In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release `20 J` of heat and `8 J` of work is done on the gas. If initial internal energy of the gas was `30 J`, what will be the final internal energy?

Options

A. 2J
B. 18J
C. 42J
D. 58J

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 (positive if added, negative if released),
- \(W\) is the work done on the system (positive if done on the system, negative if done by the system).

### Step-by-Step Solution:

1. **Identify the given values:**
- Heat released by the gas, \(Q = -20 \, \text{J}\) (since heat is released, it is negative).
- Work done on the gas, \(W = -8 \, \text{J}\) (since work is done on the gas, it is also negative).
- Initial internal energy, \(U_i = 30 \, \text{J}\).

2. **Apply the first law of thermodynamics:**
\[ \Delta U = Q + W \]
Substituting the values we have:
\[ \Delta U = -20 \, \text{J} + (-8 \, \text{J}) = -20 \, \text{J} - 8 \, \text{J} = -28 \, \text{J} \]

3. **Calculate the final internal energy:**
The change in internal energy is related to the initial and final internal energy:
\[ \Delta U = U_f - U_i \]
Rearranging gives:
\[ U_f = \Delta U + U_i \]
Substituting the values:
\[ U_f = -28 \, \text{J} + 30 \, \text{J} = 2 \, \text{J} \]

4. **Final Answer:**
The final internal energy of the gas is \(U_f = 2 \, \text{J}\).

### Summary of the Steps:
- Identify the signs of heat and work based on the problem statement.
- Use the first law of thermodynamics to find the change in internal energy.
- Calculate the final internal energy using the initial internal energy and the change.