At 27^(@)C, two moles of an ideal mono-atomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (a) final temperature of the gas (b) change in its internal energy and (c ) the work done by the gas during the process. (R = 8.31 J/mol K)

At 27^(@)C, two moles of an ideal mono-atomic gas occupy a volume V. T

1.	At 27^(@)C, two moles of an ideal mono-atomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (a) final temperature of the gas (b) change in its internal energy and (c ) the work done by the gas during the process. (R = 8.31 J/mol K)
Answer» Solution :(a) In case of adiabatic change, `PV^(R ) =` const. with `PV=nRT` So that `T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)` [with `gamma=(5//3)`] i.e., `300xx v^(2//3) = T(2V)^(2//3) or T=300//(2)^(2//3) = 189K` (b) As `DeltaU=nC_(v)DeltaT = (nR DeltaT)/(gamma-1) [ "as" C_(v)=(R )/((gamma-1))]` So, `DeltaU=(2xx8.31 xx (189-300))/(((5)/(3)-1)) = -2767.23J` Negative SIGN means internal energy DECREASES. (c ) Accordingto FIRST law of thermodynamics, `DeltaQ=DeltaU+DELTAW` and as for adiabatic change `DeltaQ= 0 rArr DeltaW = - DeltaW = 2767.23J`