Question

The equation of state of $n$ moles of a non-ideal gas can be approximated by the equation
$\left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T$
where $a$ and $b$ are constant characteristics of the gas. Which of the following can represent the equation of a quasistatic adiabat for this gas (assume that, $C_{V}$ is the molar heat capacity at constant volume is independent of temperature)?

• A. $T(V-n b)^{R / C_{V}}=$ constant
• B. $T(V-n b)^{C_{v} / R}=$ constant
• C. $\left(T+\frac{a b}{V^{2} R}\right)(V-n b)^{R / C_{V}}=$ constant
• D. $\left(T+\frac{n^{2} a b}{V^{2} R}\right)(V-n b)^{C_{V} / R}=$ constant

Answer 1

Answer: (A)

Given a van der Waal's gas with state equation,
$\left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T$
For adiabatic process, $d W=0$
So, from first law of thermodynamics, we have
$d U =d Q+d W $
$\Rightarrow d U-d W =0 $
$\Rightarrow d U+p d V=0$
$\Rightarrow \left(\frac{\partial U}{\partial T}\right)_{V} d T+\left(\frac{\partial U}{\partial V}\right)_{T} $
$d V+\left(p+\frac{n^{2} a}{V^{2}}\right) d V=0$
$\Rightarrow C_{V} d T+\left(\frac{n k T}{V-n b}\right) d V=0 $
$\Rightarrow C_{V} \frac{d T}{T}+\left(\frac{n k}{V-n b}\right) d V=0$
Integrating above equation, we get
$\Rightarrow C_{V} \int \frac{d T}{T}+n k \int \frac{d V}{V-n b} =$ constant
$\Rightarrow \log T^{C_{V}}+\log (V-n b)^{n k} =C $
$\Rightarrow (V-n b)^{R} \cdot T^{C_{V}} =e^{C} $
$\Rightarrow (V-n b)^{R / C_{V}} \cdot T=e^{C / C_{V}} =$ constant
$\therefore $ Equation for quasistatic adiabatic process is
$(V-n b)^{R / C_{V}} \cdot T=$ constant.