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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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Most popular questions from this chapter

Chapter 13 examined the variation of pressure with altitude in the Earth's atmosphere, assuming constant temperature-a model known as the isothermal atmosphere. A better approximation is to treat the pressure variations with altitude as adiabatic. Assume that air can be treated as a diatomic ideal gas with effective molar mass \(M_{\text {air }}=28.97 \mathrm{~g} / \mathrm{mol}\) a) Find the air pressure and temperature of the atmosphere as functions of altitude. Let the pressure at sea level be \(p_{0}=101.0 \mathrm{kPa}\) and the temperature at sea level be \(20.0^{\circ} \mathrm{C}\) b) Determine the altitude at which the air pressure and density are half their sea-level values. What is the temperature at this altitude, in this model? c) Compare these results with the isothermal model of Chapter \(13 .\)

Show that the adiabatic bulk modulus, defined as \(B=-V(d P / d V),\) for an ideal gas is equal to \(\gamma P\).

A monatomic ideal gas expands isothermally from \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) to \(\left\\{p_{2}, V_{2}, T_{1}\right\\} .\) Then it undergoes an isochoric process, which takes it from \(\left\\{p_{2}, V_{2}, T_{1}\right\\}\) to \(\left\\{p_{1}, V_{2}, T_{2}\right\\}\) Finally the gas undergoes an isobaric compression, which takes it back to \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) a) Use the First Law of Thermodynamics to find \(Q\) for each of these processes. b) Write an expression for total \(Q\) in terms of \(p_{1}, p_{2}, V_{1},\) and \(V_{2}\).

In a diesel engine, the fuel-air mixture is compressed rapidly. As a result, the temperature rises to the spontaneous combustion temperature for the fuel, causing the fuel to ignite. How is the temperature rise possible, given the fact that the compression occurs so rapidly that there is not enough time for a significant amount of heat to flow into or out of the fuel-air mixture?

What is the total mass of all the oxygen molecules in a cubic meter of air at normal temperature \(\left(25^{\circ} \mathrm{C}\right)\) and pressure \(\left(1.01 \cdot 10^{5} \mathrm{~Pa}\right) ?\) Note that air is about \(21 \%\) (by volume) oxygen (molecular \(\mathrm{O}_{2}\) ), with the remainder being primarily nitrogen (molecular \(\mathrm{N}_{2}\) ).
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