... must be converted to Kelvin temperatures, K.500andK 300 21 TT Insertion of the given values yieldsK )300 -KK)(500molJmol)(29.500.3( Q ))K )300 (K)500)((KmolJ1010.4(2223 J.1097.1417.101: ... solved for,1 ,0350.00650.00350.00580.0121fff and the lengths are cm7.00cm)0 .30( andcm23.0cm)0 .30( 21 ff17.88: a) The lost volume, 2.6 L, is the difference between the expanded volume ... numerical values,C.4.21K)kgJkg)(390100.0(K)kgJkg)(4190178.0(kgK)Jkg)( 1307 50.0(kg)J10kg)(334018.0( )CK)(255kgJkg)( 1307 50.0(3T(The...