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$E=\frac 12mv^2$ is valid for translational kinetic energy and the speed of the centre of mass. Vibrational or rotational energy does not count, as an object may vibrate or rotate even if it's centre of mass has zero speed. As each translational degree of freedom has the mean energy $E=\frac 12kT$, the mean translational kinetic energy is $E_\mathrm{transl}...


4

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios: $$ \begin{cases} T \propto V\\ T \propto p \end{cases} \implies p \propto \frac 1 V $$ which, as Zenix commented, is a math form of Boyle's law.


3

Well, an ideal gas has no attraction force between its molecules and does not have a boiling point, as you cannot make it liquid. Also, the sum of molecule own volumes is supposed negligible wrt the gas volume. Water does not form anything close to an ideal gas because of its hydrogen bonds. You have to heat water to 100 °C to make it boiling to overcome ...


1

This is clearly a homework question. Yet, since OP has given good effort to solve the question, I'd like to give OP some insight to solve these kinds of questions, taking current question as an example. Suppose the mixture contains $x\;\pu {g}$ of $\ce{CH4}$ and $y\;\pu {g}$ of $\ce{C2H4}$. The combustion reactions of $\ce{CH4}$ and $\ce{C2H4}$ can be given ...


1

For deriving ideal gas law we assume that there exist no intermolecular forces between molecules and their size are negligible.As the size or intermolecular interactions increase there would be a increase in deviation from ideal gas law.Obviously here interactions is the major factor.(H-bonding in water)


1

You may consider the vdW equation as the deviation from the ideal gas state equation $pV_\mathrm{m} = RT$. If you look at the vdW equation $$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$ you can easily see why the corrections have their signs. The pressure term correction has the positive sign, as the real gas has due molecule cohesion smaller volume ...


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