Note that answer does not depend on the process,

(which cannot be isobaric btw, because in that case

expansion is incompatible with decrease of temperature)

DU=DQ-DW=1200-2120=920J

DU=DQ-DW=1200-2120=920J

Change in T = (energy) / (moles) / (heat capacity)

=-920J / 5 mol / 20.7863 J/mol-K =

= -8.852K

Change in T = (energy) / (moles) / (heat capacity)

=-920J / 5 mol / 20.7863 J/mol-K =

= -8.852K

Change of internal energy

E = Q A = 1200 2120 = -920J

Internal energy of n moles of monatomic gas:

E = nCvT = 3/2 nRT

E = 3/2 nRT

Answer:

T = 2/3 (Q-A)/nR = -14.74 C

T = To T = 113.25 C

Note that answer does not depend on the process,

(which cannot be isobaric btw, because in that case

expansion is incompatible with decrease of temperature)

Answer:

T = 2/3 (Q-A)/nR = -14.74 C

T = To T = 113.25 C

This post is last updated on hrtanswers.com at Date : 1st of September – 2022