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)

• A. \(0^\circ\text{C}\)
• B. \(\frac{p_0}{aeR}\)
• C. \(\frac{ap_0}{eR}\)
• D. infinity

Hint:

When dealing with maxima/minima in thermodynamics, express the variable (like T) in terms of one parameter (like v) and differentiate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first find the pressure \( p \) as a function of volume \( v \) using the differential equation, then use the ideal gas law \( pv = RT \) to find temperature as a function of \( v \), and finally maximize it.
Step 2: Key Formula or Approach:
For 1 mole of ideal gas, \( T = \frac{pv}{R} \).
Integrate \(\frac{dp}{p} = -a \, dv\).
Step 3: Detailed Explanation:
\[ \int \frac{dp}{p} = \int -a \, dv \implies \ln p = -av + C \] Using boundary condition \( p = p_0 \) at \( v = 0 \):
\[ \ln p_0 = 0 + C \implies C = \ln p_0 \] \[ \ln \left(\frac{p}{p_0}\right) = -av \implies p = p_0 e^{-av} \] Now, expression for Temperature:
\[ T = \frac{pv}{R} = \frac{p_0 v e^{-av}}{R} \] To find maximum temperature, set \(\frac{dT}{dv} = 0\):
\[ \frac{dT}{dv} = \frac{p_0}{R} \left[ e^{-av} + v(-a)e^{-av} \right] = 0 \] \[ 1 - av = 0 \implies v = \frac{1}{a} \] Substitute \( v = 1/a \) in the temperature equation:
\[ T_{\text{max}} = \frac{p_0 (1/a) e^{-a(1/a)}}{R} = \frac{p_0}{a R e} \] Step 4: Final Answer:
The maximum temperature attained is \(\frac{p_0}{aeR}\).