Question

At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal enegy, and (iii) the work done by the gas during this process.

Answer 1

Correct Answer - A::B
(i) `T_1=27+273=300K,gamma=5/3(for monoatomic gas)`
`V_1=V`
`V_2=2V`
`T_2=?`
Since the gas expands adiabatically.
`T_1V_1^(gamma-1)=T_2V_2^(gamma-1)`
`rArrT_2=T_1((V_1)/(V_2))^(gamma-1)=300[V/(2V)]^(5//3-1)=189K`
(ii) `W=(-nR(T_2-T_1))/(gamma-1)=(-2xx8.31(189-300))/(5//3-1)`
`=(+8.13xx111xx3)/2=+2767J`
Change in internal Energy
According to first law of thermodynamics
`:. DeltaU=DeltaU+DeltaW But DeltaQ=0`
(as the process is adiabatic)
`:. DeltaU=-DeltaW=-2767J`
(iii) `W=2767J`