Question

mathemarical form of thermodynamics is ∆u=2+w it is true or false​

Answer 1

Answer:

At constant temperature, ΔT=0

∴ΔU=nC

v

ΔT

∴ΔU=q+w=0

q=−w

When work done is zero, w=0

∴ΔU=q+w

∴ΔU=q

In gaseous system, w=PΔV

∴ΔU=q+w

∴ΔU=q+PΔV

When work is done by the system, w is negative.

∴ΔU=q−w

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