Following figure shows two processes A and B for a gas. If \[\Delta {{Q}_{A}}\]and \[\Delta {{Q}_{B}}\]are the amount of heat absorbed by the system in two cases, and \[\Delta {{U}_{A}}\]and \[\Delta {{U}_{B}}\]are changes in internal energies, respectively, then : [JEE Main 9-4-2019 Morning] |
A) \[\Delta {{Q}_{A}}=\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}\]
B) \[\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}\]
C) \[\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}>\Delta {{U}_{B}}\]
D) \[\Delta {{Q}_{A}}<\Delta {{Q}_{B}};\Delta {{U}_{A}}<\Delta {{U}_{B}}\]
Correct Answer: B
Solution :
[b] |
Initial and final states for both the processes are same. |
\[\therefore \]\[\Delta {{U}_{A}}=\Delta {{U}_{B}}\] |
Work done during process A is greater than in process B. |
By First Law of thermodynamics |
\[\Delta Q=\Delta U+W\]\[\Rightarrow \]\[\Delta {{Q}_{A}}>\Delta {{Q}_{B}}\] |
Option [b] |
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