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. `-213` J
D. `+231 J`

Explanation

To find the change in internal energy (ΔU) of the gas during its isothermal expansion, we will use the first law of thermodynamics, which is given by the equation:

\[ \Delta U = q + w \]

where:
- \( \Delta U \) = change in internal energy
- \( q \) = heat exchanged (absorbed or released)
- \( w \) = work done on or by the system

### Step 1: Identify the given values
- The gas absorbs thermal energy, so \( q = +800 \, \text{J} \).
- The gas expands from a volume of \( V_1 = 10 \, \text{dm}^3 \) to \( V_2 = 20 \, \text{dm}^3 \).
- The external pressure \( P_{\text{ext}} = 1 \, \text{atm} \).

### Step 2: Calculate the work done (w)
The work done by the gas during expansion against a constant external pressure is given by:

\[ w = -P_{\text{ext}} \cdot (V_2 - V_1) \]

First, we need to convert the volumes from dm³ to liters (though they are numerically the same, it's good to note):

- \( V_1 = 10 \, \text{dm}^3 = 10 \, \text{L} \)
- \( V_2 = 20 \, \text{dm}^3 = 20 \, \text{L} \)

Now, calculate the change in volume:

\[ V_2 - V_1 = 20 \, \text{L} - 10 \, \text{L} = 10 \, \text{L} \]

Now substitute into the work formula:

\[ w = -1 \, \text{atm} \cdot 10 \, \text{L} \]

To convert atm·L to joules, we use the conversion factor \( 1 \, \text{atm} \cdot \text{L} \approx 101.3 \, \text{J} \):

\[ w = -10 \, \text{L} \cdot 1 \, \text{atm} \cdot 101.3 \, \text{J/atm·L} = -1013 \, \text{J} \]

### Step 3: Apply the first law of thermodynamics
Now substitute the values of \( q \) and \( w \) into the first law equation:

\[ \Delta U = q + w \]

\[ \Delta U = 800 \, \text{J} + (-1013 \, \text{J}) \]

\[ \Delta U = 800 \, \text{J} - 1013 \, \text{J} \]

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

### Final Answer
The change in internal energy \( \Delta U \) is:

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

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