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.

Explanation

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.