A closed system undergoes a change of state by process `1rightarrow2`for which`Q_12=10J` and `W_12=-5j`. The system is now returned to its initial state by a different path `2rightarrow1` for which `Q_21is-3J`. The work done by the gas in the process `2rightarrow1`is

Options

A. 8 J
B. Zero
C. `-2 J`
D. `+5 j`

Answer

To solve the problem step by step, 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 1: Calculate the change in internal energy (\(\Delta U\)) for the process from state 1 to state 2.

Given:
- \(Q_{12} = 10 \, J\)
- \(W_{12} = -5 \, J\)

Using the first law of thermodynamics:

\[ \Delta U = Q_{12} - W_{12} \]

Substituting the values:

\[ \Delta U = 10 \, J - (-5 \, J) = 10 \, J + 5 \, J = 15 \, J \]

### Step 2: Use the internal energy to find the work done (\(W_{21}\)) for the process from state 2 to state 1.

In the process from state 2 to state 1, we know:
- \(Q_{21} = -3 \, J\)

We also know that the change in internal energy (\(\Delta U\)) remains the same when returning to the initial state (from state 2 to state 1):

\[ \Delta U = Q_{21} - W_{21} \]

Substituting the known values:

\[ 15 \, J = -3 \, J - W_{21} \]

### Step 3: Rearranging the equation to solve for \(W_{21}\).

Rearranging gives:

\[ W_{21} = -3 \, J - 15 \, J \]

Calculating this:

\[ W_{21} = -18 \, J \]

### Final Answer:
The work done by the gas in the process from state 2 to state 1 is \(-18 \, J\).

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