# Enthalpy change for a van der waal gas

One mole of Van der Waal gas is compressed from $\pu {20 dm^3}$ to $\pu{10 dm^3}$ at $\pu{300 K}$. Given $a = \pu{3.60 dm^6 mol^-2}$ and $b = \pu{0.44 dm^3 mol^-1}$. Find $\Delta H$.

$\Delta H = -\mu C_p \Delta P + C_p \Delta T$

Since for VDW gas $\displaystyle \mu C_p = {2a\over RT}- b$.

The process took place at $\pu {300 K}$ which means $\Delta T = 0$.

$\therefore \Delta H = -\mu C_p \Delta P + C_p \Delta T = \left(-{2a \over RT} + b\right)\Delta P = \pu{0.4399 kJ/mol}$.

But the answer is $\pu {-30.5 J/mol}$

I think my mistake is taking $\Delta T = 0$ but if that were the case what should I take as $\Delta T$, I don't even know the process.

Is the author's answer correct ?

• @ChesterMiller I wonder what is wrong with what I am doing ? that will be more helpful. – A---B Apr 2 '17 at 2:58
• You can start by telling us where you got our equation for $\Delta H$ from. – Chet Miller Apr 2 '17 at 3:14
• The units for a appear to be wrong for a vdw gas, should they not be $\pu{dm^6\cdot bar \cdot mol^{-2}}$ assuming molar volumes.? – porphyrin Apr 2 '17 at 10:36
• @porphyrin Yes, I noticed that too. Also, that value of a seems very low to me (if the correct units involve Pa). – Chet Miller Apr 2 '17 at 10:55
• @Chester Miller, yes, but if we assume numerical value is ok then 3.6 is approx a for $\ce{CO2}$ but then b is almost ten times too big for this gas. In fact it seems to be too large for most any gas in $\pu{dm^3 mol^{-1}}$ . – porphyrin Apr 2 '17 at 12:53

The equation for $\Delta H$ for a van der Waals gas should be: $$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes). For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$

Explanation: In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by: $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$ where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).

The enthalpy H of the gas is equal to U + PV: $$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$ Combining the terms on the right hand side gives: $$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$ But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$ So, substituting into the equation for H gives: $$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$ But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure. So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$

• How did you get this equation ? – A---B Apr 2 '17 at 13:19
• @A---B See my explanation added to the Answer. – Chet Miller Apr 2 '17 at 15:08
• Thank you. It might take me some time to digest all this. – A---B Apr 2 '17 at 15:12

In addition to the method detailed by @Chester Miller, I offer an alternative derivation. It is often the case in thermodynamics that there are two or more ways of solving the same problem.

To calculate the change in enthalpy at constant temperature we can start with $$dH=TdS+VdP$$ and then find the change with volume and integrate in V at constant T. Thus

$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial S}{\partial V}\right)_T+ V\left(\frac{\partial P}{\partial V}\right)_T$$ using Maxwell’s equations

$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$ gives

$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T$$ which nicely give derivatives in P and V . Integrating means first working out the derivatives since $$\int dH=\int \left[ T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T\right ] dV$$ For the van der Waals gas $$P=\frac{RT}{V-b}- \frac{a}{V^2}$$ the derivatives are $$T\left(\frac{\partial P}{\partial T}\right)_V= \frac{RT}{V-b}$$ and
$$V\left(\frac{\partial P}{\partial V}\right)_T = -\frac{VRT}{(V-b)^2}+\frac{2a}{V^3}$$

substituting and working out the indefinite integral gives

$$H=-\frac{2a}{V}+\frac{bRT}{V-b} + const$$

Now the constant is the standard state value, chosen, say, to be at infinite volume, but we do not need to know what the standard state is as we only need the difference in H.

Now take the difference to get the change in enthalpy

$$\Delta H = -2a \left (\frac{1}{V_2} -\frac{1}{V_1} \right) +RTb\left ( \frac{1}{V_2-b} -\frac{1}{V_1-b} \right)$$

( As an aside, as the gas expands the work done can also be calculated using $w=-\int pdV$. This is $$w = -\int_{V_1}^{V_2} \left (\frac{RT}{V-b}-\frac{a}{V^2}\right )dv = -RT\ln \left(\frac{V_2-b}{V_1-b} \right )-a\left(\frac{1}{V_2}-\frac{1}{V_1} \right)$$ As the work done on an perfect gas is $w_{per}=-RT\ln(V_2/V_2)$ we can see that is the volume is large compared to b the work done by the van der Waals gas is less than that for the perfect gas as a result of intermolecular interactions.)

• Thanks for the answer. I prefer this derivation little more over that one. – A---B Apr 4 '17 at 14:55