A gas performs 0.320 kJ work on surroundings and absorbs 120 J of heat from the surroundings.Hence change in internal energy is

Options

A. 120.32 J
B. `-200` J
C. 400 J
D. 200 J

Explanation

To find the change in internal energy (\( \Delta U \)) of the gas, we can use the first law of thermodynamics, which is given by the equation:

\[ \Delta U = q + W \]

Where:
- \( \Delta U \) = change in internal energy
- \( q \) = heat absorbed by the system
- \( W \) = work done on the system

### Step 1: Identify the values given in the problem
- Work done by the gas on the surroundings = \( 0.320 \, \text{kJ} \)
- Heat absorbed from the surroundings = \( 120 \, \text{J} \)

### Step 2: Convert all units to the same system
Since the work is given in kilojoules and heat in joules, we need to convert kilojoules to joules for consistency.

\[ 0.320 \, \text{kJ} = 0.320 \times 1000 \, \text{J} = 320 \, \text{J} \]

### Step 3: Assign the correct signs to \( q \) and \( W \)
- Since the gas absorbs heat, \( q = +120 \, \text{J} \).
- The work done by the gas on the surroundings is considered negative, so \( W = -320 \, \text{J} \).

### Step 4: Substitute the values into the equation
Now we can substitute the values into the first law of thermodynamics equation:

\[ \Delta U = q + W = 120 \, \text{J} - 320 \, \text{J} \]

### Step 5: Calculate \( \Delta U \)
\[ \Delta U = 120 \, \text{J} - 320 \, \text{J} = -200 \, \text{J} \]

### Final Answer
The change in internal energy (\( \Delta U \)) is:

\[ \Delta U = -200 \, \text{J} \]