If the work done during the isothermal revers | vedclass

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(B) For an isothermal reversible expansion of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log(\frac{V_f}{V_i})$.Sin

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) For an isothermal reversible expansion of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log(\frac{V_f}{V_i})$.Since $PV = nRT$,we can substitute $nRT$ with $P_i V_i$ (initial state):$w = -2.303 \ P_i V_i \ \log(\frac{V_f}{V_i})$.Given: $w = -184.24 \ L \ atm$,$P_i = 10 \ atm$,$V_i = 4 \ L$.Substituting the values:$-184.24 = -2.303 	imes (10 	imes 4) 	imes \log(\frac{V_f}{4})$.$-184.24 = -2.303 	imes 40 	imes \log(\frac{V_f}{4})$.$-184.24 = -92.12 	imes \log(\frac{V_f}{4})$.$\log(\frac{V_f}{4}) = \frac{-184.24}{-92.12} = 2$.$\frac{V_f}{4} = 10^2 = 100$.$V_f = 100 	imes 4 = 400 \ L$.

If the work done during the isothermal reversible expansion of an ideal gas at a pressure of $10 \ atm$ from $4 \ L$ to a final volume is $-184.24 \ L \ atm$,the final volume of the gas in $L$ is

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