In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas is

Options

A. 100 J
B. 300 J
C. 419 J
D. 24 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,
- \(W\) is the work done by the system.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Heat given to the gas, \(Q = 200 \, \text{Joules}\)
- Work done on the gas, \(W = 100 \, \text{Joules}\)

2. **Determine the Sign of Work:**
- Since work is done on the gas, we consider it as negative in our equation:
\[ W = -100 \, \text{Joules} \]

3. **Apply the First Law of Thermodynamics:**
Substitute the values into the equation:
\[ \Delta U = Q - W \]
\[ \Delta U = 200 \, \text{J} - (-100 \, \text{J}) \]

4. **Simplify the Equation:**
\[ \Delta U = 200 \, \text{J} + 100 \, \text{J} \]
\[ \Delta U = 300 \, \text{Joules} \]

5. **Conclusion:**
The change in internal energy of the gas is:
\[ \Delta U = 300 \, \text{Joules} \]

### Final Answer:
The change in internal energy of the gas is **300 Joules**.

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