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According to the first law of thermodyna...

According to the first law of thermodynamics, `DeltaU=q+w`. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression ?

A

At constant temperature, `q=-w`

B

When no work is done , `DeltaU=q`

C

In gaseous system , `DeltaU=q+PDeltaV`

D

When work is done by the system : `DeltaU=q+w`

Text Solution

Verified by Experts

The correct Answer is:
D
When work is done by the system, `DeltaU=q-w`.
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