A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 `dm^(3)` to a volume of 20 `dm^(3)`. It absorbs 800 J of thermal energy from its surroundings. The `Delta`U is

Options

A. `-312` J
B. `+123` J
C. `+312` J
D. `-213` J

Answer

To solve the problem, we will follow the steps outlined below:

### Step 1: Identify Given Information
- Initial volume (V1) = 10 dm³
- Final volume (V2) = 20 dm³
- External pressure (P_ext) = 1 atm
- Heat absorbed (q) = 800 J

### Step 2: Calculate the Change in Volume (ΔV)
\[ \Delta V = V2 - V1 = 20 \, \text{dm}^3 - 10 \, \text{dm}^3 = 10 \, \text{dm}^3 \]

### Step 3: Calculate Work Done (W)
The work done by the gas during expansion against a constant external pressure is given by:
\[ W = -P_{ext} \Delta V \]
First, we need to convert the volume from dm³ to liters (1 dm³ = 1 L):
\[ W = -1 \, \text{atm} \times 10 \, \text{L} \]

Next, we convert the work from liter-atmospheres to joules. The conversion factor is:
\[ 1 \, \text{L} \cdot \text{atm} = 101.325 \, \text{J} \]
Thus,
\[ W = -1 \times 10 \times 101.325 \, \text{J} = -1013.25 \, \text{J} \]

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = q + W \]
Substituting the values we have:
\[ \Delta U = 800 \, \text{J} + (-1013.25 \, \text{J}) \]
\[ \Delta U = 800 \, \text{J} - 1013.25 \, \text{J} = -213.25 \, \text{J} \]

### Step 5: Round the Result
Rounding to three significant figures, we get:
\[ \Delta U \approx -213 \, \text{J} \]

### Final Answer
\[ \Delta U = -213 \, \text{J} \]

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