Question

A gas mixture at 535.0°C and 109 kPa absolute enters a heat exchanger at a rate of 67.0 m3/hr. The gas leaves the heat exchanger at 215.0°C. The change in enthalpy of the gas during the cooling process is -5.00 kJ/mol. What is the heat required in kW? Assume the gas behaves ideally and that the changes in kinetic and potential energy are negligible.

Answer 1

the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.

Step-by-step explanation:

assuming ideal gas behaviour:

PV=nRT

therefore

P= 109 Kpa= 1.07575 atm

V= 67 m3/hr = 18.6111 L/s

T= 215 °C = 488 K

R = 0.082 atm L /mol K

n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)

n= 0.5 mol/s

since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:

Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW