# Determine the internal energy change \triangle u of hydrogen, in

Determine the internal energy change $\mathrm{△}u$ of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A-2c), (b) the ${c}_{v}$ value at the average temperature (Table A-2b), and (c) the ${c}_{v}$ value at room temperature (Table A-2a).
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vrangett
Step 1
Given:
- Initial temperature ${T}_{1}=200K$
- Final temperature ${T}_{2}=800K$
Required
- Determine the internal energy change of hydrogen using
a) The empirical specific heat equation
b) The ${c}_{v}$ value at average temperature
c) The ${c}_{v}$ value at room temperature
Step 2
Solution
Part a
- Using the empirical relation of ${\stackrel{―}{c}}_{p}\left(T\right)$ from table (A-2C) and relating it to ${\stackrel{―}{c}}_{v}\left(T\right)$
${\stackrel{―}{c}}_{v}\left(T\right)={\stackrel{―}{c}}_{p}-{R}_{u}=\left(a-{R}_{u}\right)+bT+c{T}^{2}+d{T}^{3}$
Where $\left(a=29.11,b=-0.1916×{10}^{-2},c=0.4003×{10}^{-5},d=-0.8704×{10}^{-9}\right)$
- The internal energy change could be defined as the following
$\mathrm{△}\stackrel{―}{u}={\int }_{1}^{2}{\stackrel{―}{c}}_{v}\left(T\right)dT={\int }_{1}^{2}\left[\left(a-{R}_{u}\right)+bT+c{T}^{2}+d{T}^{3}\right]dT$
$\mathrm{△}\stackrel{―}{u}=\left(a-{R}_{u}\right)\left({T}_{2}-{T}_{1}\right)+\frac{1}{2}b\left({T}_{2}^{2}-{T}_{1}^{2}\right)+\frac{1}{3}c\left({T}_{2}^{3}-{T}_{1}^{3}\right)+\frac{1}{4}d\left({T}_{2}^{4}-{T}_{1}^{4}\right)$
$\mathrm{△}\stackrel{―}{u}=\left(29.11-8.314\right)\left(800-200\right)+\frac{1}{2}×\left(-0.1916×{10}^{-2}\right)×\left({800}^{2}-{200}^{2}\right)+\frac{1}{3}×\left(0.4003×{10}^{-5}\right)×\left({800}^{3}-{200}^{3}\right)+\frac{1}{4}×\left(-0.8704×{10}^{-9}\right)×\left({800}^{4}-{200}^{4}\right)=12487K\frac{J}{K}mol$

Marcus Herman
Step 1
a) In this problem we need to determine the internal energy change $\mathrm{△}u$ by using three different methods.
The first method is with the empirical specific heat equation. The equation gives us ${c}_{p}$ and we need the $\stackrel{―}{{c}_{v}}$ for the calculation.
$\stackrel{―}{{c}_{p}}=a+bT+c{T}^{2}+d{T}^{3}$
$\stackrel{―}{{c}_{v}}=\stackrel{―}{{c}_{p}}-{R}_{u}$
$\stackrel{―}{{c}_{v}}=a-{R}_{u}+bT+c{T}^{2}+d{T}^{3}$
The values for the constants a,b,c and d we find in the table A-2. We will also need the constant ${R}_{u}$
$a=29.11$
$b=-0.1916×{10}^{-2}$
$c=0.4003×{10}^{-5}$
$d=-0.8704×{10}^{-9}$
${R}_{u}=8.31447\frac{kJ}{kmolK}$
Step 2
To calculate the internal energy change per mole $\mathrm{△}\stackrel{―}{u}$ we need to integrate the ${c}_{v}$ from the initial ${T}_{1}=200K$ to the final ${T}_{2}=800K$ temperature.
$\mathrm{△}\stackrel{―}{u}={\int }_{{T}_{1}}^{{T}_{2}}\stackrel{―}{{c}_{v}}\left(T\right)dT$
$\mathrm{△}\stackrel{―}{u}={\int }_{{T}_{1}}^{{T}_{2}}\left(a-{R}_{u}+bT+c{T}^{2}+d{T}^{3}\right)dT$
$\mathrm{△}\stackrel{―}{u}={\int }_{{T}_{1}}^{{T}_{2}}\left(a-{R}_{u}+bT+c{T}^{2}+d{T}^{3}\right)dT$
$\mathrm{△}\stackrel{―}{u}=\left(a-{R}_{u}\right)\cdot {\int }_{{T}_{1}}^{{T}_{2}}dT+b\cdot {\int }_{{T}_{1}}^{{T}_{2}}TdT+c\cdot {\int }_{{T}_{1}}^{{T}_{2}}{T}^{2}dT+d\cdot {\int }_{{T}_{1}}^{{T}_{2}}{T}^{3}dT$
nick1337

a) From Table A-2 C
${c}_{v}=\left(a-R\right)+bT+c{T}^{2}+d{T}^{3}$
where: $a=29.11$
$b=0.1916×{10}^{-2}$
$c=0.4003×{10}^{-5}$
$d=0.8704×{10}^{-9}$
Substituting:
$\mathrm{△}u=\left(29.11-8.314\right)+\left(0.1916×{10}^{-2}\right)\left(800-200\right)+\left(0.4003×{10}^{-5}\right)\left({800}^{2}-{200}^{2}\right)+\left(0.8704×{10}^{-9}\right)\left({800}^{3}-{200}^{3}\right)$

b) From Table B-2
At 500 K, (average Temperature)

$\mathrm{△}u={c}_{v}\left({T}_{2}-{T}_{1}\right)$

c) Table A-2a

$\mathrm{△}u={c}_{v}\left({T}_{2}-{T}_{1}\right)$