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