THERMAL CYU 30

Ratio of Maximum to Minimum Temperature

The cyclic process is a circle in the $p-V$ indicator diagram. Based on the intercepts shown ($V$ from 1 to 3, $p$ from 1 to 3), the center of the circle is at $(2V_0, 2p_0)$ and the radius is $r = V_0$ (in volume units) and $r = p_0$ (in pressure units).

We can parameterize the pressure and volume as:

$$p = p_0(2 + \cos\theta)$$ $$V = V_0(2 + \sin\theta)$$

Since the gas is ideal, $T \propto pV$. We need to maximize and minimize the product $y = (2 + \cos\theta)(2 + \sin\theta)$.

$$y = 4 + 2(\sin\theta + \cos\theta) + \sin\theta\cos\theta$$

Let $x = \sin\theta + \cos\theta = \sqrt{2}\sin(\theta + \frac{\pi}{4})$. The range of $x$ is $[-\sqrt{2}, \sqrt{2}]$.

We know that $\sin\theta\cos\theta = \frac{x^2 – 1}{2}$. Substituting this back:

$$y = 4 + 2x + \frac{x^2 – 1}{2} = \frac{x^2}{2} + 2x + \frac{7}{2}$$

This is a parabola opening upwards. The maximum and minimum values occur at the boundaries of $x$.

• Maximum Temperature ($x = \sqrt{2}$):
  $y_{max} = 4 + 2\sqrt{2} + \frac{(\sqrt{2})^2 – 1}{2} = 4 + 2\sqrt{2} + 0.5 = 4.5 + 2\sqrt{2}$
• Minimum Temperature ($x = -\sqrt{2}$):
  $y_{min} = 4 – 2\sqrt{2} + 0.5 = 4.5 – 2\sqrt{2}$

The ratio of temperatures is:

$$\text{Ratio} = \frac{4.5 + 2\sqrt{2}}{4.5 – 2\sqrt{2}} = \frac{9 + 4\sqrt{2}}{9 – 4\sqrt{2}}$$

We can rewrite the numerator and denominator as perfect squares:

$$9 + 4\sqrt{2} = (2\sqrt{2} + 1)^2$$ $$9 – 4\sqrt{2} = (2\sqrt{2} – 1)^2$$