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Three moles of an ideal gas at 27^(@)C are compressed reversibly and isothermally from a volume of 10 dm^(3) to 5 dm^(3). Calculate the work done on the gas. |
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Answer» Solution :`T = 27^(@)C = 27 + 273 = 300 K, V_(1) = 10 dm^(3), V_(2) = 5 dm^(3)` `R = 8.314 J//K//Mol, n = 3 "MOLE"` `R = 8.314 w = -2.303 nRT log ((V_(2))/(V_(1))) = -2.303 xx 3 xx 8.314 xx 300 xx log[(5)/(1)]` `= -2.303 xx 3 xx 8.314 xx 300 xx bar(1).6990` `= -2.303 xx 3 xx 8.314 xx 300 xx (-0.3010) "" because -1 + 0.6990 = -3010`. `= +5,186.96 "joule" = 5.187 kJ` |
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