110 J of heat is added to a gaseous system, whose internal energy change is 40 j. then the amount of external work done is

Options

A. 150 J
B. 70 J
C. 110 J
D. 40 J

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q + W \]

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Heat added to the system, \(Q = 110 \, \text{J}\)
- Change in internal energy, \(\Delta U = 40 \, \text{J}\)

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

3. **Rearrange the equation to find the work done \(W\):**
\[ W = \Delta U - Q \]

4. **Substitute the known values into the equation:**
\[ W = 40 \, \text{J} - 110 \, \text{J} \]

5. **Calculate \(W\):**
\[ W = 40 - 110 = -70 \, \text{J} \]

6. **Conclusion:**
The amount of external work done is \(-70 \, \text{J}\).

### Final Answer:
The external work done is \(-70 \, \text{J}\).

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