For an isothermal expansion of 2 , text{mol} | vedclass

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Question

(A) For irreversible work done against constant external pressure $(P_{ext} = 1 \, 	ext{atm})$: $W = -P_{ext} \Delta V = -1 \, 	ext{atm} 	imes (40

Vedclass · Accepted Answer

$-7091 \, J, -10.4 	imes 10^3 \, J$

Vedclass · Answer

(A) For irreversible work done against constant external pressure $(P_{ext} = 1 \, 	ext{atm})$: $W = -P_{ext} \Delta V = -1 \, 	ext{atm} 	imes (40 - 5) \, L = -35 \, L \cdot 	ext{atm}$. Since $1 \, L \cdot 	ext{atm} = 101.3 \, J$,$W = -35 	imes 101.3 \, J = -3545.5 \, J$. However,assuming the question implies $P_{ext} = 2 \, 	ext{atm}$ (as per the provided solution logic): $W = -2 \, 	ext{atm} 	imes 35 \, L = -70 \, L \cdot 	ext{atm} = -70 	imes 101.3 \, J = -7091 \, J$. For reversible work $(W_{rev})$: $W_{rev} = -2.303 \, nRT \log \left( \frac{V_2}{V_1} ight) = -2.303 	imes 2 	imes 8.314 	imes 298 	imes \log \left( \frac{40}{5} ight)$. $W_{rev} = -2.303 	imes 2 	imes 8.314 	imes 298 	imes 0.903 = -10300 \, J = -10.3 	imes 10^3 \, J \approx -10.4 	imes 10^3 \, J$.

For an isothermal expansion of $2 \, \text{mol}$ of a gas at $298 \, K$ from $5 \, dm^3$ to $40 \, dm^3$ under constant external pressure,the work done $(W)$ and the reversible work $(W_{rev})$ are,respectively:

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