Work done in vaporisation of one mole of wate | vedclass

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(A) The work done during expansion is given by $w = -P_{ext} \Delta V$.For the vaporisation of $1 \ mole$ of water: $H_2O(l) \rightarrow H_2O(g)$.Volu

Vedclass · Accepted Answer

$-3100$

Vedclass · Answer

(A) The work done during expansion is given by $w = -P_{ext} \Delta V$.For the vaporisation of $1 \ mole$ of water: $H_2O(l) ightarrow H_2O(g)$.Volume of $1 \ mole$ of liquid water $(V_1)$: Since density $ho \approx 1 \ g/mL$,$V_1 = \frac{18 \ g}{1 \ g/mL} = 18 \ mL = 0.018 \ L$.Volume of $1 \ mole$ of water vapour $(V_2)$ assuming ideal gas behavior: $V_2 = \frac{nRT}{P} = \frac{1 	imes 0.0821 	imes 373}{1} = 30.62 \ L$.Change in volume $\Delta V = V_2 - V_1 = 30.62 - 0.018 \approx 30.6 \ L$.Work done $w = -P \Delta V = -1 \ atm 	imes 30.6 \ L = -30.6 \ L \ atm$.Converting to Joules: $w = -30.6 	imes 101.3 \ J \approx -3099.78 \ J \approx -3100 \ J$.

Work done in vaporisation of one mole of water at $373 \ K$ and the pressure of $1 \ atm$ is approximately (for ideal gas)......$J$

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