0.022 kg of `C_O2` is compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa when the work obtained is 1200 J, calculate the final pressure.

Answer

To solve the problem of finding the final pressure after isothermal and reversible compression of CO2, we can follow these steps:

### Step 1: Convert mass of CO2 to moles
Given:
- Mass of CO2 = 0.022 kg = 22 g
- Molar mass of CO2 = 44 g/mol

To find the number of moles (n):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{22 \text{ g}}{44 \text{ g/mol}} = 0.5 \text{ moles} \]

**Hint:** Remember that to convert grams to moles, you divide by the molar mass.

### Step 2: Use the work done formula for isothermal compression
The work done on the gas during isothermal compression can be expressed as:
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
However, we can also relate the pressures using:
\[ W = -nRT \ln\left(\frac{P_i}{P_f}\right) \]
Where:
- \(W = 1200 \text{ J}\) (work done on the gas)
- \(R = 8.314 \text{ J/(mol K)}\) (universal gas constant)
- \(T = 298 \text{ K}\) (temperature)
- \(P_i = 100 \text{ kPa}\) (initial pressure)

### Step 3: Rearranging the work formula
Rearranging the equation for work gives:
\[ 1200 = -0.5 \times 8.314 \times 298 \ln\left(\frac{100}{P_f}\right) \]
Calculating the constant:
\[ 0.5 \times 8.314 \times 298 = 1235.07 \]
Thus, we have:
\[ 1200 = -1235.07 \ln\left(\frac{100}{P_f}\right) \]

### Step 4: Solve for the logarithm
Dividing both sides by -1235.07:
\[ \ln\left(\frac{100}{P_f}\right) = -\frac{1200}{1235.07} \approx -0.97 \]

### Step 5: Exponentiate to solve for \(P_f\)
Taking the exponential of both sides:
\[ \frac{100}{P_f} = e^{-0.97} \]
Calculating \(e^{-0.97}\):
\[ e^{-0.97} \approx 0.378 \]
Thus:
\[ P_f = \frac{100}{0.378} \approx 263.4 \text{ kPa} \]

### Final Answer
The final pressure \(P_f\) after isothermal compression is approximately:
\[ \boxed{263.4 \text{ kPa}} \]

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