Answer:
a) The statement is true, b) [tex]K_{tr.atom} = 6.073 \times 10^{-21} \frac{J}{atom}[/tex],[tex]v_{rms} \approx 477.904 \frac{m}{s}[/tex]
Explanation:
a) The standard conditions are 273.15 K and 1 atm. Let consider that gas behaves ideally, whose equation of state is:
[tex]P \cdot V = n \cdot R_{u} \cdot T[/tex]
The volume is cleared out:
[tex]V = \frac{n \cdot R_{u} \cdot T}{P}[/tex]
By replacing terms:
[tex]V = \frac{(1 mole)\cdot (0.082 \frac{L \cdot atm}{mole \cdot K} )\cdot (273.15 K)}{1 atm}\\V = 22.399 L[/tex]
Therefore, the statement is true.
b) The average kinetic energy per atom is:
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{R_{u}\cdot T}{N_{A}}[/tex]
Where [tex]N_{A}[/tex] is the Avogadro constante and is equal to [tex]6.022 \times 10^{23} \frac{atoms}{mole}[/tex].
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{(8.317 \frac{J}{mole\cdot K} )\cdot (293.15 K)}{6.022 \times 10^{23} \frac{atoms}{mole} } \\K_{tr,atom} = 6.073\times 10^{-21} \frac{J}{atom}[/tex]
The rms speed is determined by the following formula:
[tex]v_{rms} = \sqrt{\frac{3 \cdot k \cdot T}{n \cdot M_{O} \cdot \frac{1}{N_{A}} } }\\v_{rms} = \sqrt{\frac{3\cdot(1.38 \times 10^{-23} \frac{J}{K} )\cdot(293.15 K)}{(2 moles) \cdot (16 \frac{g}{mole} )(\frac{1 kg}{1000 g} ) \cdot (\frac{1 mole}{6.022 \times 10^{23} atoms} )}} \\v_{rms} \approx 477.904 \frac{m}{s}[/tex]
