Question

The equation of state of n moles of a non-ideal gas can be approximated by the equation
$(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$
where a and b are constants characteristics of the gas. Which of the following can represent the equation of a quasistatic adiabat for this gas (Assume that $C_V$ , 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. $\left(\begin{array}{c}T+\frac{ab}{V^{2}R}\end{array}\right)(V-nb{)}^{R/C_v}=$ constant
• D. $\left(\begin{array}{c}T+\frac{n^{2}ab}{V^{2}R}\end{array}\right)(V-nb{)}^{C_v/R}=$ constant

Answer 1

The correct option is A $T(V-nb{)}^{R/C_v}=$ constant

For a reversible adiabatic process, we have $dS=0$ (Entropy change = 0)

The entropy equation is $TdS=n{C}_{V}dT+T(\frac{\partial P}{\partial T}{)}_{V}dV$

From the non-ideal gas equation, $(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$

$(\frac{\partial P}{\partial T}{)}_V=\frac{nR}{V-nb}$

for $dS=0$, we have

$n{C}_{V}dT=-T(\frac{\partial P}{\partial T}{)}_{V}dV=-nRT\frac{dV}{V-nb}$

$\Rightarrow \frac{dT}{T}=-\frac{nR}{C_V}\frac{dV}{V-nb}$

$\Rightarrow ln\left(\frac{T}{T_0}\right)=-\frac{R}{C_V}ln\left(\frac{V-nb}{V_0-nb}\right)$

$\Rightarrow T(V-nb{)}^{\frac{R}{C_V}}=T_0(V_0-nb{)}^{\frac{R}{C_V}}$

i.e., $T(V-nb{)}^{\frac{R}{C_V}}=\text{constant}$