During a process ideal gas expands to compress the spring such that `10kJ` energy is stored in the spring and `2kJ` heat is released from gas to surrounding find the change in internal energy of gas

Options

A. `12kJ`
B. `10kJ`
C. `8kJ`
D. `6kJ`

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**:
- Energy stored in the spring (work done by the gas) = \(10 \text{ kJ}\)
- Heat released from the gas to the surroundings = \(2 \text{ kJ}\)

2. **Assign the correct signs**:
- Since the gas releases heat, \(Q\) will be negative:
\[ Q = -2 \text{ kJ} \]
- The work done by the gas (compressing the spring) is positive:
\[ W = 10 \text{ kJ} \]

3. **Substitute the values into the first law of thermodynamics**:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = (-2 \text{ kJ}) - (10 \text{ kJ}) \]

4. **Calculate the change in internal energy**:
\[ \Delta U = -2 \text{ kJ} - 10 \text{ kJ} = -12 \text{ kJ} \]

5. **Interpret the result**:
The negative sign indicates that the internal energy of the gas has decreased. However, the question asks for the magnitude of the change in internal energy, which is:
\[ |\Delta U| = 12 \text{ kJ} \]

### Final Answer:
The change in internal energy of the gas is \(12 \text{ kJ}\).

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