Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J. Find work done by the gas in the given process.

Explanation

To find the work done by the gas in the given process, we can use the first law of thermodynamics, which is expressed as:

\[ \Delta Q = \Delta U + W \]

Where:
- \(\Delta Q\) is the heat added to the system (gas),
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done by the system.

### Step-by-step Solution:

1. **Identify the given values**:
- Heat taken from the gas, \(\Delta Q = -80 \, \text{J}\) (negative because heat is taken out).
- Increase in internal energy, \(\Delta U = 20 \, \text{J}\).

2. **Substitute the values into the first law of thermodynamics**:
\[ \Delta Q = \Delta U + W \]
Plugging in the known values:
\[ -80 \, \text{J} = 20 \, \text{J} + W \]

3. **Rearranging the equation to solve for work done \(W\)**:
\[ W = -80 \, \text{J} - 20 \, \text{J} \]
\[ W = -100 \, \text{J} \]

4. **Interpret the result**:
The negative sign indicates that the work is done on the gas, rather than by the gas.

### Final Answer:
The work done by the gas in the given process is \(-100 \, \text{J}\) (indicating work is done on the gas).

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