For an ideal gas the instantaneous change in pressure p with volume V is given by the equation dp/dV

Q: Why do we take $n=1$ in the ideal gas equation?

The question explicitly states 'the maximum temperature one mole of gas can attain'. Therefore, the number of moles $n$ is exactly 1.

Q: How did we integrate $\frac{dp}{p}$?

The integral of $\frac{1}{x}$ with respect to $x$ is the natural logarithm, $\ln|x|$. So, $\int \frac{dp}{p} = \ln(p)$. Applying the limits from $p_0$ to $p$ gives $\ln(p) - \ln(p_0)$, which simplifies to $\ln(p/p_0)$.

Q: Why is the maximum temperature found by setting $\frac{dT}{dV} = 0$?

This is a fundamental principle of calculus. At a local maximum or minimum of a smooth continuous function, the tangent to the curve is perfectly horizontal, meaning its slope (the first derivative) is zero.

Q: What does the equation $p = p_0 e^{-aV}$ physically mean?

It describes a thermodynamic process where the pressure of the gas drops exponentially as its volume expands. It's not a standard isothermal or adiabatic process, but a specific polytropic-like path defined by the problem.

Q: When solving $\frac{dT}{dV} = 0$, why did we ignore the $e^{-aV}$ term?

An exponential function like $e^x$ or $e^{-aV}$ is strictly positive for all real numbers. It only approaches zero as the exponent goes to negative infinity. Since it can never actually equal zero for a finite volume, we can safely divide it out, leaving only the algebraic part $(1 - aV)$ to be zero.

Question

For an ideal gas the instantaneous change in pressure $p$ with volume $V$ is given by the equation $\frac{dp}{dV} = -ap$. If $p = p_0$ at $V = 0$ is the given boundary condition, then the maximum temperature one mole of gas can attain is (Here $R$ is the gas constant)

Options:

(A) $\frac{p_0}{aeR}$

(B) $\frac{ap_0}{cR}$

(C) infinity

(D) $0^\circ	ext{C}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents us with a fascinating thermodynamic process governed by a specific differential equation: $$\frac{dp}{dV} = -ap$$

This equation tells us that the rate at which pressure changes with respect to volume is proportional to the negative of the pressure itself. This is a classic signature of exponential decay. To understand the full picture, we need to solve this differential equation to find how pressure $p$ explicitly depends on volume $V$.

### The Master Equation

We start by separating the variables. We bring all the $p$ terms to one side and the $V$ terms to the other:

$$\frac{dp}{p} = -a \, dV$$

Now, we integrate both sides. The problem gives us a boundary condition: at $V = 0$, the pressure is $p_0$. So, we set our limits of integration accordingly:

$$\int_{p_0}^p \frac{dp}{p} = -a \int_0^V dV$$

Evaluating the integrals gives us the natural logarithm:

$$\ln\left(\frac{p}{p_0}\right) = -aV$$

Taking the antilog (exponential) of both sides, we arrive at the equation of state for this specific process:

$$p = p_0 e^{-aV}$$

### Finding the Temperature Function

Our ultimate goal is to find the maximum temperature. To do this, we need to express temperature $T$ as a function of volume $V$. This is where the Ideal Gas Law comes to the rescue:

$$pV = nRT$$

The problem specifies we are dealing with exactly one mole of gas, so $n = 1$. Substituting our expression for pressure into the Ideal Gas Law yields:

$$(p_0 e^{-aV})V = RT$$

Rearranging for $T$, we get a beautiful function:

$$T(V) = \frac{p_0}{R} V e^{-aV}$$

### The Calculus of Maximization

To find the maximum value of this temperature function, we turn to calculus. The maximum occurs where the derivative of temperature with respect to volume is zero. Let's differentiate $T(V)$ using the product rule:

$$\frac{dT}{dV} = \frac{p_0}{R} \left[ e^{-aV} \cdot 1 + V \cdot e^{-aV}(-a) \right] = 0$$

Factoring out the common exponential term:

$$\frac{p_0}{R} e^{-aV} (1 - aV) = 0$$

Since the exponential term $e^{-aV}$ is always positive and can never be zero for any finite volume, the term inside the parentheses must be zero:

$$1 - aV = 0 \implies V = \frac{1}{a}$$

This is our critical volume. It's the exact point during the expansion where the gas reaches its absolute peak temperature.

### Final Calculation

All that's left is to substitute this critical volume back into our temperature function to find $T_{\max}$:

$$T_{\max} = \frac{p_0}{R} \left(\frac{1}{a}\right) e^{-a(1/a)}$$

The $a$ terms in the exponent cancel out perfectly:

$$T_{\max} = \frac{p_0}{aR} e^{-1}$$

Which can be elegantly rewritten as:

$$T_{\max} = \frac{p_0}{aeR}$$

And there we have it! The maximum temperature the gas can attain.

Select Answer:

Visualized Solution

$Differential Equation of the Process$

$Integrating to find p (V)$

$Temperature as a function of Volume$

$Maximizing Temperature$

$Finding the Critical Volume$

$Calculating Maximum Temperature$

The Sigma Insight: Kinetic Theory of Gases and Gas Laws

Analyzing the Setup

The Master Equation

Finding the Temperature Function

The Calculus of Maximization

Final Calculation

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