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?

Explanation

To find the final internal energy of the gas, we can use the first law of thermodynamics, which is given by the equation:

\[ \Delta U = Q - W \]

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 1: Identify the values given in the problem
- Heat released by the gas, \(Q = -20 \, \text{J}\) (negative because heat is released).
- Work done on the gas, \(W = -8 \, \text{J}\) (negative because work is done on the gas).
- Initial internal energy, \(U_1 = 30 \, \text{J}\).

### Step 2: Calculate the change in internal energy (\(\Delta U\))
Using the first law of thermodynamics:

\[ \Delta U = Q - W \]

Substituting the values:

\[ \Delta U = (-20 \, \text{J}) - (-8 \, \text{J}) \]

\[ \Delta U = -20 \, \text{J} + 8 \, \text{J} \]

\[ \Delta U = -12 \, \text{J} \]

### Step 3: Calculate the final internal energy (\(U_2\))
The change in internal energy can also be expressed as:

\[ \Delta U = U_2 - U_1 \]

Rearranging this gives:

\[ U_2 = U_1 + \Delta U \]

Substituting the known values:

\[ U_2 = 30 \, \text{J} + (-12 \, \text{J}) \]

\[ U_2 = 30 \, \text{J} - 12 \, \text{J} \]

\[ U_2 = 18 \, \text{J} \]

### Final Answer
The final internal energy of the gas is \(U_2 = 18 \, \text{J}\).

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