NEET 2026 (Phase 1) · ChemistryPrevious Year Question
At a certain temperature $T$ (K), during a process, $500\ \text{J}$ is absorbed by the system and work of $200\ \text{J}$ is done by the system. Then the change in internal energy of the system is:
- A.$400\ \text{J}$
- B.$300\ \text{J}$✓
- C.$700\ \text{J}$
- D.$500\ \text{J}$
Correct Answer
(B) $300\ \text{J}$
Solution & Explanation
\textbf{Answer:} (B) $300\ \text{J}$ \textbf{Solution:} First law of thermodynamics: $\Delta U = q + w$. Heat absorbed by the system is positive, $q = +500\ \text{J}$. Work done by the system is negative, $w = -200\ \text{J}$. Therefore $\Delta U = 500 + (-200) = 300\ \text{J}$.
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