23P3Q06

6ai)

COMMENT: The mean KE is directly proportional to the kelvin temperature.

6aii)

0.25

COMMENT: Think $\displaystyle \frac{PV}{T}=nR$

6aiii)

Since they have the same mean KE,

$\displaystyle \begin{aligned} & \frac{1}{2}(m){{c}_{A}}^{2}=\frac{1}{2}(2m){{c}_{B}}^{2} \\ & \frac{{{c}_{A}}}{{{c}_{B}}}=\sqrt{2} \end{aligned}$

6b)

R.m.s. speed of the molecules in container A, $\displaystyle {{c}_{A}}=\sqrt{2}(940)=1329\text{ m }{{\text{s}}^{-1}}$

Mass of a molecule in container A, $\displaystyle m=\frac{{{M}_{molar}}}{{{N}_{A}}}=\frac{4.0\times {{10}^{-3}}}{6.02\times {{10}^{23}}}=6.645\times {{10}^{-27}}\text{ kg}$

$\displaystyle$

$\displaystyle \begin{aligned} \left\langle KE \right\rangle &=\frac{3}{2}kT \\ \frac{1}{2}m{{c}_{A}}^{2}&=\frac{3}{2}kT \\ (6.645\times {{10}^{-27}}){{(1329)}^{2}}&=3(1.38\times {{10}^{-23}})T \\ T&=283\text{ K} \end{aligned}$

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23P3Q06

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