A sample of gas is compressed by an average pressure of 0.50 atmosphere so as to decrease its volume from `400cm^(3)` to `200cm^(3)`. During the process 8.00 J of heat flows out to surroundings. The change in internal energy of the system is

Options

A. `+2.13 J`
B. `+10.13J`
C. `-2.13J`
D. `-10.13J`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Values
- Initial volume (V1) = 400 cm³
- Final volume (V2) = 200 cm³
- Average external pressure (P) = 0.50 atm
- Heat flow (Q) = -8.00 J (negative because heat flows out)

### Step 2: Calculate the Change in Volume (ΔV)
\[ \Delta V = V2 - V1 = 200 \, \text{cm}^3 - 400 \, \text{cm}^3 = -200 \, \text{cm}^3 \]

### Step 3: Calculate the Work Done (W)
Using the formula for work done during compression:
\[ W = -P \Delta V \]
First, we need to convert ΔV from cm³ to liters:
\[ -200 \, \text{cm}^3 = -200 \times 10^{-3} \, \text{L} = -0.2 \, \text{L} \]
Now substituting the values:
\[ W = -0.50 \, \text{atm} \times (-0.2 \, \text{L}) = 0.10 \, \text{L atm} \]

### Step 4: Convert Work Done to Joules
Using the conversion factor \(1 \, \text{L atm} = 101.3 \, \text{J}\):
\[ W = 0.10 \, \text{L atm} \times 101.3 \, \text{J/L atm} = 10.13 \, \text{J} \]

### Step 5: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
Substituting the values:
\[ \Delta U = -8.00 \, \text{J} + 10.13 \, \text{J} = 2.13 \, \text{J} \]

### Final Answer
The change in internal energy of the system is:
\[ \Delta U = 2.13 \, \text{J} \]

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