Question

The average translational energy and the rms speed of molecules in a sample of oxygen gas at $300K$ are $6.21\times {10}^{-21}J$ and $484m/s$, respectively. The corresponding values at $600K$ are nearly (assuming ideal gas behaviour).

• A. $12.42\times {10}^{-21}J,968m/s$
• B. $8.78\times {10}^{-21}J,684m/s$
• C. $6.21\times {10}^{-21}J,968m/s$
• D. $12.42\times {10}^{-21}J,684m/s$

Answer 1

The correct option is D $12.42\times {10}^{-21}J,684m/s$

The formula for average kinetic energy is
$\overline{\left(KE\right)}=\frac{3}{2}KT$
$therefore\frac{{\overline{\left(KE\right)}}_{600K}}{{\overline{\left(KE\right)}}_{300K}}=\frac{600}{300}$
$\Rightarrow {\overline{\left(KE\right)}}_{600K}=2\times 6.21\times {10}^{-21}J=12.42\times {10}^{-21}J$
Also the formula for $rms$ velocity is
$C_{rms}=\sqrt{\frac{3KT}{m}}$
$\therefore \frac{(C_{rms}{)}_{600K}}{(C_{rms}{)}_{300K}}=\sqrt{\frac{600}{300}}$
$\Rightarrow (C_{rms}{)}_{600K}=\sqrt{2}\times 484=684m/s$.