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Chapter 14: Problem 181

A rubber object is in contact with nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(298 \mathrm{~K}\) and \(250 \mathrm{kPa}\). The solubility of nitrogen gas in rubber is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3}\).bar. The mass density of nitrogen at the interface is (a) \(0.049 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(0.064 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(0.077 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(0.092 \mathrm{~kg} / \mathrm{m}^{3}\) (e) \(0.109 \mathrm{~kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
Answer: (e) 0.109 kg/m³

Step by step solution

01

Identify the given information

We are given the following information: - Temperature, T = 298 K - Pressure, P = 250 kPa - Solubility of nitrogen in rubber = 0.00156 kmol/m³.bar - Nitrogen gas molar mass, M = 28.02 g/mol (approximately)
02

Convert pressure to bars

The pressure is given in kPa, but solubility is given in kmol/m³.bar. We need to convert the pressure to bars. 1 bar = 100 kPa So, 250 kPa = 250/100 = 2.5 bars
03

Calculate the solubility at the given pressure

We need to find the solubility of nitrogen in the rubber at the given pressure. Use the given solubility formula and the pressure in bars: Solubility = (0.00156 kmol/m³.bar) * (2.5 bars) = 0.0039 kmol/m³
04

Convert solubility to mass density

Now, convert the solubility from kmol/m³ to kg/m³ using the molar mass of nitrogen gas: Mass density = (0.0039 kmol/m³) * (28.02 g/mol) * (1 kg/1000 g) = 0.109 kg/m³.
05

Match the result to the options

Compare the calculated mass density to the given options: (a) 0.049 kg/m³ (b) 0.064 kg/m³ (c) 0.077 kg/m³ (d) 0.092 kg/m³ (e) 0.109 kg/m³ The correct option is (e) 0.109 kg/m³.

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Most popular questions from this chapter

Determine the mole fraction of dry air at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\). Take the atmospheric pressure at lake level to be \(100 \mathrm{kPa}\).

Nitrogen gas at high pressure and \(298 \mathrm{~K}\) is contained in a \(2-\mathrm{m} \times 2-\mathrm{m} \times 2-\mathrm{m}\) cubical container made of natural rubber whose walls are \(4 \mathrm{~cm}\) thick. The concentration of nitrogen in the rubber at the inner and outer surfaces are \(0.067 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.009 \mathrm{~kg} / \mathrm{m}^{3}\), respectively. The diffusion coefficient of nitrogen through rubber is \(1.5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of nitrogen by diffusion through the cubical container is (a) \(8.24 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (b) \(1.35 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (c) \(5.22 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(9.71 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (e) \(3.58 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Consider a tank that contains moist air at \(3 \mathrm{~atm}\) and whose walls are permeable to water vapor. The surrounding air at \(1 \mathrm{~atm}\) pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain.

Consider a carbonated drink in a bottle at \(37^{\circ} \mathrm{C}\) and \(130 \mathrm{kPa}\). Assuming the gas space above the liquid consists of a saturated mixture of \(\mathrm{CO}_{2}\) and water vapor and treating the drink as water, determine \((a)\) the mole fraction of the water vapor in the \(\mathrm{CO}_{2}\) gas and \((b)\) the mass of dissolved \(\mathrm{CO}_{2}\) in a 200-ml drink.

Air at \(40^{\circ} \mathrm{C}\) and 1 atm flows over a \(5-\mathrm{m}\)-long wet plate with an average velocity of \(2.5 \mathrm{~m} / \mathrm{s}\) in order to dry the surface. Using the analogy between heat and mass transfer, determine the mass transfer coefficient on the plate.
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