To find the final internal energy of the gas, we can 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 (positive if heat is added, negative if heat is removed),
- \(W\) is the work done by the system (positive if work is done by the system, negative if work is done on the system).
### Step-by-step Solution:
1. **Identify the Given Values:**
- Heat given out by the gas, \(Q = -20 \, \text{J}\) (since heat is given out, it is negative).
- Work done on the gas, \(W = -10 \, \text{J}\) (since work is done on the gas, it is negative).
- Initial internal energy, \(U_i = 40 \, \text{J}\).
2. **Calculate the Change in Internal Energy (\(\Delta U\)):**
Using the first law of thermodynamics:
\[
\Delta U = Q - W
\]
Substitute the values:
\[
\Delta U = -20 \, \text{J} - (-10 \, \text{J}) = -20 \, \text{J} + 10 \, \text{J} = -10 \, \text{J}
\]
3. **Determine the Final Internal Energy (\(U_f\)):**
The change in internal energy can also be expressed as:
\[
\Delta U = U_f - U_i
\]
Rearranging gives:
\[
U_f = U_i + \Delta U
\]
Substitute the known values:
\[
U_f = 40 \, \text{J} + (-10 \, \text{J}) = 40 \, \text{J} - 10 \, \text{J} = 30 \, \text{J}
\]
4. **Final Answer:**
The final internal energy of the gas is:
\[
U_f = 30 \, \text{J}
\]
