A gas expands in an insulated container against a constant external pressure of 1.5 atm from an initial volume of 1.2 L to 4.2 L. The change in internal energy (`DeltaU`) of gas is nearly (1 L-atm = 101.3 J)

Options

A. `- 456 J`
B. `+ 456 J`
C. `- 402 J`
D. `+ 402 J`

Answer

To solve the problem of finding the change in internal energy (ΔU) of the gas during its expansion in an insulated container, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- External pressure (P) = 1.5 atm
- Initial volume (V1) = 1.2 L
- Final volume (V2) = 4.2 L

2. **Calculate the Change in Volume (ΔV):**
\[ \Delta V = V2 - V1 = 4.2 \, \text{L} - 1.2 \, \text{L} = 3.0 \, \text{L} \]

3. **Determine the Work Done (W):**
Since the gas expands against a constant external pressure, the work done on the gas can be calculated using the formula:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = -1.5 \, \text{atm} \times 3.0 \, \text{L} = -4.5 \, \text{L-atm} \]

4. **Convert Work Done to Joules:**
We know that 1 L-atm = 101.3 J, so:
\[ W = -4.5 \, \text{L-atm} \times 101.3 \, \text{J/L-atm} = -456 \, \text{J} \]

5. **Calculate the Change in Internal Energy (ΔU):**
For an insulated container, there is no heat exchange (Q = 0), so:
\[ \Delta U = Q + W = 0 + W = W \]
Thus,
\[ \Delta U = -456 \, \text{J} \]

### Final Answer:
The change in internal energy (ΔU) of the gas is approximately **-456 J**.

---