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Chapter 19: Problem 79

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?

Short Answer

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

Step by step solution

01

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.
02

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
03

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.
04

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.

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