Find the work done on 2 mol of an ideal gas | vedclass

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(A) For a reversible isothermal process,the work done $W$ is given by $W = -2.303 \ nRT \ \log_{10}\left(\frac{P_1}{P_2}\right)$.Given: $n = 2 \ mol$,

Vedclass · Accepted Answer

$19.52$

Vedclass · Answer

(A) For a reversible isothermal process,the work done $W$ is given by $W = -2.303 \ nRT \ \log_{10}\left(\frac{P_1}{P_2}ight)$.Given: $n = 2 \ mol$,$T = 27 + 273 = 300 \ K$,$P_1 = 5.05 	imes 10^6 \ Nm^{-2}$,$P_2 = 1.01 	imes 10^5 \ Nm^{-2}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10}\left(\frac{5.05 	imes 10^6}{1.01 	imes 10^5}ight)$.$W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10}(50)$.$W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes 1.699 \approx -19520 \ J = -19.52 \ kJ$.The negative sign indicates work is done by the system during expansion,but the question asks for work done $ON$ the system during compression. Note: The pressure values provided imply expansion $(P_1 > P_2)$. If compressed,$P_2 > P_1$. Assuming the magnitude is requested: $19.52 \ kJ$.

Find the work done on $2 \ mol$ of an ideal gas at $27^{\circ} C$ if it is compressed reversibly and isothermally from $5.05 \times 10^6 \ Nm^{-2}$ to $1.01 \times 10^5 \ Nm^{-2}$ pressure. (in $kJ$)

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