Answer
$\mathrm{a}.\quad 4.0\times 10^{6}\ Pa$
$\mathrm{b}.\quad 1.18\times 10^{-20}\ \mathrm{J}$
$\mathrm{c}.\quad $Pressure halves, average kinetic energy remains the same.
Work Step by Step
$PV=nRT \quad (17- 5)$
where the universal gas constant, $R=8.31\ \mathrm{J}/(\mathrm{mol}\cdot \mathrm{K})$,
Kinetic theory relates the average kinetic energy of the molecules in a gas
to the Kelvin temperature of the gas, $T$:
$(\displaystyle \frac{1}{2}mv^{2})_{\mathrm{a}\mathrm{v}}=K_{\mathrm{a}\mathrm{v}}=\frac{3}{2}kT \qquad (17- 15)$
$(k =1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K}$ is Boltzmann's constant.$)$
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$\mathrm{a}.$
Solve 17-5 for P, (temperature in kelvins):
$P=\displaystyle \frac{nRT}{V}=\frac{(3\mathrm{mol})[8.31\ \mathrm{J}/(\mathrm{mol}\cdot \mathrm{K})](273.15+295\mathrm{K})}{0.0035\mathrm{m}^{3}}$
$\mathrm{P}= 4.0\times 10^{6}\ Pa$
$b.$
$K_{\mathrm{a}\mathrm{v}}=\displaystyle \frac{3}{2}kT=\frac{3}{2}(1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K})(273.15+295 \mathrm{K})$
$=1.18\times 10^{-20}\ \mathrm{J}$
$c.$
If the volume doubles, with all other factors constant, interpreting 17-5,
$\displaystyle \mathrm{P}=\frac{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}}{\mathrm{V}}$
P is inversely proportional to V, so it will halve (decrease by a factor of 2).
Kinetic energy relies only on absolute temperature.
If T remains the same, so does $\mathrm{K}_{\mathrm{a}\mathrm{v}}.$
