Question

In the \(T\!-\!s\) diagram shown, several ideal–gas processes are sketched: \(aa'\), \(bb'\), \(cc'\), \(dd'\). Match each curve to the process it best represents. \begin{center} \includegraphics[width=0.7\textwidth]{04.jpeg} \end{center}

• A. \(aa'\) – Isentropic; \(bb'\) – Isothermal; \(cc'\) – Isobaric; \(dd'\) – Isochoric
• B. \(aa'\) – Isothermal; \(bb'\) – Isentropic; \(cc'\) – Isochoric; \(dd'\) – Isobaric
• C. \(aa'\) – Isothermal; \(bb'\) – Isentropic; \(cc'\) – Isobaric; \(dd'\) – Isochoric
• D. \(aa'\) – Isothermal; \(bb'\) – Isobaric; \(cc'\) – Isentropic; \(dd'\) – Isochoric

Hint:

On a \(T\!-\!s\) diagram: horizontal \(\Rightarrow\) isothermal, vertical \(\Rightarrow\) isentropic. Among the slanted curves from the same start, the one with \emph{larger} entropy change for a given \(T\) rise (further right) is isobaric because \(c_p > c_v\).

Answer 1

The Correct Option is B

Step 1: Signatures on a \(T\!-\!s\) plane.
- Isothermal: \(T=\text{const}\) \(\Rightarrow\) horizontal line.
- Isentropic: \(s=\text{const}\) \(\Rightarrow\) vertical line.
- Isochoric (ideal gas): \(ds=c_v\,dT/T \Rightarrow s=c_v\ln T+\text{const}\); rising curve.
- Isobaric (ideal gas): \(ds=c_p\,dT/T \Rightarrow s=c_p\ln T+\text{const}\); also rising, but for the same \(T\)-rise the entropy change is larger than isochoric since \(c_p > c_v\). Hence the isobaric curve lies \emph{further to the right} (greater \(s\)) than the isochoric one starting from the same state. Step 2: Identify the curves on the figure.
- \(aa'\) is horizontal \(\Rightarrow\) Isothermal.
- \(bb'\) is vertical \(\Rightarrow\) Isentropic.
- Of the two rising curves, the one farther to the right is Isobaric (\(dd'\)), the other is Isochoric (\(cc'\)). \(\boxed{aa': \text{isothermal},\; bb': \text{isentropic},\; cc': \text{isochoric},\; dd': \text{isobaric}}\)