Question

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:

• A. \(500\ J\)
• B. \(400\ J\)
• C. \(300\ J\)
• D. \(700\ J\)

Hint:

Heat absorbed by system is positive, while work done by system is negative in the chemistry sign convention.

Answer 1

The Correct Option is C

Step 1: Recall first law of thermodynamics.
The first law of thermodynamics is: \[ \Delta U=q+w. \] Here: \[ \Delta U=\text{change in internal energy}, \] \[ q=\text{heat absorbed by the system}, \] and: \[ w=\text{work done on the system}. \]

Step 2: Identify heat sign.
The system absorbs \(500\ J\) heat. Heat absorbed by the system is taken as positive. So: \[ q=+500\ J. \]

Step 3: Identify work sign.
Work of \(200\ J\) is done by the system. Work done by the system is taken as negative because energy leaves the system. So: \[ w=-200\ J. \]

Step 4: Calculate \(\Delta U\).
\[ \Delta U=q+w \] \[ \Delta U=500+(-200) \] \[ \Delta U=300\ J. \] Therefore, the change in internal energy is: \[ 300\ J. \]