Question

A box of volume 22.4L filled with helium at STP, will have $N_A=6.023\times {10}^{23}$ atoms. Let us look at a much simpler, unrealistic sample of gas where we just have five atoms, having the following velocity components along a fixed direction: 1337.2 m/s,1399.6 m/s,1362.4m/s and1382.1m/s. Calculate the rms speed along the given direction, and the average kinetic energy of each molecule. The mass of a helium atom is $6.646\times {10}^{-27}kg$.

• A. 1300 m/s, 0.5210 eV
• B. 1290 m/s, 0.2412 eV
• C. 1370 m/s, 0.0389 eV
• D. 2239 m/s, 0.1002 eV

Answer 1

The correct option is C

1370 m/s, 0.0389 eV

The rms speed, $v_{rms}$, is -

$v_{rms}=\sqrt{\frac{V_1^2+V_2^2+V_3^2+V_4^2+V_5^2}{5}}$

$=\sqrt{\frac{(1337.2{)}^2+(1399.6{)}^2+(1362.4{)}^2+(1368.0{)}^2+(1382.1{)}^2}{5}}m/s$

$\approx 1370m/s.$

That is pretty fast, don't you think?

Now, average kinetic energy $(K.E)=\frac{1}{2}m{v}_{rms}^2$

$\Rightarrow (K.E)=\frac{1}{2}\times (6.646\times {10}^{-27})\times {1370}^2$

$=\frac{1}{2}\times (6.646\times {10}^{-27})\times {1370}^2\times (6.242\times {10}^{18})eV$

$\approx 0.0389eV.$