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Show that for an ideal gas mixture maintained at a constant temperature and pressure, the molar concentration \(C\) of the mixture remains constant, but this is not necessarily the case for the density \(\rho\) of the mixture.

Short Answer

Expert verified
Answer: The molar concentration of an ideal gas mixture remains constant at constant temperature and pressure because it only depends on these two variables (C=P/(RT)). On the other hand, the density of the mixture may change due to variations in the molar masses and amounts of the individual components within the mixture (ρ = (Σ (Mi * ni))/V).

Step by step solution

01

Recall the Ideal Gas Law

The ideal gas law is given by the equation: PV = nRT where P represents pressure, V represents volume, n represents the amount in moles of the gas, R is the ideal gas constant (about 8.314 J/mol·K), and T is the temperature in Kelvin.
02

Define Molar Concentration and Density

Molar concentration (C) is defined as the number of moles of a substance in a given volume (V). Mathematically, it can be expressed as: C = n/V Density (ρ) is defined as the mass (m) of a substance in a given volume (V). Mathematically, it can be expressed as: ρ = m/V
03

Relate Molar Concentration and Density to Ideal Gas Law

Now we will use the ideal gas law to relate the molar concentration and density of the mixture by rewriting the law in terms of these quantities. First, we can solve the ideal gas law for the number of moles (n) in the gas mixture: n = PV/(RT) Now we substitute this expression into the molar concentration (C) and density (ρ) equations: C = (PV)/(RTV) = P/(RT) ρ = (m/V) However, the mixture is composed of various components, and we can express the mass (m) as the sum of the product of their individual molar masses (Mi) and their respective number of moles (ni): m = Σ (Mi * ni) So, the density equation becomes: ρ = (Σ (Mi * ni))/V
04

Show Molar Concentration Remains Constant

Since we are maintaining constant temperature and pressure for the gas mixture, the molar concentration (C) remains constant: C = P/(RT), where P and T are constant.
05

Show Density May Not Remain Constant

Unlike molar concentration, the density of the gas mixture can be affected by changes in the molar masses (Mi) and the number of moles (ni) of the individual components. So, even if the pressure and temperature remain constant, the density (ρ) may change as the composition of the gas mixture changes: ρ = (Σ (Mi * ni))/V In conclusion, the molar concentration (C) of an ideal gas mixture remains constant at constant pressure and temperature, as it only depends on these two variables. In contrast, the density (ρ) of the mixture may change, as it depends on the molar masses and amounts of the individual components within the mixture.

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Most popular questions from this chapter

What is diffusion velocity? How does it affect the mass-average velocity? Can the velocity of a species in a moving medium relative to a fixed reference point be zero in a moving medium? Explain.

Consider a glass of water in a room at \(20^{\circ} \mathrm{C}\) and $100 \mathrm{kPa}$. If the relative humidity in the room is 70 percent and the water and the air are at the same temperature, determine \((a)\) the mole fraction of the water vapor in the room air, (b) the mole fraction of the water vapor in the air adjacent to the water surface, and \((c)\) the mole fraction of air in the water near the surface. Answers: (a) \(1.64\) percent, (b) \(2.34\) percent, (c) \(0.0015\) percent

Consider one-dimensional mass transfer in a moving medium that consists of species \(A\) and \(B\) with \(\rho=\) \(\rho_{A}+\rho_{B}=\) constant. Mark these statements as being True or False. (a) The rates of mass diffusion of species \(A\) and \(B\) are equal in magnitude and opposite in direction. (b) \(D_{A B}=D_{B A^{-}}\) (c) During equimolar counterdiffusion through a tube, equal numbers of moles of \(A\) and \(B\) move in opposite directions, and thus a velocity measurement device placed in the tube will read zero. (d) The lid of a tank containing propane gas (which is heavier than air) is left open. If the surrounding air and the propane in the tank are at the same temperature and pressure, no propane will escape the tank, and no air will enter.

Air at \(52^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), and 20 percent relative humidity enters a \(5-\mathrm{cm}\)-diameter tube with an average velocity of $6 \mathrm{~m} / \mathrm{s}$. The tube inner surface is wetted uniformly with water, whose vapor pressure at \(52^{\circ} \mathrm{C}\) is \(13.6 \mathrm{kPa}\). While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to \(10 \mathrm{kPa}\). Detemine \((a)\) the average mass transfer coefficient in $\mathrm{m} / \mathrm{s}\(, \)(b)$ the log-mean driving force for mass transfer in molar concentration units, \((c)\) the water evaporation rate in $\mathrm{kg} / \mathrm{h}\(, and \)(d)$ the length of the tube.

Saturated water vapor at $25^{\circ} \mathrm{C}\left(P_{\text {stt }}=3.17 \mathrm{kPa}\right)\( flows in a pipe that passes through air at \)25^{\circ} \mathrm{C}$ with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a \(9-\mathrm{mm}\) internal-diameter tube that extends $10 \mathrm{~m}$ into the air. The diffusion coefficient of vapor through air is \(2.5 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The amount of water vapor lost to the atmosphere through this individual tube by diffusion is (a) \(1.7 \times 10^{-6} \mathrm{~kg}\) (b) \(2.3 \times 10^{-6} \mathrm{~kg}\) (c) \(3.8 \times 10^{-6} \mathrm{~kg}\) (d) \(5.0 \times 10^{-6} \mathrm{~kg}\) (e) \(7.1 \times 10^{-6} \mathrm{~kg}\)

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