91Ó°ÊÓ

Convert the given units into appropriate SI units. - Volume: 1 \, \text{ft}^3 = 0.0283168 \, \text{m}^3 \implies 25 \, \text{ft}^3 = 0.708 \, \text{m}^3.- Temperature: \(250 \, ^\circ F = 250 \times \frac{5}{9} + 273.15 = 394.261 \, K\).- Pressure: 1 \, \text{lbf/in}^2 = 6895 \, \text{Pa} \implies 440 \, \text{lbf/in}^2 = 3033800 \, \text{Pa}.

Using the ideal gas law: \( PV = nRT \). Rearrange to find \( n \), the number of moles: \[ n = \frac{PV}{RT} \]Where:- \( P = 3033800 \, \text{Pa} \)- \( V = 0.708 \, \text{m}^3 \)- \( R = 8.314 \, \text{J/mol} \cdot K \) (universal gas constant)- \( T = 394.261 \, \text{K} \)Find \( n \):\( n = \frac{(3033800)(0.708)}{(8.314)(394.261)} = 652.52 \, \text{mol}\).The molar mass of ethane (\( C_2H_6 \)) is approximately \(30.07 \, \text{g/mol}\). Calculate the mass:\( m = n \times \text{molar mass} = 652.52 \times 30.07 = 19614.49 \, \text{g} = 19.61 \, \text{kg} \)

The compressibility factor \( Z \) can be found using a compressibility chart and requires the reduced pressure \( P_r \) and reduced temperature \( T_r \).- Critical Pressure of \( C_2H_6 \): \( P_c = 48.8 \, atm = 4940 \, kPa \)- Critical Temperature of \( C_2H_6 \): \( T_c = 305.32 \, K \)Calculate reduced properties:- \( P_r = \frac{P}{P_c} = \frac{3033.8 \, kPa}{4940 \, kPa} = 0.614 \)- \( T_r = \frac{T}{T_c} = \frac{394.261 \, K}{305.32 \, K} = 1.291 \)From the compressibility chart, \( Z \approx 0.85 \) (interpolation may be necessary).Therefore, actual moles \( n_{actual} = \frac{n}{Z} = \frac{652.52}{0.85} = 767.67 \, \text{mol}\). Find the actual mass: \[ m_{actual} = n_{actual} \times ext{molar mass} = 767.67 \times 30.07 = 23074.65 \, \text{g} = 23.07 \, \text{kg} \]

Determine the percentage error when using the ideal gas law.\[ \text{Error} \, (\%) = \frac{|m_{actual} - m_{ideal}|}{m_{actual}} \times 100 \]Substitute the mass values:\( \text{Error} \, (\%) = \left| \frac{23.07 - 19.61}{23.07} \right| \times 100 = 15.00\% \)