Question #1a490

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Answer 1 · 2015-06-25T12:06:49

$U_{tot} = 49.16 kJ$ $U_("tot")=49.16"kJ"$

In an ideal gas we can consider the internal energy to be due to the kinetic energy of the particles.

There are 3 degrees of freedom for a monotomic gas: they can move in the x, y, and z directions.

Each degree of freedom contributes $\frac{1}{2} k T$ $1/2kT$ of energy which makes $\frac{3}{2} k T$ $3/2kT$ in total.

So total internal energy $U$ $U$ is given by:

$U = \frac{3}{2} k T$ $U=3/2kT$

$k$ $k$ is the gas constant per mole and is The Boltzmann Constant.

$k = 1.38 \times 10^{- 23} m^{2} . k g . s^{- 2} . K^{- 1}$ $k=1.38xx10^(-23)m^(2).kg.s^(-2).K^(-1)$

$T$ $T$ is the absolute temperature.

A diatomic gas like nitrogen $N_{2}$ $N_2$ has 5 degrees of freedom. 3 for translational movement (x,y and z) and 2 for rotation.

This makes 5 in all. So for a diatomic gas:

$U = \frac{5}{2} k T$ $U=5/2kT$

In our example we have 4.5 moles of gas so:

$U_{tot} = \frac{5}{2} \times L \times 4.5 \times k T$ $U_("tot")=5/2xxLxx4.5xxkT$

$L$ $L$ is the Avogadro Constant = $6.02 \times 10^{23} m o l^{- 1}$ $6.02xx10^(23)mol^(-1)$

$U_("tot")=5/2xx6.02xxcancel(10^(23))xx4.5xx1.38xxcancel(10^(-23))xx526$

$U_("tot")=49160.2"J"$

$=49.16"kJ"$