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1.00 mol of an ideal gas at 300 K expands at constant temperature from 10.0 bar to 2.00 bar.

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$1.00$ mol of an ideal gas at $300$ K expands at constant temperature from $10.0$ bar to $2.00$ bar. Find its starting and final volumes. Then find the work (a) in one step against a constant outside pressure of $2.00$ bar, and (b) reversibly. Use $R = 0.08314$ L bar K $^{-1}$ mol $^{-1}$ $= 8.314$ J K $^{-1}$ mol $^{-1}$ .

Need a reminder? The key ideaShow

Work against a constant outside pressure: $w = -p_\text{ex}\Delta V$ , with $1$ L bar $= 100$ J and $1$ L atm $= 101.3$ J. Free expansion: $w = 0$ . Reversible isothermal ideal gas: $w = -2.303\,nRT\log\dfrac{V_f}{V_i}$ , and $\Delta U = 0$ , so $q = -w$ .

The worked solution

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1.00 mol of an ideal gas at 300 K expands at constant temperature from 10.0 bar to 2.00 bar.

The worked solution

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