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0

As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible. So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.

3

I would approach this a little differently. I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars: $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or, $$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$ This would provide an equation for expressing a in terms of b. ...

1

What this ultimately boils down to is that this is a system of equations which we're trying to solve. Since we have two unknowns, we need two equations. Luckily, we have 3. $$17.76 atm = \frac{RT}{V_m - b} - \frac{a}{V_m^2}$$ where our three equations come from having three different known values of the molar volume. As is typical with systems of ...

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The equation can be rewritten by substituting P = RT/V from the Ideal Gas Equation. This gives us (A√RT/(V√2πm) showing that increase in temperature leads to increase in rate of effusion. Absolute Temperature simply refers to the temperature on the Kelvin scale of absolute temperature.

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A dimensional analysis for your calculation... $$\begin{array}\\ P &= dgh \\ &= (0.0135951\ \mathrm{kg/cm}^3)(9.81\ \mathrm{m/s^2)}(0.76\ \mathrm{m}) \\ &= 0.101 \ \mathrm{kg\cdot m^2 \cdot cm^{-3} \cdot s^{-2}} \end{array}$$

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Just complementing the first answer of @Buck Thorn : Full rewriting of the van Der Waals equation in terms of $Z=f(p,V_\mathrm{m})$: $$\left(p+\frac{a}{{V_\mathrm{m}}^2}\right)\left(V_\mathrm{m}-b\right)=RT$$ $$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$ $$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} -... 2 The van der Waals equation is$$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$which for p \gg a/V_m^2 ^\ast can be rewritten as$$pV_m=RT+bp$$or$$pV=nRT+nbp \tag{1} The problem says that the tangent of this curve evaluated at the same $T$ has an intercept (the value of $pV$ when $p$ goes to zero) of $\pu{40 Latm}$ when $n=2$, which means ...

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