Asked by S
A monatomic ideal gas expands from 1.00m^3 to 2.50m^3 st a constant pressure of 2.00x10^5 Pa. Find the change in the internal energy of the gas.
Answers
Answered by
Elena
Internal energy is U=3•n•R•T/2 => chage in internal energy is ΔU=3•n•R•ΔT/2.
From the ideal gas law PV=nRT =>
n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) =>
ΔU=3•n•R•ΔT/2=3•p(V2-V1)/2=3•2•10⁵•(2.5-1)/2=4.5•10⁵ J
From the ideal gas law PV=nRT =>
n•R•ΔT=nR(T2-T1)=p2•V2 =p1•V1=p(V2-V1) =>
ΔU=3•n•R•ΔT/2=3•p(V2-V1)/2=3•2•10⁵•(2.5-1)/2=4.5•10⁵ J
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.