Heat And Thermodynamics Ques 121
- A cubical box of side $1 m$ contains helium gas (atomic weight 4 ) at a pressure of $100 N / m^{2}$. During an observation time of $1 s$, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take, $R=\frac{25}{3} J / mol-K$ and $k=1.38 \times 10^{-23} J / K$.
$(2002,5 M)$
(a) Evaluate the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
(c) Evaluate the total mass of helium gas in the box.
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Solution:
Formula:
- Volume of the box $=1 m^{3}$
Pressure of the gas $=100 N / m^{2}$
Let $T$ be the temperature of the gas. Then,
(a) Time between two consecutive collisions with one wall $=\frac{1}{500} s$. This time should be equal to $\frac{2 l}{v _{rms}}$, where $l$ is the side of the cube.
$$ \begin{array}{rlrl} & & \frac{2 l}{v _{rms}} & =\frac{1}{500} \\ \text { or } & \sqrt{v _{rms}} & =1000 m / s \quad(\text { as } l=1 m) \\ \text { or } & \sqrt{\frac{3 R T}{M}} & =1000 \\ \therefore & T=\frac{(1000)^{2} M}{3 R} & =\frac{(10)^{6}\left(4 \times 10^{-3}\right)}{3(25 / 3)}=160 K \end{array} $$
(b) Average kinetic energy per atom
$$ \begin{aligned} =\frac{3}{2} k T & =\frac{3}{2}\left(1.38 \times 10^{-23}\right)(160) J \\ & =3.312 \times 10^{-21} J \end{aligned} $$
(c) From $p V=n R T=\frac{m}{M} R T$
We get mass of helium gas in the box, $m=\frac{p V M}{R T}$
Substituting the values, we get
$$ m=\frac{(100)(1)(4)}{(25 / 3)(160)}=0.3 g $$
BETA
