- #1
zenterix
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- Homework Statement
- 3.4 (a) Calculate the work done upon expansion of 1 mol of gas quasi-statically and isothermally from volume ##v_i## to a volume ##v_f##, when the equation of state is
$$\left ( P+\frac{a}{v^2}\right )(v-b)=RT$$
where ##a## and ##b## are the van der Waals constants.
(b) If ##a=1.4\times 10^9\mathrm{N\cdot m^4/mol}## and ##b=3.2\times 10^{-5}\mathrm{m^3/mol}##, how much work is done when the gas expands from a volume of 10 liters to a volume of 22.4 liters at ##20^{\circ} \text{C}##?
- Relevant Equations
- ##W=-\int PdV##
My question is about units.
For the first part, we can solve the state equation for ##P## as a function of ##V## and ##T##. We obtain
$$P(V,T)=\frac{RT}{V-b}-\frac{a}{V^2}\tag{1}$$
The units question already comes up here because as far as I can tell the right-hand side doesn't have the correct units.
$$\mathrm{\frac{J}{mol\cdot K}\cdot K\cdot \frac{1}{m^3-\frac{m^3}{mol}}-\frac{N\cdot m^4}{mol}\cdot \frac{1}{(m^3)^2}}\tag{2}$$
$$\mathrm{\frac{N\cdot m}{mol}\cdot \frac{mol}{m^3\cdot mol -m^3}-\frac{N\cdot m}{mol\cdot m^3}}\tag{3}$$
$$\mathrm{\frac{N\cdot m}{m^3\cdot mol-m^3}-\frac{N}{mol\cdot m^2}}\tag{4}$$
However, if we move past this for now, work in our quasi-static isothermal process is
$$W=-\int_{V_i}^{V_f} PdV$$
$$=(...)$$
$$=RT\ln{\frac{V_i-b}{V_f-b}}-\frac{a}{V_f}+\frac{a}{V_i}\tag{5}$$
Then, if we plug in the values given in (b) we get
$$W=\mathrm{8.31\frac{J}{mol\cdot K}\cdot 293 K\cdot \ln{\left ( \frac{10L-3.2\cdot 10^{-5} \frac{m^3}{mol}}{22.4L - 1.4\cdot 10^9 \frac{N\cdot m^4}{L\cdot mol}} \right )}-\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{22.4L}+\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{10L}}$$
The numerical result of this calculation is ##7.75\times 10^{10}## which matches the answer at the end of the book.
My question is again about the units (which, as expected are probably not working given that they weren't working to begin with).
Specifically, the units of what is in the logarithm.
In the logarithm numerator and denominator individually, we have liters minus ##\mathrm{m^3/mol}##.
Does this subtraction make sense?
Now, for all the units to work out, we need the logarithm to have no units at all. Then, the other units all work out to be ##\frac{J}{mol}##. Now, these units are not correct, because we expect just ##J##.
After having written this all out, I think the problem is in the original state equation.
I think there is one or more hidden ##n##'s that were replaced with ##1\text{mol}## by the problem statement.
That being said, when I look back at the main text of the book I am reading, the van der Waals state equation does indeed seem to be correct in the problem statement. Which brings me back to my main question: how do the units work in this state equation?
For the first part, we can solve the state equation for ##P## as a function of ##V## and ##T##. We obtain
$$P(V,T)=\frac{RT}{V-b}-\frac{a}{V^2}\tag{1}$$
The units question already comes up here because as far as I can tell the right-hand side doesn't have the correct units.
$$\mathrm{\frac{J}{mol\cdot K}\cdot K\cdot \frac{1}{m^3-\frac{m^3}{mol}}-\frac{N\cdot m^4}{mol}\cdot \frac{1}{(m^3)^2}}\tag{2}$$
$$\mathrm{\frac{N\cdot m}{mol}\cdot \frac{mol}{m^3\cdot mol -m^3}-\frac{N\cdot m}{mol\cdot m^3}}\tag{3}$$
$$\mathrm{\frac{N\cdot m}{m^3\cdot mol-m^3}-\frac{N}{mol\cdot m^2}}\tag{4}$$
However, if we move past this for now, work in our quasi-static isothermal process is
$$W=-\int_{V_i}^{V_f} PdV$$
$$=(...)$$
$$=RT\ln{\frac{V_i-b}{V_f-b}}-\frac{a}{V_f}+\frac{a}{V_i}\tag{5}$$
Then, if we plug in the values given in (b) we get
$$W=\mathrm{8.31\frac{J}{mol\cdot K}\cdot 293 K\cdot \ln{\left ( \frac{10L-3.2\cdot 10^{-5} \frac{m^3}{mol}}{22.4L - 1.4\cdot 10^9 \frac{N\cdot m^4}{L\cdot mol}} \right )}-\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{22.4L}+\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{10L}}$$
The numerical result of this calculation is ##7.75\times 10^{10}## which matches the answer at the end of the book.
My question is again about the units (which, as expected are probably not working given that they weren't working to begin with).
Specifically, the units of what is in the logarithm.
In the logarithm numerator and denominator individually, we have liters minus ##\mathrm{m^3/mol}##.
Does this subtraction make sense?
Now, for all the units to work out, we need the logarithm to have no units at all. Then, the other units all work out to be ##\frac{J}{mol}##. Now, these units are not correct, because we expect just ##J##.
After having written this all out, I think the problem is in the original state equation.
I think there is one or more hidden ##n##'s that were replaced with ##1\text{mol}## by the problem statement.
That being said, when I look back at the main text of the book I am reading, the van der Waals state equation does indeed seem to be correct in the problem statement. Which brings me back to my main question: how do the units work in this state equation?
Last edited:
). So does an exponentiation.