`500cm^(3)` of a sample of an ideal gas is compressed by an average pressure of `0.1 atm` of `250 cm^(3)`. During this process, `10J` of heat flows out to the surroundings. Calculate the change in internal enegry of the system.

Options

A. `-2.16 J`
B. 12.156 J
C. -7.46 J
D. 101.3 J

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q + W \]

Where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat added to the system (negative if heat is lost)
- \(W\) = work done on the system (negative if work is done by the system)

### Step 1: Write down the given data
- Initial volume, \(V_i = 500 \, \text{cm}^3\)
- Final volume, \(V_f = 250 \, \text{cm}^3\)
- Average pressure, \(P = 0.1 \, \text{atm}\)
- Heat flowing out, \(Q = -10 \, \text{J}\) (since heat is lost)

### Step 2: Calculate the change in volume
\[ \Delta V = V_f - V_i = 250 \, \text{cm}^3 - 500 \, \text{cm}^3 = -250 \, \text{cm}^3 \]
Convert to liters:
\[ \Delta V = -250 \, \text{cm}^3 = -0.25 \, \text{L} \]

### Step 3: Calculate the work done on the system
Using the formula for work done:
\[ W = -P \Delta V \]
Convert pressure from atm to Joules:
\[ 1 \, \text{atm} = 101.3 \, \text{J/L} \]
Now substituting the values:
\[ W = -0.1 \, \text{atm} \times (-0.25 \, \text{L}) = 0.025 \, \text{L} \cdot \text{atm} \]
Convert \(W\) to Joules:
\[ W = 0.025 \, \text{L} \cdot \text{atm} \times 101.3 \, \text{J/L} = 2.5325 \, \text{J} \]

### Step 4: Apply the first law of thermodynamics
Now substitute \(Q\) and \(W\) into the first law equation:
\[ \Delta U = Q + W \]
\[ \Delta U = -10 \, \text{J} + 2.5325 \, \text{J} \]
\[ \Delta U = -7.4675 \, \text{J} \]

### Final Answer
The change in internal energy of the system is approximately:
\[ \Delta U \approx -7.47 \, \text{J} \]