To find the change in internal energy (ΔU) of the gas, we can follow these steps:
### Step 1: Understand the relationship between ΔU, Q, and W
The change in internal energy (ΔU) is given by the equation:
\[
\Delta U = Q + W
\]
where:
- \(Q\) is the heat transfer,
- \(W\) is the work done by the system.
### Step 2: Identify the type of process
Since the gas is expanding in a well-insulated container, it is an adiabatic process. In an adiabatic process, there is no heat transfer, so:
\[
Q = 0
\]
Thus, the equation simplifies to:
\[
\Delta U = W
\]
### Step 3: Calculate the work done (W)
The work done by the gas during expansion against a constant external pressure is given by:
\[
W = -P_{\text{ext}} \Delta V
\]
where:
- \(P_{\text{ext}}\) is the external pressure,
- \(\Delta V\) is the change in volume.
### Step 4: Determine the change in volume (ΔV)
Given:
- Initial volume \(V_1 = 2.50 \, \text{L}\)
- Final volume \(V_2 = 4.50 \, \text{L}\)
Calculate the change in volume:
\[
\Delta V = V_2 - V_1 = 4.50 \, \text{L} - 2.50 \, \text{L} = 2.00 \, \text{L}
\]
### Step 5: Substitute values into the work formula
Given the external pressure \(P_{\text{ext}} = 2.5 \, \text{atm}\):
\[
W = -P_{\text{ext}} \Delta V = -2.5 \, \text{atm} \times 2.00 \, \text{L}
\]
### Step 6: Convert the work to Joules
The work calculated will be in liter-atmospheres, which we need to convert to Joules. The conversion factor is:
\[
1 \, \text{L} \cdot \text{atm} = 101.325 \, \text{J}
\]
Now calculate:
\[
W = -2.5 \, \text{atm} \times 2.00 \, \text{L} = -5.00 \, \text{L} \cdot \text{atm}
\]
Convert to Joules:
\[
W = -5.00 \, \text{L} \cdot \text{atm} \times 101.325 \, \text{J/L} \cdot \text{atm} = -506.625 \, \text{J}
\]
### Step 7: Final result for ΔU
Since \(\Delta U = W\):
\[
\Delta U = -506.625 \, \text{J} \approx -507 \, \text{J}
\]
### Conclusion
The change in internal energy (ΔU) of the gas is approximately:
\[
\Delta U \approx -507 \, \text{J}
\]
