A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy `DeltaU` of the gas in joules will be

Options

A. 1136.25 J
B. `-500 J`
C. -505 J
D. +505 J

Explanation

To solve the problem, we will follow the steps outlined in the video transcript, applying the principles of thermodynamics to calculate the change in internal energy (ΔU) of the gas.

### Step-by-Step Solution:

1. **Understand the First Law of Thermodynamics**:
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (ΔQ) plus the work done on the system (ΔW).
\[ \Delta U = \Delta Q + \Delta W \]

2. **Identify the Type of System**:
The problem states that the gas is in a well-insulated container, which means it is an isolated system. For an isolated system, no heat is exchanged with the surroundings.
\[ \Delta Q = 0 \]

3. **Calculate the Work Done (ΔW)**:
The work done by the gas during expansion can be calculated using the formula:
\[ \Delta W = -P \Delta V \]
where \(P\) is the external pressure and \(\Delta V\) is the change in volume.

4. **Determine the Change in Volume (ΔV)**:
The initial volume (\(V_i\)) is 2.50 L and the final volume (\(V_f\)) is 4.50 L. Thus, the change in volume is:
\[ \Delta V = V_f - V_i = 4.50 \, \text{L} - 2.50 \, \text{L} = 2.00 \, \text{L} \]

5. **Substitute Values into the Work Done Formula**:
Given that the external pressure \(P\) is 2.5 atm, we can substitute the values into the work done formula:
\[ \Delta W = -P \Delta V = -2.5 \, \text{atm} \times 2.00 \, \text{L} \]

6. **Convert Work from atm·L to Joules**:
To convert the work from atm·L to Joules, we use the conversion factor \(1 \, \text{atm} \cdot \text{L} = 101.325 \, \text{J}\):
\[ \Delta W = -2.5 \times 2.00 \, \text{L} \times 101.325 \, \text{J/(atm·L)} = -5.0 \, \text{atm·L} \times 101.325 \, \text{J/(atm·L)} = -506.625 \, \text{J} \]

7. **Calculate the Change in Internal Energy (ΔU)**:
Since \(\Delta Q = 0\), we have:
\[ \Delta U = \Delta Q + \Delta W = 0 + (-506.625 \, \text{J}) = -506.625 \, \text{J} \]
Rounding this to three significant figures gives us:
\[ \Delta U \approx -505 \, \text{J} \]

### Final Answer:
The change in internal energy (ΔU) of the gas is approximately \(-505 \, \text{J}\).

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