If the work done by 2 mol of an ideal gas du | vedclass

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(B) The formula for work done in isothermal reversible expansion is $W_{rev} = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$.Given: $n = 2 \ mol$,$V_1

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$227$

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(B) The formula for work done in isothermal reversible expansion is $W_{rev} = -2.303 nRT \log \left(\frac{V_2}{V_1}ight)$.Given: $n = 2 \ mol$,$V_1 = 5 \ L$,$V_2 = 50 \ L$,$W = -189.1 \ L \ atm$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.Substituting the values: $-189.1 = -2.303 	imes 2 	imes 0.082 	imes T 	imes \log \left(\frac{50}{5}ight)$.$-189.1 = -2.303 	imes 2 	imes 0.082 	imes T 	imes 1$.$T = \frac{-189.1}{-2.303 	imes 2 	imes 0.082} \approx 500 \ K$.Converting to Celsius: $T(^{\circ}C) = 500 - 273 = 227^{\circ}C$.

If the work done by $2 \ mol$ of an ideal gas during isothermal reversible expansion from $5 \ L$ to $50 \ L$ is $-189.1 \ L \ atm$,the temperature of the gas (in $^{\circ}C$) is

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