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. `150j`
B. `70j`
C. `110j`
D. `40j`

Answer

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

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

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

### Step-by-Step Solution:

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

2. **Apply the First Law of Thermodynamics:**
According to the first law of thermodynamics, we can rearrange the equation to find the work done:
\[ \Delta W = \Delta Q - \Delta U \]

3. **Substitute the Known Values:**
Now, we substitute the values we have into the equation:
\[ \Delta W = 110 \, \text{J} - 40 \, \text{J} \]

4. **Calculate the Work Done:**
Performing the subtraction gives:
\[ \Delta W = 70 \, \text{J} \]

5. **Conclusion:**
The amount of external work done by the system is \(70 \, \text{J}\).

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

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