Calculate `Delta U`, q and w when 2.0 mol of an ideal gas at `25^(@)C` are compressed isothermally and reversibly from 1.0 bar to 10.0 bar.

Explanation

To solve the problem, we need to calculate the change in internal energy (ΔU), heat exchanged (q), and work done (w) during the isothermal compression of an ideal gas. Here are the steps to solve the question:

### Step 1: Understand the conditions of the process
Since the gas is being compressed isothermally, the temperature remains constant. The initial pressure (P1) is 1.0 bar, and the final pressure (P2) is 10.0 bar. The number of moles (n) of the gas is 2.0 mol, and the temperature (T) is given as 25°C, which is equivalent to 298 K.

### Step 2: Calculate the change in internal energy (ΔU)
For an ideal gas undergoing an isothermal process, the change in internal energy (ΔU) is given by:
\[ \Delta U = n C_V \Delta T \]
Since the temperature (T) is constant, ΔT = 0. Therefore:
\[ \Delta U = 0 \]

### Step 3: Apply the first law of thermodynamics
The first law of thermodynamics states:
\[ \Delta U = q + w \]
Since we found that ΔU = 0, we can rearrange the equation to find q:
\[ 0 = q + w \implies q = -w \]

### Step 4: Calculate the work done (w)
For an isothermal process, the work done on the gas can be calculated using the formula:
\[ w = -nRT \ln\left(\frac{P_2}{P_1}\right) \]
Where:
- n = 2.0 mol
- R = 8.314 J/(K·mol) (universal gas constant)
- T = 298 K
- P1 = 1.0 bar = 100 kPa (1 bar = 100 kPa)
- P2 = 10.0 bar = 1000 kPa

Substituting the values into the equation:
\[ w = -2.0 \times 8.314 \times 298 \times \ln\left(\frac{10.0}{1.0}\right) \]

Calculating the logarithm:
\[ \ln(10) \approx 2.303 \]

Now substituting this value:
\[ w = -2.0 \times 8.314 \times 298 \times 2.303 \]
Calculating the work done:
\[ w \approx -2.0 \times 8.314 \times 298 \times 2.303 \approx -11,412.0 \text{ J} \approx -11.412 \text{ kJ} \]

### Step 5: Calculate the heat exchanged (q)
Using the relationship we derived from the first law:
\[ q = -w = 11.412 \text{ kJ} \]

### Summary of Results
- ΔU = 0
- w = -11.412 kJ
- q = 11.412 kJ