**Mixture and Heat Requirement Calculation****Scenario:**We have two gas mixtures being mixed and heated:1. **Mixture 1:** 300 L/h of a 15.0 mole% C3H8 and 85.0% n-C4H10 mixture at 0°C and 1.10 atm2. **Mixture 2:** 300 L/h of a 55.0 mole% C3H8 and 45.0% n-C4H10 mixture at 25°C and 1.10 atmThese mixtures are combined and heated to 227°C at constant pressure.**Objective:**Calculate the heat requirement for this process in kJ/h using the following data assuming ideal-gas behavior.**Enthalpy Data Table:**| Temperature (°C) | \( \hat{H} \) for Propane (J/mol) | \( \hat{H} \) for Butane (J/mol) ||------------------|----------------------------------|---------------------------------|| 0 | 0 | 0 || 25 | 1772 | 2394 || 227 | 20,685 | 27,442 |**Task:**Determine the heat requirement utilizing the enthalpy data provided for propane and n-butane, taking into account their respective proportions in each mixture and the ideal-gas behavior assumption. The solution should provide the heat needed in kJ/h.

Given: Two mixtures (1&2) containing propane and butane of same volumetric flowrate are mixed in a mixer as shown in figure It is required to calculate heat requirement for mixing two mixtures till 227 °C It is assumed that the gases are behaving ideally In that case mole percentage=volume percentage For mixture 1 Volume of propane =xP1*V1=0.15*300 Lh=45 Lh Volume of butane =xB1*V1=0.85*300 Lh=255 Lh For mixture 2 Volume of propane =xP2*V2=0.55*300 Lh=165 Lh Volume of butane =xB2*V2=0.45*300 Lh=135 Lh Total volume after mixing At standard conditions 1 mole of gas occupies 22.414 L Number of moles of propane in mixture 3 (nP3)= (45+165) Lh22.414 Lmol =9.369 mol Number of moles of butane in mixture 3 (nB3)= (255+135) Lh22.414 Lmol =17.4 mol Number of moles of propane in mixture 2 (nP2)= (165) Lh22.414 Lmol =7.36 mol Number of moles of butane in mixture 2 (nB2)= (135) Lh22.414 Lmol =6.023 molNumber of moles of propane in mixture 1 (nP1)= (45) Lh22.414 Lmol =2.007 mol Number of moles of butane in mixture 1 (nB1)= (255) Lh22.414 Lmol =11.3768 mol Heat required (Q)= (Final enthalpy)-(Initial enthalpy) Q=(9.369 mol*20685 Jmol+17.4 mol*27442 Jmol)-(7.36 mol*1772 Jmol+6.023*2394)+0⇒Q=643827.583 J=643.827583 kJ