Question
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
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