Question

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

Answer: The mass densities of hydrogen gas inside the nickel wall are 0.341 g/L on the 5 atm side and 0.205 g/L on the 3 atm side. The mass densities of hydrogen gas outside the nickel wall are also 0.341 g/L on the 5 atm side and 0.205 g/L on the 3 atm side.

Step-by-step solution

Write down the Ideal Gas Law

The Ideal Gas Law is given by the formula: \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvins.

Convert temperature to Kelvin

The temperature of the gas is given as \(85^{\circ} \mathrm{C}\). We need to convert this to Kelvin by adding 273.15. \(T = 85 + 273.15 = 358.15 \mathrm{K}\).

Use the Ideal Gas Law to find the number of moles

We will use the Ideal Gas Law to find the number of moles of hydrogen gas on both sides of the wall. First, we have to rearrange the equation to solve for n: \(n=\frac{PV}{RT}\). Since we don't have the volume, we will work with the molar volume, \(V_m=\frac{V}{n}\). Rearranging the Ideal Gas Law, we get: \(PV_m=RT\).

Calculate the molar volume for both sides

Using the rearranged Ideal Gas Law, we will find the molar volume for both sides. For the 5 atm side: \(V_{m1} = \frac{RT}{P_1} = \frac{(0.0821\ \mathrm{L\cdot atm/mol\cdot K})(358.15\ \mathrm{K})}{5\ \mathrm{atm}} = 5.904\ \mathrm{L/mol}\). For the 3 atm side: \(V_{m2} = \frac{RT}{P_2} = \frac{(0.0821\ \mathrm{L\cdot atm/mol\cdot K})(358.15\ \mathrm{K})}{3\ \mathrm{atm}} = 9.840\ \mathrm{L/mol}\).

Find the mass densities inside the nickel wall

Since hydrogen is H\(_2\), the molar mass of hydrogen gas is 2.016 g/mol. Now we can find the mass densities for both sides by using the formula: \(\rho = \frac{m}{V} = \frac{n\cdot M}{V}\). Remember that we are working with the molar volume, so we will have \({\rho = \frac{M}{V_m}}\). For the 5 atm side: \(\rho_1 = \frac{2.016\ \mathrm{g/mol}}{5.904\ \mathrm{L/mol}} = 0.341\ \mathrm{g/L}\). For the 3 atm side: \(\rho_2 = \frac{2.016\ \mathrm{g/mol}}{9.840\ \mathrm{L/mol}} = 0.205\ \mathrm{g/L}\). (b) Calculate the mass densities of hydrogen gas outside the nickel wall on both sides.

Find the number of moles outside

To find the mass densities outside the nickel wall, we must assume the wall itself does not change the molar volume. The gas outside the wall will also follow the Ideal Gas Law, so \(n\) and \(V_m\) can be used for the gas outside as well.

Find the mass densities outside the nickel wall

Using the same formula as in Step 5, we can find the mass densities outside the nickel wall using the same calculated molar volume values. For the 5 atm side: \(\rho_{out1} = \frac{2.016\ \mathrm{g/mol}}{5.904\ \mathrm{L/mol}} = 0.341\ \mathrm{g/L}\). For the 3 atm side: \(\rho_{out2} = \frac{2.016\ \mathrm{g/mol}}{9.840\ \mathrm{L/mol}} = 0.205\ \mathrm{g/L}\). As a result, the mass densities of hydrogen gas inside the nickel wall are 0.341 g/L on the 5 atm side and 0.205 g/L on the 3 atm side. The mass densities of hydrogen gas outside the nickel wall are also 0.341 g/L on the 5 atm side and 0.205 g/L on the 3 atm side.