Question

Calculate the densities (in \(\mathrm{g} / \mathrm{L}\) ) of the following gases at \(78^{\circ} \mathrm{F}\) and \(13.6 \mathrm{psi}\) (a) xenon (b) methane (c) acetylene \(\mathrm{C}_{2} \mathrm{H}_{2}\)

Short answer

Based on the given conditions (78°F and 13.6 psi), calculate the densities of the following gases: (a) Xenon (Xe) (b) Methane (CH4) (c) Acetylene (C2H2) Answer: (a) Xenon: ≈ 4.931 g/L (b) Methane: ≈ 0.611 g/L (c) Acetylene: ≈ 1.042 g/L

Step-by-step solution

Convert temperature from Fahrenheit to Kelvin

First, convert the given temperature (78°F) to Celsius using the formula: $$ C = (F - 32) \times \frac{5}{9} $$ Plugging in the given Fahrenheit temperature: $$ C = (78 - 32) \times \frac{5}{9} \approx 25.56^{\circ}\mathrm{C} $$ Now, convert the Celsius temperature to Kelvin: $$ K = C + 273.15 $$ $$ K = 25.56 + 273.15 \approx 298.71\, \mathrm{K} $$

Convert pressure from psi to atm

Convert the given pressure (13.6 psi) to atmospheres (atm) using the conversion factor 14.696 psi = 1 atm: $$ P_{\text{atm}} = P_{\text{psi}} \times \frac{1\, \text{atm}}{14.696\, \text{psi}} $$ $$ P_{\text{atm}} = 13.6\, \text{psi} \times \frac{1\, \text{atm}}{14.696\, \text{psi}} \approx 0.926\, \mathrm{atm} $$

Calculate the densities using the Ideal Gas Law

The Ideal Gas Law is given by: $$ PV = nRT $$ Rearrange the equation to solve for density (\(\rho = \frac{n}{V}M\), where \(M\) is molar mass) we get: $$ \rho = \frac{PM}{RT} $$ Now, input the values of \(P\), \(R\), and \(T\) just calculated, and the molar mass (\(M\)) of the gases: (a) Xenon (Xe) has a molar mass of \(131.29\, \mathrm{g/mol}\) : $$ \rho_{\text{Xe}} = \frac{(0.926\, \mathrm{atm})(131.29\, \mathrm{g/mol})}{(0.0821\, \mathrm{L\cdot atm/(mol\cdot K)})(298.71\, \mathrm{K})} $$ $$ \rho_{\text{Xe}} \approx 4.931\, \mathrm{g/L} $$ (b) Methane (CH4) has a molar mass of \(16.04\, \mathrm{g/mol}\) : $$ \rho_{\text{CH4}} = \frac{(0.926\, \mathrm{atm})(16.04\, \mathrm{g/mol})}{(0.0821\, \mathrm{L\cdot atm/(mol\cdot K)})(298.71\, \mathrm{K})} $$ $$ \rho_{\text{CH4}} \approx 0.611\, \mathrm{g/L} $$ (c) Acetylene (\(\mathrm{C_2H_2}\)) has a molar mass of \(26.04\, \mathrm{g/mol}\) : $$ \rho_{\mathrm{C_2H_2}} = \frac{(0.926\, \mathrm{atm})(26.04\, \mathrm{g/mol})}{(0.0821\, \mathrm{L\cdot atm\atop (mol\cdot K)})(298.71\, \mathrm{K})} $$ $$ \rho_{\mathrm{C_2H_2}} \approx 1.042\, \mathrm{g/L} $$ Thus, the densities of the gases at \(78^{\circ} \mathrm{F}\) and \(13.6 \mathrm{psi}\) are: (a) Xenon: \(\approx 4.931\, \mathrm{g/L}\) (b) Methane: \(\approx 0.611\, \mathrm{g/L}\) (c) Acetylene: \(\approx 1.042\, \mathrm{g/L}\)