Question:medium
At a certain temperature \(K\), during a process, \(500\ J\) is absorbed by the system and work of \(200\ J\) is done by the system. Then change in internal energy of the system is:
At a certain temperature \(K\), during a process, \(500\ J\) is absorbed by the system and work of \(200\ J\) is done by the system. Then change in internal energy of the system is:
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Heat absorbed by system is positive, while work done by system is negative in the chemistry sign convention.
Updated On: May 28, 2026
- \(500\ J\)
- \(400\ J\)
- \(300\ J\)
- \(700\ J\)
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the change in internal energy ($\Delta U$) of a thermodynamic system. It provides two specific energy exchanges: heat ($q$) absorbed and work ($w$) performed. This requires an understanding of the First Law of Thermodynamics, which is essentially the law of conservation of energy applied to thermodynamic systems. We must be particularly careful with the sign conventions used in chemistry.
Step 2: Key Formula or Approach:
The First Law of Thermodynamics is expressed as: $\Delta U = q + w$. According to the IUPAC convention (standard for Chemistry):
Heat (q): If heat is absorbed/gained by the system, $q$ is positive ($+$). If heat is released, $q$ is negative ($-$).
Work (w): If work is done {on} the system (compression), $w$ is positive ($+$). If work is done {by} the system (expansion), $w$ is negative ($-$).
Step 3: Detailed Explanation:
Analyze Heat (q): The problem states $500 J$ is "absorbed by the system." This represents an influx of energy, so $q = +500 J$.
Analyze Work (w): The problem states $200 J$ of work is "done by the system." Since the system is spending energy to do work on its surroundings, the internal energy should decrease by this amount. Thus, $w = -200 J$.
Calculating $\Delta U$: Plug the values into the formula: \[ \Delta U = q + w \] \[ \Delta U = (+500 J) + (-200 J) \] \[ \Delta U = 500 - 200 = 300 J \]
The final result of $300 J$ means that although the system did some work, it gained more heat than it spent, resulting in a net increase of $300 J$ in its total internal energy.
Step 4: Final Answer:
The change in internal energy of the system is $300 J$.
The question asks for the change in internal energy ($\Delta U$) of a thermodynamic system. It provides two specific energy exchanges: heat ($q$) absorbed and work ($w$) performed. This requires an understanding of the First Law of Thermodynamics, which is essentially the law of conservation of energy applied to thermodynamic systems. We must be particularly careful with the sign conventions used in chemistry.
Step 2: Key Formula or Approach:
The First Law of Thermodynamics is expressed as: $\Delta U = q + w$. According to the IUPAC convention (standard for Chemistry):
Heat (q): If heat is absorbed/gained by the system, $q$ is positive ($+$). If heat is released, $q$ is negative ($-$).
Work (w): If work is done {on} the system (compression), $w$ is positive ($+$). If work is done {by} the system (expansion), $w$ is negative ($-$).
Step 3: Detailed Explanation:
Analyze Heat (q): The problem states $500 J$ is "absorbed by the system." This represents an influx of energy, so $q = +500 J$.
Analyze Work (w): The problem states $200 J$ of work is "done by the system." Since the system is spending energy to do work on its surroundings, the internal energy should decrease by this amount. Thus, $w = -200 J$.
Calculating $\Delta U$: Plug the values into the formula: \[ \Delta U = q + w \] \[ \Delta U = (+500 J) + (-200 J) \] \[ \Delta U = 500 - 200 = 300 J \]
The final result of $300 J$ means that although the system did some work, it gained more heat than it spent, resulting in a net increase of $300 J$ in its total internal energy.
Step 4: Final Answer:
The change in internal energy of the system is $300 J$.
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