Three moles of an ideal monatomic gas are at a temperature of 396 K. Then 2438 J of heat is added to the gas, and 897 J of work is done on it. What is the final temperature of the gas?
2 answers
what grade are you in???
ΔQ=ΔU+W
ΔU= ΔQ-W=2438-897 =1541 J
ΔU=νcΔT=ν(iR/2) ΔT
ΔT=2 ΔU/νiR =2•1541/3•3•8.31=41.2 K
ΔT= T₂-T₁
T₂=T₁+ΔT=396+41.2=437.2 K
ΔU= ΔQ-W=2438-897 =1541 J
ΔU=νcΔT=ν(iR/2) ΔT
ΔT=2 ΔU/νiR =2•1541/3•3•8.31=41.2 K
ΔT= T₂-T₁
T₂=T₁+ΔT=396+41.2=437.2 K
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