In a given process work done on a gas is 40 J and increase in its internal energy is 10J. Find heat given or taken to/from the gas in this process.

Explanation

To solve the problem, we will use the first law of thermodynamics, which is given by the equation:

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

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Work done on the gas, \(\Delta W = -40 \, \text{J}\) (since work is done on the gas, we take it as negative).
- Increase in internal energy, \(\Delta U = 10 \, \text{J}\).

2. **Apply the first law of thermodynamics:**
\[ \Delta Q = \Delta U + \Delta W \]

3. **Substitute the known values into the equation:**
\[ \Delta Q = 10 \, \text{J} + (-40 \, \text{J}) \]

4. **Calculate \(\Delta Q\):**
\[ \Delta Q = 10 \, \text{J} - 40 \, \text{J} = -30 \, \text{J} \]

5. **Interpret the result:**
The negative sign indicates that heat is taken from the gas.

### Final Answer:
The heat given or taken from the gas in this process is \(-30 \, \text{J}\), which means that 30 J of heat is taken from the gas.