n moles of an ideal gas undergoes a process A and B as shown in the figure. The maximum temperature

Q: Why isn't the temperature maximum at point A or B?

Temperature is proportional to the product $pV$. At point A, $pV = 2p_0V_0$. At point B, $pV = 2p_0V_0$. However, in the middle of the line, the product $pV$ is higher (e.g., at the midpoint, $pV = 2.25p_0V_0$). Thus, the temperature peaks somewhere in between.

Q: Could we have expressed temperature as a function of pressure instead of volume?

Absolutely! You could write $V$ in terms of $p$ and substitute it into $pV = nRT$ to get $T(p)$. Differentiating $T$ with respect to $p$ would yield the exact same maximum temperature.

Q: What does the condition $\frac{dT}{dV} = 0$ physically mean here?

It represents the exact moment during the expansion where the gas stops heating up and starts cooling down. It's the peak of the thermal energy curve for this specific process.

Q: Is the maximum temperature always at the exact midpoint of a straight line $p-V$ process?

Yes, for any straight line process with a negative slope in the $p-V$ diagram, the maximum temperature (and maximum $pV$ product) always occurs exactly at the geometric midpoint of the line segment connecting the intercepts.

Q: Why do we draw isotherms as hyperbolas?

An isotherm represents a constant temperature process. According to Boyle's Law, $pV = \text{constant}$. Mathematically, an equation of the form $xy = c$ graphs as a rectangular hyperbola.

Question

$n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be

Options:

(A) $\frac{9}{4}\frac{p_0V_0}{nR}$

(B) $\frac{3}{2}\frac{p_0V_0}{nR}$

(C) $\frac{9}{2}\frac{p_0V_0}{nR}$

(D) $\frac{9p_0V_0}{nR}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are tracking the state of an ideal gas as it undergoes a specific thermodynamic process. The path it takes is represented as a straight line on a $p-V$ diagram, starting from state $A(V_0, 2p_0)$ and ending at state $B(2V_0, p_0)$. Our mission is to find the absolute maximum temperature the gas reaches during this journey.

At first glance, you might think the maximum temperature occurs at one of the endpoints. However, if we calculate the product $pV$ at both $A$ and $B$, we get $2p_0V_0$ for both. Since $T \propto pV$, the temperature is identical at the start and the end! This means the gas must heat up, reach a peak temperature, and then cool back down. Let's find that peak.

### The Master Equation
To track the temperature continuously, we need a mathematical function that relates pressure and volume along this specific path. Since the path is a straight line, we can use the two-point form of a linear equation:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

Substituting our thermodynamic coordinates $(V, p)$:

$$p - 2p_0 = \frac{p_0 - 2p_0}{2V_0 - V_0}(V - V_0)$$

Simplifying the slope gives us $-\frac{p_0}{V_0}$. Distributing this and rearranging the terms, we obtain the pressure as a function of volume:

$$p = -\frac{p_0}{V_0}V + 3p_0$$

### The Calculus of Thermodynamics
Now, we bring in the Ideal Gas Law, $pV = nRT$. By multiplying our pressure equation by $V$, we can express the temperature $T$ entirely as a quadratic function of $V$:

$$pV = -\frac{p_0}{V_0}V^2 + 3p_0V$$

$$nRT = -\frac{p_0}{V_0}V^2 + 3p_0V$$

$$T = \frac{1}{nR}\left(-\frac{p_0}{V_0}V^2 + 3p_0Vight)$$

This is a downward-opening parabola, confirming our intuition that the temperature rises to a maximum and then falls. To find this maximum, we apply the standard calculus technique of setting the derivative to zero:

$$\frac{dT}{dV} = \frac{1}{nR}\left(-\frac{2p_0}{V_0}V + 3p_0ight) = 0$$

Solving for $V$, we find that the maximum temperature occurs exactly at:

$$V = \frac{3}{2}V_0$$

### Final Calculation and Geometric Intuition
Finally, we substitute this optimal volume back into our temperature equation to find the peak value:

$$T_{max} = \frac{1}{nR}\left(-\frac{p_0}{V_0}\left(\frac{3}{2}V_0ight)^2 + 3p_0\left(\frac{3}{2}V_0ight)ight)$$

$$T_{max} = \frac{1}{nR}\left(-\frac{9}{4}p_0V_0 + \frac{9}{2}p_0V_0ight)$$

**$$T_{max} = \frac{9}{4}\frac{p_0V_0}{nR}$$**

Geometrically, this result is beautiful. Isotherms (lines of constant temperature) are rectangular hyperbolas defined by $pV = 	ext{constant}$. The maximum temperature occurs at the exact point where our straight-line process is perfectly tangent to the highest possible isotherm. For any straight line with a negative slope, this point of tangency is always exactly at its geometric midpoint!

Animated Solution for Physics - Thermodynamics: $n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be

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Visualized Solution

$p - V$ Diagram Analysis

Mathematical Tools

Coordinate Substitution

Pressure as a Function of Volume

Temperature Function

Maximizing Temperature

Final Calculation

Geometric Interpretation

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup Imagine you are tracking the state of an ideal gas as it undergoes a specific thermodynamic process

The Master Equation To track the temperature continuously, we need a mathematical function that relates pressure and volume along this specific path

The Calculus of Thermodynamics Now, we bring in the Ideal Gas Law, $p V = n R T$

Final Calculation and Geometric Intuition

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