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Consider a nickel wall separating hydrogen gas that is maintained on one side at \(5 \mathrm{~atm}\) and on the opposite at \(3 \mathrm{~atm}\). If the temperature is constant at \(85^{\circ} \mathrm{C}\), determine \((a)\) the mass densities of hydrogen gas in the nickel wall on both sides and \((b)\) the mass densities of hydrogen outside the nickel wall on both sides.

Short Answer

Expert verified
Question: Calculate the mass densities of hydrogen gas inside the nickel wall and outside the nickel wall on both sides, given that the temperature is 85°C, and the pressures are 5 atm and 3 atm, respectively. Answer: The mass densities of hydrogen gas inside and outside the nickel wall are 0.33208 g/L for the side with 5 atm pressure, and 0.19924 g/L for the side with 3 atm pressure.

Step by step solution

01

Convert temperature to Kelvin

The given temperature is 85°C, so we first convert this to Kelvin by adding 273.15: \(T = 85 + 273.15 = 358.15 \mathrm{K}\)
02

Calculate the number density of hydrogen gas using the ideal gas law

We can use the ideal gas law equation to find the number density (n/V) of hydrogen gas on both sides of the nickel wall. The ideal gas law equation is: \(PV = nRT\) Where: P - pressure V - volume n - number of moles R - universal gas constant T - temperature in Kelvin Rearrange the equation to find n/V: \((n/V) = P/(RT)\) On the side with 5 atm pressure: \((n/V)_1 = \frac{5 \mathrm{~atm}}{(0.0821 \mathrm{\frac{L·atm}{mol·K}})(358.15 \mathrm{K})} = 0.16604 \mathrm{\frac{mol}{L}}\) On the side with 3 atm pressure: \((n/V)_2 = \frac{3 \mathrm{~atm}}{(0.0821 \mathrm{\frac{L·atm}{mol·K}})(358.15 \mathrm{K})} = 0.09962 \mathrm{\frac{mol}{L}}\)
03

Calculate mass density of hydrogen gas inside the nickel wall

Now, we can find the mass density of hydrogen gas inside the nickel wall on both sides using the molar mass of hydrogen (M_H = 2 g/mol): On the side with 5 atm pressure: \(\rho_1 = (n/V)_1 · M_H = 0.16604 \mathrm{\frac{mol}{L}} · (2 \mathrm{\frac{g}{mol}}) = 0.33208 \mathrm{\frac{g}{L}}\) On the side with 3 atm pressure: \(\rho_2 = (n/V)_2 · M_H = 0.09962 \mathrm{\frac{mol}{L}} · (2 \mathrm{\frac{g}{mol}}) = 0.19924 \mathrm{\frac{g}{L}}\) So, the mass densities of hydrogen gas inside the nickel wall are: - Side with 5 atm pressure: 0.33208 g/L - Side with 3 atm pressure: 0.19924 g/L
04

Calculate mass density of hydrogen gas outside the nickel wall

For the mass densities of hydrogen outside the nickel wall on both sides, we will assume that the gas is ideal and follows the same ideal gas law equation. Since the conditions are the same as inside the wall, the mass densities will also be the same. So, the mass densities of hydrogen gas outside the nickel wall are: - Side with 5 atm pressure: 0.33208 g/L - Side with 3 atm pressure: 0.19924 g/L

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and amount of an ideal gas through the equation \( PV = nRT \). In this equation, \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles of gas, \( R \) for the universal gas constant, and \( T \) for temperature in Kelvin.

Understanding this concept is crucial when we aim to calculate the mass density of a gas, which is the mass per unit volume (\( \rho = m/V \)) of the gas. By rearranging the ideal gas law equation to \( (n/V) = P/(RT) \), we can obtain the number density of gas (\( n/V \) - the number of moles per unit volume), and further calculate the mass density by multiplying it with the molar mass of the gas. This law assumes that the gas particles are in constant random motion and do not interact with each other, which is a good approximation under most conditions but can deviate at very high pressures or low temperatures.

It's important for students to remember that when dealing with gases, temperature should always be in Kelvin, which can be converted from Celsius by adding 273.15 to the Celsius temperature. The ideal gas law is particularly useful for predicting how a gas will behave under different conditions of temperature and pressure.
Hydrogen Gas
Hydrogen gas (\( H_2 \) is the simplest and most abundant chemical substance in the universe, primarily found as a gas with two hydrogen atoms bonded together.

In the context of the ideal gas law and mass density calculations, it's essential to know the molar mass of hydrogen gas, which is approximately 2 grams per mole (\( 2 \frac{g}{mol} \)). This low molar mass makes hydrogen one of the lightest gases, which in turn affects its mass density under given conditions of temperature and pressure.

The behavior of hydrogen gas under different conditions can be used as a model for an ideal gas because it's a diatomic gas and, in many ways, comes close to fulfilling the assumptions of the ideal gas law, especially at standard temperature and pressure. For many calculations, hydrogen gas can be assumed to behave ideally, which simplifies computation. Additionally, knowing the properties of hydrogen gas is crucial in various applications such as fuel cells and the production of synthetic materials.
Pressure and Temperature Effects
Pressure and temperature are among the most significant factors influencing the behavior of gases, according to the ideal gas law. As we've seen from the equation \( PV = nRT \), changing either pressure or temperature will affect the volume and mass density of a gas.

For instance, increasing the pressure of a gas, while keeping the temperature constant, will often decrease its volume and increase the mass density, since more gas particles will be packed into the same space. Conversely, increasing the temperature will typically result in an increase in volume and a decrease in mass density, assuming pressure is kept constant. This is due to the gas particles moving more vigorously and requiring more space.

In the textbook example of hydrogen gas separated by a nickel wall at different pressures, we clearly see the effect of pressure on the gas's mass density. Higher pressure on one side results in a greater mass density when compared to the side with lower pressure. Notably, because temperature remains consistent in this scenario, it demonstrates the role pressure plays independently of temperature effects.

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Most popular questions from this chapter

In a manufacturing facility, \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\) wet brass plates coming out of a water bath are to be dried by passing them through a section where dry air at 1 atm and \(25^{\circ} \mathrm{C}\) is blown parallel to their surfaces at \(4 \mathrm{~m} / \mathrm{s}\). If the plates are at \(15^{\circ} \mathrm{C}\) and there are no dry spots, determine the rate of evaporation from both sides of a plate.

Express the mass flow rate of water vapor through a wall of thickness \(L\) in terms of the partial pressure of water vapor on both sides of the wall and the permeability of the wall to the water vapor.

Consider a 15-cm-internal-diameter, 10-m-long circular duct whose interior surface is wet. The duct is to be dried by forcing dry air at \(1 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\) through it at an average velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through a chilled room, and it remains at an average temperature of \(15^{\circ} \mathrm{C}\) at all times. Determine the mass transfer coefficient in the duct.

A 2-m-deep 20-m \(\times 20-\mathrm{m}\) heated swimming pool is maintained at a constant temperature of \(30^{\circ} \mathrm{C}\) at a location where the atmospheric pressure is \(1 \mathrm{~atm}\). If the ambient air is at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity and the effective sky temperature is \(0^{\circ} \mathrm{C}\), determine the rate of heat loss from the top surface of the pool by ( \(a\) ) radiation, \((b)\) natural convection, and (c) evaporation. ( \(d\) ) Assuming the heat losses to the ground to be negligible, determine the size of the heater.

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.

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