Four moles of an ideal gas at 2.5 atm and `27^(@)C` are compressed isothermally to half of its volume by an external pressure of 3 atm. Calculate w, q and `DeltaU`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify Given Data
- Number of moles (n) = 4 moles
- Initial pressure (P1) = 2.5 atm
- Final pressure (P_ext) = 3 atm
- Initial temperature (T) = 27°C = 300 K (after converting to Kelvin)
- The gas is compressed isothermally to half its volume.

### Step 2: Calculate Initial Volume (V_initial)
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- R = 0.0821 L·atm/(K·mol)

We can rearrange the equation to find the initial volume:
\[ V_{initial} = \frac{nRT}{P_{initial}} \]

Substituting the values:
\[ V_{initial} = \frac{4 \, \text{moles} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{2.5 \, \text{atm}} \]

Calculating:
\[ V_{initial} = \frac{98.52}{2.5} = 39.408 \, \text{L} \]

### Step 3: Calculate Final Volume (V_final)
Since the gas is compressed to half its volume:
\[ V_{final} = \frac{V_{initial}}{2} = \frac{39.408}{2} = 19.704 \, \text{L} \]

### Step 4: Calculate Work Done (W)
The work done on the gas during isothermal compression can be calculated using the formula:
\[ W = -P_{ext} \Delta V \]
Where:
\[ \Delta V = V_{final} - V_{initial} \]

Calculating \(\Delta V\):
\[ \Delta V = 19.704 \, \text{L} - 39.408 \, \text{L} = -19.704 \, \text{L} \]

Now substituting into the work equation:
\[ W = -3 \, \text{atm} \times (-19.704 \, \text{L}) = 59.112 \, \text{L·atm} \]

### Step 5: Convert Work to Joules
To convert L·atm to Joules, we use the conversion factor:
\[ 1 \, \text{L·atm} = 101.3 \, \text{J} \]

Thus:
\[ W = 59.112 \, \text{L·atm} \times 101.3 \, \text{J/L·atm} = 5985.5 \, \text{J} \]

### Step 6: Calculate Heat (q)
For an isothermal process in an ideal gas, the change in internal energy (\( \Delta U \)) is zero:
\[ \Delta U = 0 \]

Using the first law of thermodynamics:
\[ \Delta U = q + W \]
Since \( \Delta U = 0 \):
\[ 0 = q + W \]
Thus:
\[ q = -W = -5985.5 \, \text{J} \]

### Summary of Results
- Work done (W) = 5985.5 J
- Heat exchanged (q) = -5985.5 J
- Change in internal energy (\( \Delta U \)) = 0 J