A perfect gas goes from state A to state B by absorbing 8xx10^(5)J of heat and doing 6.5xx10^(5)J of external work. It is now transferred between the same two states in another process in which it absorbs 10^(5)J of heat. In second process. Find the work done in the second process.

A perfect gas goes from state A to state B by absorbing 8xx10^(5)J of

1.	A perfect gas goes from state A to state B by absorbing 8xx10^(5)J of heat and doing 6.5xx10^(5)J of external work. It is now transferred between the same two states in another process in which it absorbs 10^(5)J of heat. In second process. Find the work done in the second process.
Answer» SOLUTION :According to the FIRST LAW of thermodynamics `DeltaU=DeltaQ-DeltaW` In the first process `DeltaU=8xx10^(5)-6.5xx10^(5)=1.5xx10^(5)J` Now `DeltaU`, being a state function, remains the same in the second process, `DeltaW=DeltaQ-DeltaU=1 xx10^(5)-1.5xx10^(5)` `DeltaW=-0.5xx10^(5)J` The negative sign shows that work is doen on the GAS.