Answer C and D please!! **Ammonia Vapor and the SRK Equation of State**Ammonia vapor is compressed to 50 bar and 110°C. Use the SRK equation of state to determine the vapor mass density (g/L). For ammonia, \( T_c = 405.5 \, K \), \( P_c = 111.3 \, \text{atm} \), and \( \omega = 0.257 \).**a. Determine \( T_r \), \( a \), \( b \), \( m \), and \( \alpha \).** - Calculate the reduced temperature \( T_r \) and the parameters \( a \), \( b \), \( m \), and \( \alpha \) that are used in the SRK equation.**b. Plot the pressure as a function of specific volume over a range of 0.05 – 5 L/mol.** - Make sure to properly label the axes. Plot a sufficient number of points to get a smooth curve. You must use Excel to prepare your plot. Axes should be properly labelled, and the plot must have a title.**c. Determine the specific volume of ammonia vapor at 50 bar and 110°C (L/mol).** - Calculate the specific volume using the conditions provided.**d. Determine the vapor mass density (g/L).** - Using the results from the previous calculations, find the vapor mass density. Note: Ensure that the calculations are carried out systematically, and the plot generated represents the data accurately with an appropriate title and labels for the axes.

Given data: T = 110 0C P = 50 bar Tc = 405.5 K Pc = 111.3 atm or 112.77 bar ω = 0.257Reduced properties of the ammonia is calculated as follows: T = 110 0C or 383 K Tr = 0.944 (383/405.5) Pr = 0.443 (50/112.77) The value of α for SRK equation is calculated as follows: αTr,ω= 1+0.480+1.574ω-0.176ω21-Tr122αTr,ω= 1+0.480+1.574×0.257-0.176×0.25721-0.9442αTr,ω= 1+0.480+1.574×0.257-0.176×0.2572×0.0284032 αTr,ω = 1.05 The value of q and β is calculated as follows: q = ψαTr,ωῺTr= 0.42748 × 1.050.08664×0.944q = 5.49 β = ῺPrTr= 0.08664 × 0.4430.944β = 0.0406The compressibility factor of ammonia vapor is calculated as follows: Z = 1+β-qβZ-βZ+εβZ+σβ For, SRK EOS,ε = 0σ = 0Thus,Z = 1+β-qβZ-βZZ+β Z = 1+0.0406-5.49*0.0406Z-0.0406ZZ+0.0406 Z Z2+0.0406Z= 1.0406Z2+0.0406Z-0.2229Z+0.00905 Z3+0.0406Z2= 1.0406Z2 + 0.04224 Z-0.2229Z+0.00905Z3 - Z2 + 0.18066 Z - 0.00905 = 0Z =0.784(c) The molar volume of ammonia vapor at given conditions is calculated as follows: V⏜ = ZRTP= 0.784 × 83.14 bar.cm3mol.K × 383 K50 barV⏜= 499.29 cm3mol × 10-3 L1 cm3V⏜ = 0.49929 Lmol(d) Molar mass of ammonia vapor = 17 g/mol Vapor mass density of ammonia is calculated as follows: ρ = 1V⏜= 10.49929 Lmol= 2 molL × 17 gmolρ= 34 gL