Question Problem 5 [10 points] Van Der Waals, suggested modelling the behaviour of a gas in a sealed container by accounting for the sizes of the molecules and attractive forces between them. You know that Volume \( V \), measured in \( m^{3} \) and temperature \( T \) in \( K \) of water vapour in a container exhibit the relationship \[ \left(10^{5}+\frac{5.5}{V^{2}}\right) V=8.3 T \] If the average velocity of molecules \( v_{a v g} \) in \( \mathrm{m} / \mathrm{s} \) is given by \[ v_{a v g}=\sqrt{3300 T} \] (A) Find \( v_{a v g} \) as a function of \( V \). (B) Find the average velocity of a water molecule at \( T=528 \mathrm{~K} \) and the Volume at \( T= \) \( 528 K \) (C) Use the above and the tangent line approximation to find the volume when \( v_{\text {avg }} \) increases by \( 1 \mathrm{~m} / \mathrm{s} \).

042P7K The Asker · Chemistry

Transcribed Image Text: Problem 5 [10 points] Van Der Waals, suggested modelling the behaviour of a gas in a sealed container by accounting for the sizes of the molecules and attractive forces between them. You know that Volume \( V \), measured in \( m^{3} \) and temperature \( T \) in \( K \) of water vapour in a container exhibit the relationship \[ \left(10^{5}+\frac{5.5}{V^{2}}\right) V=8.3 T \] If the average velocity of molecules \( v_{a v g} \) in \( \mathrm{m} / \mathrm{s} \) is given by \[ v_{a v g}=\sqrt{3300 T} \] (A) Find \( v_{a v g} \) as a function of \( V \). (B) Find the average velocity of a water molecule at \( T=528 \mathrm{~K} \) and the Volume at \( T= \) \( 528 K \) (C) Use the above and the tangent line approximation to find the volume when \( v_{\text {avg }} \) increases by \( 1 \mathrm{~m} / \mathrm{s} \).

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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/6(A) Using equation (A), we have:\( \mathrm{{v}_{{a}}{v}{g}=\sqrt{{{3{,}300}{T}}}} \) ExplanationWe want to express this in terms of V, so we need to eliminate T using the given relationship between V and T:PV = nRTwhere P is the pressure, n is the number of moles of gas, and R is the ideal gas constant. Explanation:Please refer to solution in this step.Step2/6For water vapour, we can use the Van der Waals equation: \( \mathrm{{\left({P}+{a}{\left(\frac{{n}}{{V}}\right)}^{{2}}\right)}{\left({V}-{n}{b}\right)}={n}{R}{T}} \)where a and b are constants that depend on the properties of the gas. Rearranging, we get:P = (nRT)/(V - nb) - a(n/V)^2. Substituting this expression for P into the ideal gas law, we get:\( \mathrm{\frac{{{n}{R}{T}}}{{{V}-{n}{b}}}-{a}{\left(\frac{{n}}{{V}}\right)}^{{2}}={R}\frac{{T}}{{V}}} \)Simplifying, we get:\( \mathrm{{n}{R}{T}={\left({R}\frac{{T}}{{V}}\right)}{\left({V}-{n}{b}\right)}+{a}{\left(\frac{{n}}{{V}}\right)}^{{2}}{\left({V}-{n}{b}\right)}} \)Explanation:Please refer to solution in this step.Step3/6Expanding and canceling terms, we get:nRT = RT - nRTb/V + a(n^2/V^2)(V - nb)Solving for T, we get:T = (RT)/(nR - b/V) - (a/V)(n/V). Substituting this expression for T into equation (A), we get:\( \mathrm{{v}_{{a}}{v}{g}=\sqrt{{{3{,}300}{\left[\frac{{{R}{T}}}{{{n}{R}-\frac{{b}}{{V}}}}-{\left(\frac{{a}}{{V}}\right)}{\left(\frac{{n}}{{V}}\right)}\right]}}}} \) Simplifying, we get:\( \mathrm{{v}_{{a}}{v}{g}=\sqrt{{{\left({3{,}300}\frac{{R}}{{n}}\right)}{\left(\frac{{T}}{{{1}-\frac{{b}}{{V}}}}\right)}-{\left({3{,}300}\frac{{a}}{{n}}{V}\right)}}}} \)Explanation:Please refer to solution in this step.Step4/6( ... See the full answer