1 kg of a gas does 20 kJ of work and receives 16 kJ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of 9 kJ. The work done by the gas in the second expansion is :

Options

A. 32 kJ
B. 5 kJ
C. `-4kJ`
D. 13 kJ

Answer

To solve the problem step by step, we will use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). The equation can be expressed as:

\[ \Delta U = Q - W \]

### Step 1: Analyze the first expansion
1. We know that during the first expansion, the gas does 20 kJ of work (W = 20 kJ) and receives 16 kJ of heat (Q = 16 kJ).
2. We can use the first law of thermodynamics to find the change in internal energy (ΔU) for this process.

\[ \Delta U = Q - W \]
\[ \Delta U = 16 \text{ kJ} - 20 \text{ kJ} \]
\[ \Delta U = -4 \text{ kJ} \]

### Step 2: Analyze the second expansion
1. In the second expansion, the gas receives 9 kJ of heat (Q = 9 kJ) while transitioning from the same initial state to the same final state.
2. Since ΔU is a state function, it remains the same for both processes. Therefore, ΔU for the second expansion is also -4 kJ.

### Step 3: Apply the first law of thermodynamics again
1. We can again use the first law of thermodynamics to find the work done (W) in the second expansion.

\[ \Delta U = Q - W \]
\[ -4 \text{ kJ} = 9 \text{ kJ} - W \]

### Step 4: Solve for W
1. Rearranging the equation to solve for W:

\[ W = 9 \text{ kJ} + 4 \text{ kJ} \]
\[ W = 13 \text{ kJ} \]

### Conclusion
The work done by the gas in the second expansion is **13 kJ**.

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