Question

Derive a relation for the slope of the \(v=\) constant lines on a \(T-P\) diagram for a gas that obeys the van der Waals equation of state.

Short answer

The relationship for the slope of lines with constant volume (v) on a temperature (T) versus pressure (P) diagram for a gas adhering to the van der Waals equation is: $$\frac{dT}{dP} = \frac{v-b}{R}$$ This indicates that the slope of the constant volume lines is directly proportional to the quantity (v-b), and inversely proportional to the gas constant (R).

Step-by-step solution

Write the van der Waals equation

Let's start by writing out the van der Waals equation of state for a gas: $$\left(P+\frac{a}{v^2}\right)(v-b) = RT$$ where \(P\) is the pressure, \(v\) is the specific volume (volume per unit mass), \(T\) is the temperature, \(R\) is the gas constant, and \(a\) and \(b\) are constants.

Solve the van der Waals equation for temperature

Our task is to find the slope of \(v\)-constant lines in the \(T-P\) diagram. Thus, we should express \(T\) as a function of \(P\) and \(v\). To do this, let us solve the van der Waals equation for \(T\): $$T = \frac{P(v-b) + \frac{a}{v^2}(v-b)}{R}$$

Differentiate the temperature equation with respect to pressure

Now we need to find the slope of the lines determined by this equation, which means computing the derivative \(\frac{dT}{dP}\) while keeping \(v\) constant. So, let's differentiate the above equation with respect to \(P\): $$\frac{dT}{dP} = \frac{\partial}{\partial P} \left(\frac{P(v-b) + \frac{a}{v^2}(v-b)}{R}\right)$$ As \(v\) is constant, and \(a\), \(b\), and \(R\) are constants, the differentiation is fairly simple: $$\frac{dT}{dP} = \frac{v-b}{R}$$

State the answer

We have found the relation for the slope of the \(v\)-constant lines on a \(T-P\) diagram for a gas obeying the van der Waals equation of state: $$\frac{dT}{dP} = \frac{v-b}{R}$$ This implies that the slope of constant volume lines is directly proportional to \((v-b)\) and inversely proportional to the gas constant \(R\).