Question

The equation of state of $n$ moles of a non-ideal gas can be approximated by the equation
$\left(p+\frac{n^{2}a}{V^2}\right)(V-nb)=nRT$
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. $T(V-nb{)}^{R/C_V}=$ constant
• B. $T(V-nb{)}^{C_v/R}=$ constant
• C. $\left(T+\frac{ab}{V^{2}R}\right)(V-nb{)}^{R/C_V}=$ constant
• D. $\left(T+\frac{n^{2}ab}{V^{2}R}\right)(V-nb{)}^{C_V/R}=$ constant

Answer 1

Answer: (A)

Given a van der Waal's gas with state equation,
$\left(p+\frac{n^{2}a}{V^2}\right)(V-nb)=nRT$
For adiabatic process, $dW=0$
So, from first law of thermodynamics, we have
$dU=dQ+dW$
$\Rightarrow dU-dW=0$
$\Rightarrow dU+pdV=0$
$\Rightarrow {\left(\frac{\partial U}{\partial T}\right)}_{V}dT+{\left(\frac{\partial U}{\partial V}\right)}_T$
$dV+\left(p+\frac{n^{2}a}{V^2}\right)dV=0$
$\Rightarrow C_{V}dT+\left(\frac{nkT}{V-nb}\right)dV=0$
$\Rightarrow C_V\frac{dT}{T}+\left(\frac{nk}{V-nb}\right)dV=0$
Integrating above equation, we get
$\Rightarrow C_V\int \frac{dT}{T}+nk\int \frac{dV}{V-nb}=$ constant
$\Rightarrow \log T^{C_V}+\log (V-nb{)}^{nk}=C$
$\Rightarrow (V-nb{)}^R\cdot T^{C_V}=e^C$
$\Rightarrow (V-nb{)}^{R/C_V}\cdot T=e^{C/C_V}=$ constant
$\therefore$ Equation for quasistatic adiabatic process is
$(V-nb{)}^{R/C_V}\cdot T=$ constant.