For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation `(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. `0^(@)` C
B. `(ap_(0))/(eR)`
C. `(P_(0))/(aR)`
D. infinity

Answer

To solve the problem, we need to analyze the given equation and apply some thermodynamic principles. Let's break down the solution step by step.

### Step 1: Understand the given equation
We start with the equation that describes the change in pressure with respect to volume for an ideal gas:
\[ \frac{dp}{dv} = -ap \]
This indicates that the pressure decreases as the volume increases, where \( a \) is a constant.

### Step 2: Rearranging and integrating
We can rearrange the equation to separate variables:
\[ \frac{dp}{p} = -a \, dv \]
Now, we integrate both sides. The left side integrates to \(\ln p\) and the right side integrates to \(-av\):
\[ \int \frac{dp}{p} = \int -a \, dv \implies \ln p = -av + C \]
where \( C \) is the constant of integration.

### Step 3: Exponentiating to solve for pressure
Exponentiating both sides gives us:
\[ p = e^{-av + C} = e^C e^{-av} \]
Let \( e^C = p_0 \) (the pressure at \( v = 0 \)), so we can write:
\[ p = p_0 e^{-av} \]

### Step 4: Relate pressure to temperature
For an ideal gas, we know that:
\[ pV = nRT \]
Since we are dealing with one mole of gas (\( n = 1 \)), we can write:
\[ p = \frac{RT}{V} \]
Now, substituting our expression for \( p \) into this equation:
\[ p_0 e^{-av} = \frac{RT}{V} \]

### Step 5: Solve for temperature
Rearranging gives us:
\[ RT = p_0 e^{-av} V \]
Now, we need to express \( T \) in terms of \( V \):
\[ T = \frac{p_0 e^{-av} V}{R} \]

### Step 6: Maximize temperature with respect to volume
To find the maximum temperature, we need to differentiate \( T \) with respect to \( V \) and set the derivative equal to zero:
\[ \frac{dT}{dV} = \frac{p_0}{R} \left( e^{-av} + V(-a)e^{-av} \right) = 0 \]
Factoring out \( e^{-av} \) (which is never zero), we get:
\[ p_0 \left( 1 - aV \right) = 0 \]
Thus, setting \( 1 - aV = 0 \) gives:
\[ V = \frac{1}{a} \]

### Step 7: Substitute back to find maximum temperature
Substituting \( V = \frac{1}{a} \) back into the temperature equation:
\[ T_{\text{max}} = \frac{p_0 e^{-a \cdot \frac{1}{a}} \cdot \frac{1}{a}}{R} = \frac{p_0 e^{-1}}{aR} \]

### Final Answer
Thus, the maximum temperature one mole of gas can attain is:
\[ T_{\text{max}} = \frac{p_0}{aR e} \]