   Chapter 2.5, Problem 8E ### Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
ISBN: 9781111580704

#### Solutions

Chapter
Section ### Fundamentals of Chemical Engineeri...

1st Edition
Kevin D. Dahm + 1 other
ISBN: 9781111580704
Textbook Problem

# If the heat capacity of an ideal gas is CP* = 30 + 0.05T, with CP* in J/mol · K and T representing the temperature in K, what is the change in internal energy and the change in enthalpy for five moles of this gas when its heated from P = 1 bar and T = 300 K to P = 5 bar and T = 600 K?

Interpretation Introduction

Interpretation:

The change in internal energy and the change in enthalpy.

Concept Introduction:

The expression of change in internal energy (dU_).

dU_=CV*(dT)

Here, constant volume is CV*, and change in temperature is dT.

The expression to obtain the value of CV*.

CV*=CP*R

Here, constant pressure heat capacity for ideal gas is CP* and gas constant is R.

The expression to obtain the change in enthalpy (dH_).

dH_=CP*dT

Explanation

Given information:

Initial Temperature is T1=300 K.

Final Temperature is T2=600 K.

Initial pressure is P1=1 bar.

Final pressure is P2=5 bar.

Write the expression of change in internal energy (dU_).

dU_=CV*(dT)        (1)

Write the expression to obtain the value of CV*.

CV*=CP*R        (2)

Substitute 30+0.05T for CP* and 8.314JKmol for R in Equation (2).

CV*=30+0.05T(8.314JmolK)=(21.686+0.05T)Jmol

Substitute 21.686+0.05T for CV* and multiply both sides by N in Equation (1).

dU=N(21.686+0.05T)dT        (3)

Here, number of moles is N.

Integrate Equation (3) and substitute 5 moles for N.

U1U2dU=(5moles)300K600K(21.686+0.05T)dTU2U1=(5moles){21.686[T]300K600K+0.05[T22]300K600K}U2U1=(5moles){21

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