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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. ...


3

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work $$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext} \mathrm{d}V$$ Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$. Conceptually, the ...


2

You are correct that there are aspects to consider: The increase of pressure equates to more particles per unit of volume. It is like filling a metro with more and more people during rush hour, the propability of collisions increases. The increase of temperature equates to a higher kinetic energy of the molecules, so if they hit each other, than they do ...


2

When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture. And when the pressure is extremely low, the volume increases but not as much as calculated with the ...


2

There exists an equation that you must know: $$n = \frac{m}{M}.$$ Here $n$ is the amount of substance of a given sample of pure product, $m$ is the mass of the same sample, and $M$ is the molar mass of the pure substance. Here, your sample has a weight equal to $\pu{9 g}$. The molar mass of $\ce{H2}$ is $2\times \pu{1 g/mol} = \pu{2 g/mol}$. So the amount ...


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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