Search any question & find its solution
Question:
Answered & Verified by Expert
Match the following:
\(\begin{array}{|c|c|c|}
\hline & \text { List I } & \text { List II } \\
\hline \text { (A) } & \Delta U=W_{\mathrm{ad}} & \text { I. Isothermal reversible expansion } \\
\hline \text { (B) } & \Delta U=q-W & \text { II. Wall is adiabatic } \\
\hline(\mathrm{C}) & \Delta U=-q & \text { III. Thermally conducting walls } \\
\hline \text { (D) } & \Delta U=0 & \text { IV. Isolated system } \\
\hline & & \text { V. Closed system } \\
\hline
\end{array}\)
The correct answer is
Options:
\(\begin{array}{|c|c|c|}
\hline & \text { List I } & \text { List II } \\
\hline \text { (A) } & \Delta U=W_{\mathrm{ad}} & \text { I. Isothermal reversible expansion } \\
\hline \text { (B) } & \Delta U=q-W & \text { II. Wall is adiabatic } \\
\hline(\mathrm{C}) & \Delta U=-q & \text { III. Thermally conducting walls } \\
\hline \text { (D) } & \Delta U=0 & \text { IV. Isolated system } \\
\hline & & \text { V. Closed system } \\
\hline
\end{array}\)
The correct answer is
Solution:
2596 Upvotes
Verified Answer
The correct answer is:
\(\begin{array}{cc} A & B & C & D \\ II & V & III & I \end{array}\)
The correct match is
\(\mathrm{II} \rightarrow \mathrm{A}, \mathrm{V} \rightarrow \mathrm{B}, \mathrm{III} \rightarrow \mathrm{C}, \mathrm{I} \rightarrow \mathrm{D} .\)
Their explanations are as follows:
(D) \(\rightarrow\) I
In an isothermal reversible expansion
\(\begin{aligned}
\Delta T & =0 \\
\Delta U & =C_V, \Delta T=0
\end{aligned}\)
(A) \(\rightarrow\) II
If wall is adiabatic then
\(q=0\)
From first law of thermodynamics
\(\begin{aligned}
\Delta U & =q+W \\
\therefore \quad \Delta U & =W_{\text {adiabatic }}
\end{aligned}\)
(C) \(\rightarrow\) III
For thermally conducting walls
\(\Delta U=-q\)
(B) \(\rightarrow \mathbf{V}\)
For close system
\(q \neq 0, W \neq 0 \text { and } W=- \text { ve }\)
From first law of thermodynamics
\(\Delta U=q-W\)
\(\mathrm{II} \rightarrow \mathrm{A}, \mathrm{V} \rightarrow \mathrm{B}, \mathrm{III} \rightarrow \mathrm{C}, \mathrm{I} \rightarrow \mathrm{D} .\)
Their explanations are as follows:
(D) \(\rightarrow\) I
In an isothermal reversible expansion
\(\begin{aligned}
\Delta T & =0 \\
\Delta U & =C_V, \Delta T=0
\end{aligned}\)
(A) \(\rightarrow\) II
If wall is adiabatic then
\(q=0\)
From first law of thermodynamics
\(\begin{aligned}
\Delta U & =q+W \\
\therefore \quad \Delta U & =W_{\text {adiabatic }}
\end{aligned}\)
(C) \(\rightarrow\) III
For thermally conducting walls
\(\Delta U=-q\)
(B) \(\rightarrow \mathbf{V}\)
For close system
\(q \neq 0, W \neq 0 \text { and } W=- \text { ve }\)
From first law of thermodynamics
\(\Delta U=q-W\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.