The volume of a gas decreases from `500 c c` to `300 c c` when a sample of gas is compressed by an average pressure of 0.6 atm. During this process 10 J of heat is liberated. The change in internal energy is

Options

A. `-2.16 J`
B. `12.156 J`
C. `2.16 J`
D. `101.3 J`

Explanation

To solve the problem, we can 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 on the system (negative if work is done by the system).

### Step 1: Identify the values given in the problem
- Initial volume (\(V_i\)) = 500 cc
- Final volume (\(V_f\)) = 300 cc
- Average pressure (\(P\)) = 0.6 atm
- Heat liberated (\(Q\)) = -10 J (since heat is released)

### Step 2: Calculate the work done on the system
The work done on the gas during compression can be calculated using the formula:

\[ W = -P \Delta V \]

Where:
- \(\Delta V = V_f - V_i\)

Calculating \(\Delta V\):

\[ \Delta V = 300 \, \text{cc} - 500 \, \text{cc} = -200 \, \text{cc} \]

Now, convert cc to liters for the work calculation. Since \(1 \, \text{cc} = 0.001 \, \text{L}\):

\[ \Delta V = -200 \, \text{cc} = -0.2 \, \text{L} \]

Now, we can calculate the work done:

\[ W = -P \Delta V = -0.6 \, \text{atm} \times (-0.2 \, \text{L}) \]

To convert atm·L to Joules, we use the conversion factor \(1 \, \text{atm} \cdot \text{L} = 101.325 \, \text{J}\):

\[ W = 0.6 \times 0.2 \times 101.325 = 12.159 \, \text{J} \]

### Step 3: Apply the first law of thermodynamics
Now we can find the change in internal energy:

\[ \Delta U = Q + W \]

Substituting the values we have:

\[ \Delta U = -10 \, \text{J} + 12.159 \, \text{J} \]

Calculating \(\Delta U\):

\[ \Delta U = 2.159 \, \text{J} \]

### Final Answer
The change in internal energy (\(\Delta U\)) is approximately **2.16 J**.

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