Question

Three different processes that can occur in an ideal monoatomic gas are shown in the \(P\) vs \(V\) diagram. The paths are labelled as \(A\rightarrow B\), \(A\rightarrow C\) and \(A\rightarrow D\). The change in internal energies during these process are taken as \(E_{AB}\), \(E_{AC}\) and \(E_{AD}\) and the work done as \(W_{AB}\), \(W_{AC}\) and \(W_{AD}\).
The correct relation between these parameters are :

• A. \(E_{AB}\lt E_{AC}\lt E_{AD},\ W_{AB}\gt 0,\ W_{AC}\gt W_{AD}\)
• B. \(E_{AB}=E_{AC}=E_{AD},\ W_{AB}\gt 0,\ W_{AC}=0,\ W_{AD}\lt 0\)
• C. \(E_{AB}\gt E_{AC}\gt E_{AD},\ W_{AB}\lt W_{AC}\lt W_{AD}\)
• D. \(E_{AB}=E_{AC}\lt E_{AD},\ W_{AB}\gt 0,\ W_{AC}=0,\ W_{AD}\lt 0\)

Hint:

Internal energy of an ideal gas depends only on temperature, and work is the signed area under the path, so check where B, C, D lie on the isotherms and whether volume increases or decreases.

Answer 1

The Correct Option is B

In the diagram points B, C and D lie on the same upper isotherm \(T_1\), while A lies on the lower isotherm \(T_2\) (\(T_1\gt T_2\)).
For an ideal gas \(\Delta E=nC_v\Delta T\), so \(E_{AB}=E_{AC}=E_{AD}=nC_v(T_1-T_2)\).
Work done: \(A\rightarrow B\) is an isobaric expansion (volume increases), so \(W_{AB}=P\Delta V\gt 0\).
\(A\rightarrow C\) is at constant volume, so \(W_{AC}=0\).
\(A\rightarrow D\) ends at a smaller volume than A (D is to the left of A), so the gas is compressed and \(W_{AD}\lt 0\).
Hence \(E_{AB}=E_{AC}=E_{AD}\), \(W_{AB}\gt 0\), \(W_{AC}=0\), \(W_{AD}\lt 0\).
Answer: (2) \(E_{AB}=E_{AC}=E_{AD},\ W_{AB}\gt 0,\ W_{AC}=0,\ W_{AD}\lt 0\)