-2
$\begingroup$

If a closed rigid wall container filled with 20 gram of Helium gas moving at a constant speed of 10 m/sec is stopped abruptly to stand still, find the increment in temperature of this He sample.

Way I have solved

I used Average kinetic energy formula and $nfRT/2$(f is degree of freedom here).

My sir told me that $nfRT/2$ = Translational kinetic energy + Rotational kinetic energy.

So I used the formula $nfRT/2$= ${\frac{1mv^2}{2}}$ + Rotational kinetic energy.

In the question , Rotational kinetic energy is not given.Rotational kinetic energy will be 0 here since He is has 3 degrees of freedom.

I get is Kinetic energy = $1/2 * 20/1000 * 10^2 =1

1 = $n3T/2$

Should I take n =1?Because it not given.

Have I solved it correctly and is there any other way to solve it.

$\endgroup$
8
  • 1
    $\begingroup$ Try to use more specific titles, as there are many questions about thermodynamics here. By 20 gm you probably mean g as gram.Include your complete computation procedure, and preferably use MathJax for formula formatting. Note that the goal of the site is not a task proof reading nor solving, but helping to understand principles. It also prefers elaborated question. Do you have troubles to understand or apply a particular principle ? $\endgroup$ – Poutnik Nov 20 '20 at 8:14
  • $\begingroup$ Always think thoroughly about possible answers to your question and search for them before asking, together with providing of your intermediate results of the failure. $\endgroup$ – Poutnik Nov 20 '20 at 8:18
  • $\begingroup$ Yes.one is that whether n will =1? Second , what I have written , is it correct? Is there any other to approach this problem. $\endgroup$ – srijan Sri Nov 20 '20 at 8:19
  • $\begingroup$ @Poutnik This is not there online.Made by sir itself. $\endgroup$ – srijan Sri Nov 20 '20 at 8:19
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Poutnik Nov 20 '20 at 8:21
1
$\begingroup$

The total mechanical energy (other kinds of energy are ignored here ) of moving gas is the sum of potential energy and translational, rotational and vibrational kinetic energy :

$$E = E_\mathrm{pot} + E_\mathrm{trans} + E_\mathrm{vibr} + E_\mathrm{rot}$$

These energies can be separated as the energy of the bulk volume of a gas and the energy of gas molecules in the inertial frame where the centre of the gass mass is in the rest:

$$\small E = (E_\mathrm{pot,bulk} + E_\mathrm{pot,mol}) + (E_\mathrm{trans,bulk} + E_\mathrm{trans,mol}) \\ + (E_\mathrm{vibr,bulk} + E_\mathrm{vibr,mol}) + (E_\mathrm{rot,bulk} + E_\mathrm{rot,mol})$$

We can for our case consider $E_\mathrm{pot,bulk} = 0$, $E_\mathrm{vibr,bulk} = 0$, $E_\mathrm{rot,bulk} = 0$, $E_\mathrm{pot,mol} = 0$, assuming zero bulk gas and molecular potential energy and no gas vibration nor rotation.

$$E = E_\mathrm{trans,bulk} + E_\mathrm{trans,mol} + E_\mathrm{vibr,mol} + E_\mathrm{rot,mol}$$

For the translational kinetic energy of the bulk gas is the notorious formula:

$$E_\mathrm{trans,bulk} = \frac 12 \cdot m \cdot v^2$$

We can with just a little error consider helium as an ideal gas. For 1 mole of an ideal gas, each degree of freedom of motion has energy $E = \frac 12RT$, where $R \approx \pu{8.314 JK^-1mol^-1}$ is the universal gas constant and $T$ is absolute temperature. Their kinetic energy is then:

$$E_\mathrm{kin,mol}=\frac{f_\mathrm{trans} + f_\mathrm{vibr} + f_\mathrm{rot}}2RT$$

Atoms of monoatomic gasses like helium have 3 degrees of freedom $f$ for translational motion, but do not rotate nor vibrate, so $f_\mathrm{vibr} = f_\mathrm{rot} = 0$

Therefore, the kinetic energy of helium molecules of the total mass $m$ and molar mass $M$ , forming helium thermal energy is: $$E_\mathrm{therm}=\frac {3mRT}{2M}$$

So we have finally the energy of the gaseous helium as:

$$E = \frac 12 \cdot m \cdot v^2 + \frac{3m}{2M}RT = \frac m2 \cdot ( v^2 + \frac{3RT}{M})$$

As the helium mass is constant and as we honour the law of energy conservation, the following equations must be true ( with indexes 1,2 as initial and final states:

$$v_1^2 + \frac{3RT_1}{M} = v_2^2 + \frac{3RT_2}{M}$$

$$ \frac{3R}{M}(T_2 - T_1) = v_1^2 - v_2^2$$

As the final speed $v_2 = 0$, we can write:

$$ \Delta T = \frac {M \cdot v_1^2}{3R} $$

The helium molar mass $M \approx \pu{0.0040026 kg mol^-1} \approx \pu{0.004 kg mol^-1}$

$$ \Delta T = \frac {M \cdot v_1^2}{3R} = \frac {(\pu{0.004 kg mol-1} ){(\pu{10 ms^-1})}^2}{\pu{3 \cdot 8.314 JK^-1mol^-1}} \approx \pu{+0.016 K}$$


For formatting of chem/math formulas/expressions, use can use MathJax, also look here and here. See also MathJax on Math SE.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.