5 , mol of an ideal gas expands isothermally | vedclass

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(B) For an isothermal irreversible expansion of an ideal gas,the work done is given by: $w = -P_{ext} \Delta V$. Since $V = \frac{nRT}{P}$,we have $\D

Vedclass · Accepted Answer

$-11.224 \, kJ$

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(B) For an isothermal irreversible expansion of an ideal gas,the work done is given by: $w = -P_{ext} \Delta V$. Since $V = \frac{nRT}{P}$,we have $\Delta V = nRT (\frac{1}{P_2} - \frac{1}{P_1})$. Given $P_{ext} = P_2 = 1 \, atm$,$n = 5 \, mol$,$T = 300 \, K$,$P_1 = 10 \, atm$,and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$. Substituting the values: $w = -nRT P_{ext} (\frac{1}{P_2} - \frac{1}{P_1}) = -nRT (1 - \frac{P_2}{P_1})$. $w = -5 	imes 8.314 	imes 300 	imes (1 - \frac{1}{10}) \, J$. $w = -12471 	imes 0.9 \, J = -11223.9 \, J = -11.224 \, kJ$.

$5 \, mol$ of an ideal gas expands isothermally and irreversibly from a pressure of $10 \, atm$ to $1 \, atm$ against a constant external pressure of $1 \, atm$. The work done $(w_{irr})$ at $300 \, K$ is:

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