Question

A cylinder of nitrogen gas has a volume of $15,000c{m}^3$and a pressure of$100$ atm.
a. What is the thermal energy of this gas at room temperature$\left({20}^{\circ }C\right)$?
b. What is the mean free path in the gas?
c. The valve is opened and the gas is allowed to expand slowly and isothermally until it reaches a pressure of $1.0$atm . What is the change in the thermal energy of the gas?

Short answer

a)The Thermal energy of gas is$E=379KJ.$
b)The Mean free path in tha gas is$\lambda =2.25nm.$
c)The change in thermal energy of the gas is Zero.

Step-by-step solution

Step: 1   Finding the Thermal Energy of gas at  room temperature(20∘C): (part a)

Because nitrogen is a diatomic gas, the thermal energy is calculated as follows:

$E=\frac{5}{2}N{K}_{B}T$

Now we know that , $pV=N{K}_{B}T$

From equation localid="1648381889785" $\left(1\right)$and$\left(2\right)$we get that

$E=\frac{5}{2}pV$

Given that$p=100\mathrm{atm}=1.01\times {10}^7\mathrm{N}/{\mathrm{m}}^2$and $V=15000{\mathrm{cm}}^3=0.015{\mathrm{m}}^3,$, then substituting these to the formula:

localid="1648390595486" $E=\frac{5}{2}\times 1.01\times {10}^7\times 0.015=379\mathrm{kJ}.$

Step: 2  Finding the Mean free path: (part b)

The mean free path is given by

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$

Now we have

$\frac{N}{V}=\frac{p}{K_{B}T}=\frac{1.01\times {10}^7}{1.38\times {10}^{-23}\times 293}=2.5\times {10}^{27}$

and $r$for nitrogen is $r=1\times {10}^{-10}$,then the mean free path will be:

$\begin{array}{r}\lambda =\frac{1}{4\sqrt{2}\pi \times {10}^{27}\times {\left(1\times {10}^{-10}\right)}^2}\\ =2.25\times {10}^{-9}\mathrm{m}\\ =2.25\mathrm{nm}.\end{array}$

Step: 3  Change in Thermal Energy of gas: (part c)

But even though the formula shows that thermal energy is proportional to the number of gas molecules, degrees of freedom, and gas temperature, none of these factors change in this isothermal expansion.

As a result, the total amount of thermal energy will remain unchanged. Thermal energy will remain unchanged.