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Strictly speaking, the pressure inside must be certainly higher inside due to the restoring forces generated by the stretching of the plastic bag. It is this force that causes the gases to gush out when punctured or opened. Bottom line: the assumption of the pressure inside being equal to atmospheric pressure can be considered correct with a considerably ...


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above question can be solved by the method given in the picture


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The derivation is not that hard. If my memory is correct this how we have done that: From Boyle's Law: $$P_1V_1 = P_2V_2 \label{eqn:1}\tag{1}$$ From Charles' Law: $$\frac{V_1}{T_1} = \frac{V_2}{T_2} \label{eqn:2}\tag{2}$$ Multiplying \eqref{eqn:1} and \eqref{eqn:2} gives: $$\frac{P_1V_1^2}{T_1} = \frac{P_2V_2^2}{T_2} \label{eqn:3}\tag{3}$$ If $V_1=V_2$...


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As was pointed out in the comments, the problem arises when you assume that the two starting points are equivalent. @IvanNeretin pointed out that Boyle's Law is only useful for situations where the temperature doesn't change (isothermal) and Charles' Law is only useful in situations where the pressure doesn't change (isobaric). This means that even ...


0

You can use the formula for the ideal gas $$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$ As pressure, temperature and density are given, you get the mean molar mass. $$M_\mathrm{mean}=\frac{\rho RT}{p}$$ From that, you get molar fractions. $$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$ $$x_{\ce{O3}}=1-x_{\ce{O2}}$$ ...


1

The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers Wikipedia is correct. Expressed in, not expressed as. $3/2$ can be expressed in simple whole numbers $3$ and $2$.


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You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write $$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$ where $$\begin{align}n_{g,1}&=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&=n_{H_2O}+n_{Ar}\end{align}$$ You also ...


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In the picture above, the C is the critical point. There seems to be some easy methods to calculate the critical temperature, but not so with the critical pressure. For bubble point it would give me a good indication of where the max range is. But I realised that the critical temperature point won't help me finding the min and max range for dew point ...


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