2 = P1 * V1 / P2 V2 = (34.14 kPa * 1.000 L) / 60.7 kPa V2 = 0.562 L
04

Conclusion

At a volume of 0.562 L, the pressure inside the cylinder will be equal to the vapor pressure of pentane, and the first drop of liquid pentane will appear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
When analyzing problems involving gases, it's essential to familiarize ourselves with the ideal gas law, which is represented by the equation \( PV = nRT \). This equation helps us connect the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas.

Why do we consider gases to be 'ideal'? In an ideal gas, the particles are assumed to be point particles that do not attract or repel each other, and the only space they occupy is their individual volumes. This model allows for a simplified mathematical treatment of gases that can often accurately describe their behavior under standard conditions.

Let's apply this to the exercise at hand. We calculated the amount of pentane in moles and then used the ideal gas law to find the initial pressure of the gas under given conditions. This method is a cornerstone in solving for unknown variables in gas-related problems and is incredibly useful in predicting how a gas will behave when subjected to changes in temperature, pressure, or volume.
Boyle's Law
Boyle's Law is a special case of the ideal gas law that applies when the temperature (T) and the amount (n) of gas are both constant. The law is presented as \( P1V1 = P2V2 \), indicating that the pressure of a gas is inversely proportional to its volume.

The law takes its name from the Anglo-Irish chemist Robert Boyle, who published the original law in 1662. Boyle’s observations hold a critical place in the history of chemistry because they were among the first to define how elements behave under different physical conditions.

In our exercise, we use Boyle's Law to predict the change in volume of pentane gas as its pressure changes. Knowing that the pressure increases to the vapor pressure of pentane at the point of phase transition, we can calculate the volume at which the gas begins to condense into liquid. This enables us to understand and anticipate the effects of manipulating pressure on the volume of a gas at a constant temperature.
Phase Transition
Phase transition refers to the transformation between different states of matter, such as going from a gas to a liquid or from a solid to a gas. These transitions occur when energy in the form of heat is either absorbed or released by a substance, leading to changes in temperature and pressure conditions.

At the vapor pressure, which is the pressure exerted by a vapor when it's in equilibrium with its liquid form at a given temperature, the phase transition from gas to liquid starts to occur. This concept is crucial in the given exercise, where we determine the volume at which gaseous pentane starts to condense when subjected to its vapor pressure.

Understanding phase transitions is significant for explaining phenomena such as boiling, melting, and condensation. These transitions are not just academic - they are vital in various applications, including industrial processes, meteorology, and even daily cooking.

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Most popular questions from this chapter

Which of the following gases has the highest root-mean-square speed? a) nitrogen at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) b) argon at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) c) argon at 2 atm and \(30^{\circ} \mathrm{C}\) d) oxygen at 2 atm and \(30^{\circ} \mathrm{C}\) e) nitrogen at 2 atm and \(15^{\circ} \mathrm{C}\)

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Suppose \(6 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of 3.00 atm (called state 1 ), undergoes the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?

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