Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

Given:

• Mass of O₂ = 24 g
• Initial pressure = \(P_1 = 1.6\) bar
• Final pressure = \(P_2 = 1\) bar
• Temperature = \(T = 298\) K

To find: Maximum work (\(W_{\text{max}}\))

Formula: \(W_{\text{max}} = -2.303 \cdot nRT \cdot \log_{10} \frac{P_1}{P_2}\)

Calculation:

• Number of moles of O₂ (\(n\)) = \(\frac{24 \, \text{g}}{32 \, \text{g mol}^{-1}} = 0.75 \, \text{mol}\)
• Gas constant (\(R\)) = \(8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}\)

Now, using the formula:

\[W_{\text{max}} = -2.303 \cdot nRT \cdot \log_{10} \frac{P_1}{P_2}\]

Substituting the values:

\[W_{\text{max}} = -2.303 \cdot 0.75 \, \text{mol} \cdot 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1} \cdot 298 \, \text{K} \cdot \log_{10} \frac{1.6}{1}\]

Calculating further:

\[W_{\text{max}} = -2.303 \cdot 0.75 \cdot 8.314 \cdot 298 \cdot 0.2041\]

Finally:

\[W_{\text{max}} = -873.4 \, \text{J}\]

The maximum work done is \(-873.4 \, \text{J}\).