In an isothermal and reversible process, $1 \cdot 6 \times 10^{-2} \mathrm{~kg} \mathrm{O}_{2}$ expands from…

In an isothermal and reversible process, $1 \cdot 6 \times 10^{-2} \mathrm{~kg} \mathrm{O}_{2}$ expands from $10 \mathrm{dm}^{3}$ to $100 \mathrm{dm}^{3}$ at $300 \mathrm{~K}$, work done in the process is $(\mathrm{R}=8.314 \mathrm{~J})$
  1. $-1436 \mathrm{~J}$
  2. $-5744 \mathrm{~J}$
  3. $-4308 \mathrm{~J}$
  4. $-2872 \mathrm{~J}$

Solution

$\mathrm{V}_{1}=10 \mathrm{dm}^{3} \quad, \mathrm{~V}_{2}=100 \mathrm{dm}^{3}, \quad \mathrm{~T}=300 \mathrm{~K}$, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~m}_{\mathrm{O}_{2}}=1.6 \times 10^{-2} \mathrm{~kg}=16 \mathrm{~g}$ $\therefore \quad \mathrm{n}=\frac{\mathrm{m}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{16}{32}=0.5 \mathrm{~mol}$ For isothermal and reversible process, $\begin{aligned} W_{\max } &=-2.303 \mathrm{nRT} \log _{10} \frac{V_{2}}{V_{1}} \\ \therefore W_{\max } &=-2.303 \times 0.5 \times 8.314 \times 300 \times \log _{10} \frac{100}{10}=-2872 \mathrm{~J} \end{aligned}$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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