The pressure of given mass of a gas in a thermodynamic system is changed in such a way that 20 joule of heat is released from the gas and 8 joule of work is done on the gas. If the initial internal energy of the gas was 30 joule then final internal energy will be

Options

A. (a)2 J
B. (b)42 J
C. (c)18 J
D. (d)58 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 (negative if heat is released),
- \(W\) is the work done by the system (negative if work is done on the system).

### Step-by-Step Solution:

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

2. **Apply the first law of thermodynamics:**
\[ \Delta U = Q - W \]

3. **Substitute the values into the equation:**
\[ \Delta U = (-20) - (-8) \]

4. **Simplify the equation:**
\[ \Delta U = -20 + 8 = -12 \, \text{J} \]

5. **Calculate the final internal energy:**
\[ U_f = U_i + \Delta U \]
\[ U_f = 30 + (-12) = 30 - 12 = 18 \, \text{J} \]

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