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(B) For a reversible isothermal expansion,the work done by the system is given by: $\omega = -nRT \ln \left(\frac{V_2}{V_1}\right)$.Given:$n = 1 \ mol

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) For a reversible isothermal expansion,the work done by the system is given by: $\omega = -nRT \ln \left(\frac{V_2}{V_1}ight)$.Given:$n = 1 \ mol$$T = 18 + 273.15 = 291.15 \ K$$V_1 = 10 \ L$$V_2 = 10 + 90 = 100 \ L$$R = 0.08206 \ L \ atm \ mol^{-1} \ K^{-1}$Substituting the values:$\omega = -1 	imes 0.08206 	imes 291.15 	imes \ln \left(\frac{100}{10}ight)$$\omega = -23.887 	imes \ln(10)$$\omega = -23.887 	imes 2.303$$\omega \approx -55.01 \ L \ atm$The magnitude of work done by the system is $55 \ L \ atm$.

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