The volume of one mole of an ideal gas with adiabatic exponent gammais varied according to V=b/T,where b is a constant. Find the amount of Heat is absorbed by the gas in this process temperature raised by DeltaT

The volume of one mole of an ideal gas with adiabatic exponent gammais

1.	The volume of one mole of an ideal gas with adiabatic exponent gammais varied according to V=b/T,where b is a constant. Find the amount of Heat is absorbed by the gas in this process temperature raised by DeltaT
Answer» `((1-gamma)/(gamma+1))RDeltaT` `R/(gamma-1)DeltaT` `((2-gamma)/(gamma-1))RDeltaT` `(RDeltaT)/(gamma-1)` Solution :`V=b/T RARR T=b/V`…(1) `PV=muRT` `=(muRb)/V` `THEREFORE PV^2=muRb`=constant Now for adiabatic CHANGE `PV^x` =constant by comparing it x=2 Now, `c=R/(gamma-1)+R/(1-x)` `=R/(gamma-1)+R/(1-2)` `=R/(gamma-1)-R` `=R[(1-gamma+1)/(gamma-1)]=R[(2-gamma)/(gamma-1)]` HEAT absorbed `DeltaQ=muCDeltaT` here `mu=1` mole `therefore DeltaQ=RDeltaT((2-gamma)/(gamma-1))`